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AC Circuits
Alternating Current (AC) Circuits
by
Prof. Dr. Osman SEVAİOĞLU
Electrical and Electronics Engineering Department
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 1
AC Circuits
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What is Direct Current (DC) ?
Definition
Switch is turned “on” at: t = 1 sec
Direct Current (DC) is a current, that
does not change in time
Current (Amp)
80
60
Switch
Current, I
R1= 5 Ohms
40
I = 60 A
DC (Constant) Current
20
+
Vs= 600 V
R2= 5 Ohms
0
1
2
3
4
5
7
6
Time (Sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 2
AC Circuits
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What is Alternating Current (AC) ?
Definition
Alternating Current (AC) is a current, that
changes in time
Non - Sinusoidal Alternating Current
Current (Amp)
Current (Amp)
Sinusoidal Alternating Current
10
Time (msec)
0
5
10
15
5.0
4.0
3.0
20
2.0
1.0
- 10
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
Time (Sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 3
AC Circuits
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Parameters of a Sinusoidal Waveform
Sinusoidal Voltage
Definition
Sinusoidal voltage is a voltage
with waveform as shown on the
RHS
^
V(t) = V sin ( wt +  )
where
V(t) is the voltage waveform,
^
V is the peak value (amplitude),
w is the angular frequency,
 is the phase shift, i.e. angle of the
voltage at t = 0, (phase angle)
Voltage (Volt)
Please note that;
Angle = wt = 2 f t
312
300
200
Phase angle 
^
V = Amplitude
100
Angle (Radians)
0
/2

3/2
f = 50 Hz
w = 2  f = 314 rad/sec
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 4
2
AC Circuits
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Parameters of a Sinusoidal Waveform
Voltage Waveform
V (Volts)
312
^
V = Amplitude = 312 Volts
300
200
100
0
Time (msec)
0
5
10
15
20
-100
-200
-312
-300
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 5
AC Circuits
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Parameters of a Sinusoidal Waveform
Period and Frequency
Full period = T = 20 msec  360 o
V (Volts)
300
200
Frequency = 1/ T = 1/(20 x 10 -3)
= 50 Hz
100
Time (msec)
0
0
5
10
15
20
25
30
35
-100
-200
-300
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 6
40
AC Circuits
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AC (Alternating Current) Circuit
Positive half cycle
Negative half cycle
I (Amp)
I (Amp)
25
I(t)
25
25
20
20
20
15
15
15
10
10
10
5
5
V(t) +
5
0
0
5
10
0,015
Load
0
-5
-5
-10
-10
-10
-15
-15
-15
-20
-20
-25
-25
-20
-25
Time (msec)
0,005
V(t) +
0
5
10
0
-5
0
I(t)
Time (msec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 7
Load
AC Circuits
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+
Elements of AC Circuits: Capacitor
Definition
Capacitor is a device that can store
electrical charge
Positive conductor
Insulating Layer
Negative conductor
The simplest configuration consists of
two parallel conducting plates
separated by an insulating layer
+
Insulating Layer provides dielectricity
(prevents current flow) between
positive and negative conductors
_
Symbolic
representation
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 8
AC Circuits
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Capacitance
Definition
Small Capacitance
Large Capacitance
Capacitors store electrical charge
__
C 1 < C2
Capacitance = C2
+
Capacitance = C1
+
Storage capacity of a capacitor is
called “capacitance”
_
Water (hydroulic) example
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 9
AC Circuits
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Capacitor-Practical Configuration
Geometry
+
Capacitor plates are packaged in a roll
form in order to have smaller size
_
_
+
Aluminum cover
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 10
AC Circuits
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Capacitor-Practical Configuration
Geometry
Capacitor cylinders are then
connected in parallel in bank form
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 11
AC Circuits
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Capacitor-Practical Configuration
Geometry
Capacitor banks
Control relay
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 12
AC Circuits
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Capacitor-Practical Configuration
Geometry
Single Phase
Three Phase
Single and three-phase capacitor
banks
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 13
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AC Circuits
MV (Medium Voltage) Shunt Capacitor Banks
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 14
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AC Circuits
MV (Medium Voltage) Shunt Capacitor Banks
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 15
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AC Circuits
MV (Medium Voltage) Shunt Capacitor Banks
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 16
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AC Circuits
LV (Low Voltage) Capacitor Banks
Protective Breakers
Contactors
Capacitor Bank
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 17
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AC Circuits
Electronic Capacitors in a Motherboard
Capacitors
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 18
AC Circuits
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Basic Relation
Basic Principle
Voltage Source
• Charge stored in a capacitor is
proportional to the voltage V
applied
or
Q=CV
where, Q is charge stored (Coulombs),
V is voltage (Volts),
C is capacitance (Farads)
+
• Charge stored in a capacitor is
proportional to the capacitance C,
Capacitance
+
C
V
_
Symbolic Representation
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 19
AC Circuits
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Definition of Farad
Definition
1 Farad is the capacitance that creates 1 Volt
voltage difference between the terminals of
the plates when charged by 1 Coulomb of
electrical charge
Capacitor
Q=CV
where, Q = 1 Coulomb,
V = 1 Volt,
C = 1 Farad
+
C = 1 Farad
_
+
Q = 1 Coulomb
1 Volt
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 20
AC Circuits
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Current in a Capacitance
I(t)
Definition
The relation;
Q(t) = C V(t)
or differentiating both sides with respect to time
C
V(t)
_
may be written in time domain as;
+
+
Q=CV
dQ(t)/dt = C dV(t)/dt
remembering that;
dQ(t)/dt = I(t)
It can be written that;
I(t) = C dV(t) / dt
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 21
AC Circuits
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Current in a Capacitance
I (Amp) Voltage Waveforms
Phase Shift between Current and
V (Volts), I (Amp)
I(t) = C d V(t) / dt
= C d/dt Vmax sin wt
= C Vmax w coswt
= Imax coswt
25
20
Vmax
15
Imax
10
where, Imax = C Vmax w
5
I(t)
0
0.005
0.010
0.015
0.020
-5
+
V(t) +
C
-10
-15
-20
_
-25
Time (Sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 22
AC Circuits
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Current in a Capacitance
The above equation may be integrated
with respect to time, yielding the
following voltage - current relation for a
Capacitor
where V(0) is the initial voltage across
the capacitor, representing the initial
voltage due to the initial charge stored in
the capacitor
I(t)
Switch
+
+
V(t) = (1/C)  I(t)dt + V(0)
Switch is turned “on” at: t = 0 sec
V(t)
C
Vc (0)
_
Definition
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 23
AC Circuits
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Example - 1
I(t)
Problem
V(t)
C
_
V(t) = 5 (1 – e -t) Volts
+
+
Determine the time waveform of the current
flowing in the circuit shown on the RHS by
assuming that the capacitor is charged by the
exponential voltage V(t) shown in the figure
C = 0.1 F
V(t) = 5 (1 - e-t) (Volts)
5.0
4.0
3.0
2.0
1.0
Vmax = Maximum voltage that can be reached = 5 Volts
Qmax = C x Vmax = Maximum charge that can be stored
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 24
AC Circuits
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Example - 1
Solution
V(t) = 5 (1 - e-t) (Volts)
I(t)
5.0
4.0
I(t) = C dV(t)/dt
3.0
V(t) = 5 (1 - e-t) Volts
Hence,
+
+
V(t)
2.0
1.0
_
I(t) = C d V(t) / dt
= C d/dt 5 (1 – e -t)
= 0.1 x 5 e -t
= 0.5 x e -t Ampers
C
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
3.0
4.0
5.0
6.0
I(t) (Amp)
t (sec)
0.5
0.4
0.3
0.2
0.1
0.0
0.0
1.0
2.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 25
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AC Circuits
Example - 1
Charge Stored in a
Capacitor
Qmax = C x Vmax = Maximum charge that can be stored
I(t)
Now, determine the time waveform of
the charge stored in the capacitor
V(t)
C = 0.1 F
_
Q(t) = C x V(t)
= 0.1 x 5 (1 - e-t)
= 0.5 x (1 - e-t) Coulombs
+
Charge stored in the capacitor starts
from zero and gradually increases to its
final value
+
Q(t) (Coulombs)
0.5
0.4
0.3
0.2
0.1
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0 t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 26
AC Circuits
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Example - 2
Current source shown in the circuit
shown on the RHS provides 10 A
constant current within the time
interval;
Switch is turned “on” at: t = 0 sec,
“off” at: t = 1.0 sec.
14.0
I(t)
12.0
10.0
+
t ε 0, 1 sec 
Capacitor is initially charged to 2 Volts
voltage
I(t) (Amp)
Switch
I(t)
8.0
4.0
2.0
Determine the voltage across the
capacitor within the time interval;
t ε 0, 1 sec
C
6.0
_
Problem
0.0
0.2
0.4
0.6
0.8
1.0
1.2
t (sec)
C=1F
Vc(0) = 2 Volts
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 27
AC Circuits
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Example - 2
+
Voltage across the capacitance can
be expressed as
14.0
I(t)
V(t) = (1/C)  I(t)dt + V(0)
C
12.0
10.0
8.0
_
Solution
I(t) (Amp)
I(t)
6.0
4.0
where,
2.0
V(0) = 2 Volts
is the initial voltage across the
capacitor
t ε 0, 1
Hence;
V(t) = (1/C)  I(t) dt + 2
= 1 x 10  dt + 2
= 10 t + 2 Volts
0.0
C=1F
Vc(0) = 2 Volts
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.4
0.6
0.8
1.0
1.2
V(t) (Volts)
t (sec)
14.0
12.0
10.0
V(t) = 10 t +2
Volts
8.0
6.0
4.0
2.0
0.0
0.0
0.2
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 28
AC Circuits
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Solution of R-C Circuits
Problem
Solve the “RC circuit” shown on the RHS for
current waveform I(t) flowing in the circuit
when the switch is turned “on” at t = 0
Solution
Switch
I(t)
R
^
+ V(t) = V sin (wt + )
+
Writing down KVL for the circuit
Switch is turned “on” at: t = 0 sec
Vc(0) = V0
C
V(t) = R I(t) + VC(t)
= R I(t) + (1/C)  I(t)dt + Vc(0)
_
Differentiating both sides wrt time once;
d/dt V(t) = R d/dt I(t) + (1/C) I(t)
or dividing both sides by R
d/dt I(t) + (1/RC) I(t) = (1/R) d/dt V(t)
d/dt Vc(0) = 0
A first order ordinary differential equation
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 29
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
I(t)
Solution
d/dt I(t) + (1/RC) I(t) = (1/R) d/dt V(t)
+
_
^
d/dt I(t) + (1/RC) I(t) = (1/R) d/dt ( V sin ( wt +  ) )
^
d/dt I(t) + (1/RC) I(t) = (V/R) w cos ( wt +  )
Voltage (Volt)
R
+
Solve the resulting first order ordinary differential equation (ODE)
^
V(t) = V sin (wt +)
Vc(0)=V0
C
V( t ) = ^
V sin ( wt +  )
^
d/dt V( t ) = V w cos( wt +  )
312
300
200
Amplitude
100
0
/2

3/2
2
Angle (Radians)
Phase angle 
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 30
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AC Circuits
Solution of the resulting First Order Ordinary Differential Equation
Solution
Solve the resulting first order ordinary differential equation (ODE)
^
Switch is turned “on” at:
t = 0 sec
I(t)
dI(t) / dt + (1/RC) I(t) = (V/R) w cos ( wt +  )
R
^
V(t) = V sin(wt + )
_
^
µ(t) dI(t)/dt + I(t) (1/RC) µ(t) = µ(t) ( V/R ) w cos ( wt +  )
^
µ(t) dI(t)/dt + I(t) d/dt µ(t) = ( V/R ) µ(t) w cos ( wt +  )
^
d/dt [µ(t) I(t)]
= ( V/R ) µ(t) w cos ( wt +  )
^
 d/dt [µ(t) I(t)] dt
= ( V/R )  µ(t) w cos ( wt +  ) dt + I(0)
^
µ(t) I(t)
= ( V/R ) w  µ(t) cos ( wt +  ) dt + I(0)
I(t)
= ^I µ(t) -1 w µ(t) cos(wt +  ) dt + µ(t)-1I(0)
+
+
Define an integration factor µ(t) = e t / RC
Multiply both sides of the above ODE by this factor;
Vc(0)=V0
C
d/dt µ(t) = (1/RC) µ(t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 31
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Reference
Subsituting the integration factor µ(t) = e t/RC into the above
solution;
^
I(t) = I e - t/RC w  e t/RC cos (wt +  ) dt + e - t/RC I(0)
^
= I e - t/RC w  e t/RC (coswt cos - sinwt sin ) dt + e - t/RC I(0)
cos (a + b) = cos a cos b – sin a sin b
Taken from the Reference: Calculus and Analytic Geometry,
Thomas, Addison Wesley, Third Ed. 1965
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 32
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Now, proceeding;
^
I(t) = I e - t/RC w  e t/RC (coswt cos - sinwt sin ) dt + e - t/RC I(0)
^
= I e - t/RC w [ cos  e t/RC coswt dt - sin  e t/RC sinwt dt ] + e - t/RC I(0)
e
ax
cos bx dx =
e ax
b sin bx + a cos bx
-----------------------------a 2 + b2
e
ax
sin bx dx =
e ax
a sin bx - b cos bx
-----------------------------a 2 + b2
Taken from the Reference:Calculus and Analytic Geometry, Thomas, Addison
Wesley, Third Ed. 1965, pp. 369
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 33
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Now, proceeding;
^
I(t) = I e - t/RC w [ cos  e t/RC coswt dt - sin  e t/RC sinwt dt ] + e - t/RC I(0)
e
t/RC
cos wt dt =
e t/RC
w sin wt + (1/RC) cos wt
---------------------------------(1/RC)2 + w2
e
t/RC
sin wt dt =
e t/RC
(1/RC) sin wt - w cos wt
---------------------------------(1/RC)2 + w2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 34
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Now, proceeding;
^
I(t) = I e - t/RC w [ cos  e t/RC coswt dt - sin  e t/RC sinwt dt ] + e - t/RC I(0)
w sin wt + (1/RC) cos wt
(1/RC) sin wt - w cos wt
^
t/RC
t/RC
t/RC
I(t) = I e
w [ cos e
---------------------------------- - sin e
---------------------------------- ] + e - t/RC I(0)
(1/RC)2 + w2
^
I(t) = I e
- t/RC
w [ cos
e t/RC
(1/RC)2 + w2
w sin wt + (1/RC) cos wt
(1/RC) sin wt - w cos wt
t/RC
---------------------------------- - sin e
---------------------------------- ] + e - t/RC I(0)
(1/RC)2 + w2
(1/RC)2 + w2
^
Iw
I(t) = ----------------- [ cos [w sin wt + (1/RC) cos wt ] - sin [(1/RC) sin wt - w cos wt ] ] + e - t/RC I(0)
(1/RC)2 + w2
^I w
I(t) = ----------------- [ (wcos - (1/RC) sin  ) sin wt + ((1/RC) cos  + w sin ) cos wt ] + e - t/RC I(0)
(1/RC)2 + w2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 35
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
^I w
I(t) = ----------------- [ (wcos - (1/RC) sin  ) sin wt + ((1/RC) cos  + w sin ) cos wt ] + e - t/RC I(0)
(1/RC)2 + w2
^
Iw
I(t) = ------------------√(1/RC)2 + w2

wcos - (1/RC) sin 
(1/RC) cos  + w sin
------------------------------- sin wt + -------------------------------- cos wt
√ (1/RC)2 + w2
√ (1/RC)2 + w2
w / √ (1/RC)2 + w2 = cos 

+ e - t/RC I(0)
(1/RC) / √ (1/RC)2 + w2 = sin 
w2 + (1/RC)2 = r 2
(1/RC)
r = √ (1/RC)2 + w2

w
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 36
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
^I w
I(t) = ------------------√(1/RC)2 + w2

wcos - (1/RC) sin 
(1/RC) cos  + w sin
------------------------------- sin wt + -------------------------------- cos wt
√ (1/RC)2 + w2
√ (1/RC)2 + w2

+ e - t/RC I(0)
^I w
I(t) = -------------------  (cos cos -sin sin  ) sin wt + (sin cos  +cos sin ) cos wt  + e - t/RC I(0)
√(1/RC)2 + w2
cos  cos  - sin  sin  = cos( +  )
sin  cos  + cos  sin  = sin ( +  )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 37
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AC Circuits
Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
^
Iw
I(t) = --------------------  cos ( +  ) sin wt + sin ( +  ) cos wt  + e - t/RC I(0)
√(1/RC)2 + w2
sin ( wt +  +  )
^
Iw
I(t) = ------------------- sin ( wt +  +  ) + e - t/RC I(0)
√(1/RC)2 + w2
Switch is turned “on” at: t = 0 sec
I(t)
R
C
Vc(0)=V0
^
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 38
+
^
+ V(t) = V sin(wt + )
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Subsituting the above term into the solution;
Switch is turned “on” at: t = 0 sec
Switch
I(t)
^
R
+
+
Iw
I(t) = ------------------- sin ( wt +  +  ) + e - t/RC I(0)
√(1/RC)2 + w2
Vc(0)=V0
C
Itr(t) =Transient Term
_
Iss(t) = Steady-State Term
^
V(t) = V sin(wt + )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 39
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Calculation of Initial Value of I(t)
Switch is turned “on” at: t = 0 sec
Initially, the system is an AC circuit operating in
steady state, with a resistance R and capacitance C
drived by an AC source
I(t)
R
^
Hence the current I(t) depends on the source
voltage V(t) and Z = R + j X
I 
= V 
I 
= V 
I 
= V /Z
+
+
Vc(0)=V0
C
^
V(t) = V sin(wt + )
_
V(t) = V sin (wt + )
/ ( R + jX )
^
/Z
^
 +Tan-1 (X / R)
-Tan-1 (X / R)
Please note that capacitive
reactance has negative angle
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 40
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Calculation of Initial Value of I(t)
I(t)
r = √ (1/RC)2 + w2
(1/RC)

X = 1 / (wC) 
X / R = 1 / (wRC)
Tan-1 (X / R) = Tan-1 (1 / (wRC) = Tan-1((1/RC) / w) =
+
+
w
R
Vc(0)=V0
C
^
V(t) = V sin(wt + )
_
w2 + (1/RC)2 = r 2
Switch is turned “on” at: t = 0 sec
^
I (t) = V / Z sin (wt + +Tan-1 (X / R)))
^
I (0) = V / Z sin ( + Tan-1 (X / R))
^
^
I (0) = V / Z sin ( +  )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 41
AC Circuits
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Subsituting the above term into the solution, the final form of the
solution waveform becomes;
^
Iw
^
I(t) = ------------------- sin ( wt +  +  ) + e - t/RC V / Z sin ( +  )
√(1/RC)2 + w2
Iss(t) = Steady-State Term
Itr(t) =Transient Term
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 42
AC Circuits
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R-C Circuits: Example
Example
Switch is turned “on” at: t = 0 sec
Now assume that the parameters of the circuit on
the RHS are as follows;
I(t)
^
V(t) = V sin wt = 312 sin wt Volts
R
+
Vc(0)=V0
C
Iw
I(t) = ------------------- sin ( wt +  +  ) + e - t/RC I(0)
√(1/RC)2 + w2
^
V(t) = V sin(wt + )
_
^
Iss(t) = Steady-State Term
+
R = 10 Ohms
C = 10 µ Farads
Itr(t) =Transient Term
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 43
AC Circuits
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20
R-C Circuits: Example
a sin wt
Steady-State Term
16
12
8
4
0
-4
0,0
0,5
1,0
1,5
2,0
2,5
3,0
-8
-12
a sin wt
+
b cos wt
= sin ( wt +  +  )
-16
-20
20
^
Iw
I(t) = --------------------  cos ( +  ) sin wt +sin ( +  ) cos wt
√(1/RC)2 R-C
+ w2 Circuits: Example
a
b
b cos wt
16
12
8
4
0
-4 0,0
0,5
1,0
1,5
2,0
2,5
3,0
-8
-12
-16
-20
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 44
AC Circuits
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20
R-C Circuits: Example
a sin wt
Steady-State Term
16
12
^
8
4
0
-4
0,0
0,5
1,0
1,5
2,0
2,5
3,0
Iw
I(t) = ------------------- sin ( wt +  +  )
√(1/RC)2 + w2
-8
-12
-16
-20
20
b cos wt
16
25
12
20
8
15
4
10
0
-4 0,0
-8
-12
5
0,5
1,0
1,5
2,0
2,5
3,0
0
-5 0
0,5
1
1,5
2
2,5
-10
-15
-16
-20
-20
-25
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 45
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AC Circuits
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R-C Circuits: Example
20
I tr (t)
Overall Solution
16
^
Iw
I(t) = ------------------- sin ( wt +  +  ) + e - t/RC I(0)
√(1/RC)2 + w2
12
8
4
0
0,0
25
0,5
1,0
1,5
2,0
2,5
3,0
I ss (t)
I ss (t) + I tr (t)
30
15
10
20
5
10
0
-10
-15
-20
-25
Itr(t) =Transient Term
40
20
-5 0
Iss(t) = Steady-State Term
0,5
1
1,5
2
2,5
3
0
-10
0,0
0,5
1,0
1,5
2,0
2,5
3,0
-20
-30
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 46
AC Circuits
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RC Circuits with DC Voltage Source - Two Simple Rules
Rule - 1
A filled capacitor acts initially as a DC
voltage source due to the stored charge
The initial voltage of this capacitor may
then be represented as a DC voltage source
in series with an uncharged capacitor
R=2
0
V(t) = (1 / C)  I(t)dt = V(0) = V0
-∞
VDC
+
V0 +
I(0)
I(0)
R=2
C =1000 µF
Vc(0)=V0
I(0) = (VDC-V0) / R
+
VDC –V0 +
_
+
C =1000 µF
Vc(0)=0
+
VDC
+
R=2
C =1000 µF
Vc(0)=0
_
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 47
AC Circuits
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RC Circuits with DC Voltage Source - Two Simple Rules
Then the initial value of current
will be;
I(0) = ( VDC – V0 ) / R
The current waveform will then be;
I(t) = I(∞) + [ I(0) - I(∞) ] e -t/
 is called the time constant of the circuit defined
Please note that an uncharged capacitor acts
effectively as short circuit, i.e. V0 = 0
 = RC = The time required a for capacitor to reach
63 % of its full charge
= 2  x 1000 µ F = 2 x 1000 x 10-6 = 0.002 sec
R
I(0) = ( VDC – V0 ) / R
I(t) (Amp)
V = VDC-V0 +
0.4
+
0.5
I(0)
0.3
SC
_
Where
as;
0.2
0.1
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 48
AC Circuits
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RC Circuits with DC Voltage Source - Two Simple Rules
Rule - 2
A fully charged capacitor acts finally as
open circuit to DC current
I(t) = C d/dt V(t) = C d/dt (constant) = 0 (OC)
The current waveform will then be;
I(t) = I(∞) + [ I(0) - I(∞) ] e -t/

Where is called the time constant of the circuit
defined as;
 = RC = The time required a for capacitor to
reach 63 % of its full charge
= 2  x 1000 µ F = 2 x 1000 x 10-6 = 2 msec
R
VDC +
I(∞) = 0
0.5
OC
I(t) (Amp)
0.4
0.3
0.2
Then the final value of current will be;
I(∞) = 0
0.1
I(∞) = 0
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 49
AC Circuits
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RC Circuits with DC Voltage Source - Two Simple Rules
R=2
Solution
VDC
+
C =1000 µF
Vc(0)=V0
_
The initial value of current will be;
I(0) = ( VDC – V0 ) / R
The final value of current will be;
I(∞) = 0
+
Substituting the above expressions into
the current expression
I(t)
I(t) (Amp)
24
I(0) = 18 A
20
16
The current waveform will then be;
I(t) = I(∞) + [ I(0) - I(∞) ] e -t/
I(t) = [( VDC – V0 ) / R ] e -t/
12
% 63 of the initial value
8
4
I(∞) = 0
0
0

0.002
0.004
0.006
0.008
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 50
0.01
AC Circuits
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Meaning of the Time Constant
Definition
Time constant  of a electric circuit is
the duration for the current to get
reduced by 63 % of its initial value
Time constant  of an RC circuit is
simply expressed as:
 = RC

The Effect of  on decay
5
4
% 63 of the initial value
3
% 63 of the inital value
2
1
0
0
0.5
1
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
t (sec)
2
2 > 1
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 51
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AC Circuits
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Example
R= 2 Ohm
Problem
V(∞) = VS = 24 Volts
V(t) (Volts)
_
Capacitor will behave as a DC source at the
beginning and as OC at the end, hence;
V(0) = V0 = 6 Volts
Vc(0) = V0 = 6 V
C = 1 mF
24.0
20.0
16.0
12.0
% 63 of the final value
8.0
4.0
0
The voltage waveform will then be;
+
Find the voltage waveform across the 1 mF
capacitor shown on the RHS, when it has an
initial voltage of 6 Volts and charged by a 24
Volts DC voltage source through a wire with 2
Ohm resistance
+ VDC= 24 Volts
I(t)
0
0.002
0.004
0.006
0.008
0.01
t (sec)
V(t) = V(∞) + [V(0) - V(∞)] e -t/
 = RC = Time Constant: The time required
V(t) = 24 + ( 6 – 24 ) e-t/0.002 = 24 -18 e-t/0.002 Volts
for a capacitor to reach 63 % of full charge
= 2 x 1000 µF = 2 x 0.001 = 0.002 sec
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 52
AC Circuits
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Energy Stored in a Capacitor
Instantaneous Energy
Stored in a Capacitor
Assuming that the instantaneous
voltage across the capacitor is Vc(t)
P(t) = VC(t) I(t)
Example
Calculate the stored energy in a 10 µF
capacitor fully charged with a 12 Volts DC
voltage
Wc= (1/2) 10 x 10-6 x 122 = 720 x 10-6 Joule
I(t)
WC(t) =  P(t) dt
=  VC(t) I(t) dt
or
V(t)
Vc(t)
C
_
= C  VC(t) dVC(t)
+
+
=  VC(t) C dVC(t) / dt dt
WC(t) = (1/2) C VC2(t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 53
AC Circuits
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Example
^
Problem
V(t) (Volts)
312
C
Angle (Radians)
^
V (t) = V sin wt
WC(t) = (1/2) C VC2(t)
^
Wc(t) = (1/2) 10 x 10-6 V 2 sin 2 wt
= 0.000005 x 3122 sin2 wt
= 0.4867 sin2 wt
= 0.4867 ( ½ – ½ cos 2wt )
0
/2

3/2
2
V(t) = 312 sin wt
- 312
0.6
Wc(t)
C = 10 µ F
0.5
0.4
0.3
0.2
sin2
cos2
wt = 1 –
wt = 1 – ( 1 + cos2wt ) / 2
= ½ – ½ cos2wt
Mean of Wc(t) > 0
0.1
Angle (Radians)
0
/2

3/2
2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 54
_
+
+
Find the instantaneous energy in the
capacitor for the voltage shown in the
figure
I(t)
V( t ) = V sin wt
AC Circuits
METU
Series Connected Capacitances
Series connected capacitances
V1(t) =
V2(t) =
+
----------=
V (t) =
=
(1/C1)  I(t)dt
(1/C2)  I(t)dt
+
------------------------------------[ (1/C1 ) + (1/C2) ]  I(t)dt
(1/Ctot )  I(t)dt
I(t)
C1
+
+
V1 (t)
V(t)
Hence,
Ctot
Series connected capacitances
are combined in the same way as
for shunt connected resistances
1
= ---------------------(1/C1 ) + (1/C2 )
C2
+
V2 (t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 55
AC Circuits
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Series and Shunt Connected Capacitances
Shunt connected capacitances
I1(t)
I2(t)
+
----------I (t)
=
=
=
=
=
C1 d V(t) / dt
C2 d V(t) / dt
+
---------------------------(C1 + C2) d V(t) / dt
Ctot
d V(t) / dt
Shunt connected capacitances are
simply added
I(t)
+
V(t)
Vc(t)
I1
I2
C1
C2
Where,
C tot = C1 + C2
is the total capacitance
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 56
AC Circuits
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Inductance
Definition
Coil
Core
Toroidal Coil
Toroidal Core
Inductance is a winding or
coil of wire around a core
Core may be either insulator
or a ferromagnetic material
+
Symbolic representation
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 57
AC Circuits
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Ferrite Core Toroidal Inductor
Definition
I
I
Ferriet core inductor has a
toroidal ferrit core inside
Ferrite core
Toroidal coil
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 58
METU
AC Circuits
Air Core Inductor
Configuration
Air core inductor has no core inside
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 59
AC Circuits
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Basic Relation
Voltage across an inductor is
proportional to the rate of change of
current
1 Henry is the value of inductance defined as
1 Henry = 1 Volt x 1 second / 1 Amp
I(t)
+ V(t)
L
+
V(t) = L d I(t) / dt
where, V(t) is the voltage across the
inductance,
I(t) is the current flowing
through,
L is the inductance (Henry)
Inductance L
Voltage Source V(t)
_
Definition
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 60
AC Circuits
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Current in an Inductance
Definition
Voltage Source V(t)
The equation;
Inductance L
I(t)
V(t) = L d I(t) / dt
can be written in inverse form as
where I(0) is the current initially flowing
in the inductor
+
I(t) = (1/L)  V(t)dt + I(0)
+
V(t)
L
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 61
AC Circuits
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Current in an Inductance
I (Amp) Voltage Waveforms
Phase Shift between Current and
V (Volts), I (Amp)
I(t) = (1/L)  V(t) dt
= (1/L)  Vmax sin wt dt
= - (Vmax / wL) coswt
= - Imax coswt
25
20
15
10
I(t)
5
0
V(t)
+
+
Inductance
Vmax
Imax
0.005
0.010
0.015
0.020
-5
-10
-15
-20
-25
Time (Sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 62
AC Circuits
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Series and Shunt Connected Inductors
Series connected inductors are added
V1(t) =
L1 d I(t) / dt
V2(t) =
L2 d I(t) / dt
+
----------- +-------------------------V(t) = (L1 + L2) d I(t) / dt
= L tot
d I(t) / dt
where
L tot = L1 + L2
is the total inductance
Series connected inductances
I(t)
+
+
V1 (t)
L1
V(t)
+
L2
V2 (t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 63
AC Circuits
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Series and Shunt Connected Inductors
Shunt connected inductances are
combined in the same way as in shunt
connected resistances
I1(t) = (1/L1)  V(t)dt
I2(t) = (1/L2)  V(t)dt
+
-------- +-------------------I(t) = [ (1/L1 ) + (1/L2) ]  V(t)dt
= (1/Ltot )
 V(t)dt
Shunt connected inductances
I(t)
+
+
V(t)
Vc(t)
I1
I2
L1
L2
Hence,
Ltot
1
= ---------------------(1/L1 ) + (1/L2 )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 64
AC Circuits
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Example - 3
I(t)
Problem
Calculate the voltage across the 100 mH
inductor with the current shown in the figure
on the RHS
I(t) = 0
t<1s
I(t) = 1/((5-1)) (t – 1) = ¼ (t - 1)
1≤t≤5s
I(t) = 1
5≤t≤9s
I(t) = -1/((5-1)) (t – 13) = -¼ (t - 13) 9 ≤ t ≤ 13 s
I(t) = 0
t ≥ 13 s
+
+
I(t)
L
L = 100 mH
I(t) (Amp)
1.0
0.5
0.0
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 65
AC Circuits
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Example - 3
I(t)
Solution
V(t) = L d I(t) / dt
+
+
I(t)
L = 100 mH
L
Differentiating the expression for current
waveform
d/ I(t) dt = d (¼ (t - 1)) /dt = ¼
V(t) (Volts)
and multiplying by the inductance L (L = 10-1 H);
0.025
V(t) = 0
V(t) = 10-1 x ¼ = 0.025 V
V(t) = 0
V(t) = -10-1 x ¼ = -0.025 V
V(t) = 0
t<1s
1≤t≤5s
5≤t≤9s
9 ≤ t ≤ 13 s
t ≥ 13 s
0.0
-0.025
0.0
2.0
4.0
6.0
8.0
10.0
12.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 66
14.0
AC Circuits
METU
Example - 4
I(t)
Problem
V(t) = 0
V(t) = -10 V
V(t) = 0
t < 0 sec
0 ≤ t ≤ 1 sec
t ≥ 1 sec
+
+
Assume that the inductor shown on the RHS
is connected to a voltage source with the
waveform shown in the figure
Determine the inductor current waveform by
assuming that the initial current in the
inductor is zero
V(t)
L
L = 100 mH
V(t) (Volts)
0.0
-10
0.0
0.5
1.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 67
AC Circuits
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Example - 4
I(t)
Solution
I(t) = (1/L)  V(t) dt + I(0)
+
+
V(t)
L
Integrating the voltage expression;
I(t) = (1/L)  V(t) dt + I(0)
= 1/(100 x 10-3)  V(t) dt
I(t) = 0
I(t) = 1/(100 x 10-3)  V(t) dt
= 10  V(t) dt
= 10  -10 dt
= -100 t
I(t) = -100 A
t < 0 sec
0 ≤ t ≤ 1 sec
t ≥ 1 sec
L = 100 mH
I(t) (A)
0.0
-100
0.0
0.5
1.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 68
AC Circuits
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Energy Stored in an Inductor
I(t)
Problem
I(t)
VL(t)
L
_
P(t) = VL(t) I(t)
+
Calculate the instantaneous energy
stored in an inductor with an inductance
L and an instantaneous voltage VL(t)
WL(t) =  P(t) dt
=  VL(t) I(t) dt
=  I(t) L dI(t) / dt dt
= L  I(t) dI(t)
or
WL(t) = ½ L I2(t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 69
AC Circuits
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Example – 5
I(t)
Problem
Find the instantaneous energy in the
inductor for the current shown in the figure
t<1s
1≤t≤5s
5≤t≤9s
9 ≤ t ≤ 13 s
t ≥ 13 s
+
I (t) = 0
I (t) = ¼ (t - 1) Amp
I (t) = 1 Amp
I (t) = - ¼ (t – 13) Amp
I (t) = 0
+
V(t)
L
L = 10 mH
I(t) (Amp)
1.0
0.5
0.0
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 70
AC Circuits
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Example – 5
I(t)
Solution
+
+
WL(t) = ½ L I2(t)
V(t)
By using the above formula
W(t) = 0 Joule
W(t) = ½ 10-2 x (¼ (t - 1))2
= 0.3125 x 10-3 x (t-1)2 Joules
W(t) = 0.01 / 2 x 12 = 0.005 Joules
W(t) = ½ 10-2 x (¼ (t - 13))2
= 0.3125 x 10-3 x (t - 13)2 Joules
W(t) = 0
L = 10 mH
L
t<1s
1≤t≤5s
5≤t≤9s
9 ≤ t ≤ 13 s
t ≥ 13 s
Please note that 1 Joule = 1 Watt x 1 sec
W(t) (Joule)
0.006
0.005
0.004
0.003
0.002
0.001
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 71
AC Circuits
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R-L Circuits
Solve the R-L circuit shown on the RHS
which consists of a resistance in series
with an inductance for current waveform
when the switch is turned “on” at time:
t = 0 sec
Writing down KVL for the circuit
V(t) = R I(t) + L dI(t) / dt
or
dI(t) / dt + (R/L) I(t) = (1/L) V(t)
R
Switch
I(t)
+
+
Solution
Switch is turned “on” at: t = 0 sec
V(t)
I(0)=I0
L
_
Problem
A first order ordinary differential equation
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 72
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Solution of the resulting First Order Ordinary Differential Equation
^ sin ( wt +  )
V( t ) = V
Solution
R
Solve the resulting first order ordinary differential equation (ODE)
I(t)
dI(t) / dt + (R/L) I(t) = (1/L) V(t)
+
^
+
dI(t) / dt + (R/L) I(t) = (1/L) V sin ( wt +  )
V(t)
I(0)=I0
L
Voltage (Volt)
_
312
300
200
Amplitude
100
Angle
0
/2

3/2
(Radians)
2
Phase angle 
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 73
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AC Circuits
Solution of the resulting First Order Ordinary Differential Equation
^
V( t ) = V sin ( wt +  )
Solution
Solve the resulting first order ordinary differential equation (ODE)
^
R
dI(t) / dt + (R/L) I(t) = (V/L) sin ( wt +  )
+
V(t)
I(0)=I0
L
_
^
µ(t) dI(t)/dt + I(t) (R/L) µ(t) = µ(t) ( V / L ) sin ( wt +  )
^
µ(t) dI(t)/dt + I(t) d/dt µ(t) = ( V / L ) µ(t) sin ( wt +  )
^
d/dt [µ(t) I(t)]
= ( V / L ) µ(t) sin ( wt +  )
^
 d/dt [µ(t) I(t)] dt
= ( V / L )  µ(t) sin ( wt +  ) dt + I(0)
^
µ(t) I(t)
= ( V / L )  µ(t) sin ( wt +  ) dt + I(0)
^
I(t)
= I µ(t) -1  µ(t) sin (wt +  ) dt + µ(t)-1I(0)
+
Define an integration factor µ(t) = e t R/L
Multiply both sides of the above ODE by this factor;
I(t)
d/dt µ(t) = (R/L) µ(t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 74
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Subsituting the integration factor µ(t) = e t R/L into
the above solution;
Switch is turned “on” at: t = 0 sec
Switch
R
^
I(t)
I(t) = I e – t R/L  e t R/L sin (wt +  ) dt + e – t R/L I(0)
+
+
V(t)
e
t R/L
sin wt dt =
e t R/L
(R/L) sinwt - w coswt
---------------------------------(R/L)2 + w 2
Taken from the Reference: Calculus and Analytic Geometry,
Thomas,Addison Wesley, Third Ed. 1965, pp. 369
L
_
Let  = 0 (for simplicity)
I(0)=I0
^
V( t ) = V sin ( wt +  )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 75
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Solution of the resulting First Order Ordinary Differential Equation
Solution (Continued)
Subsituting the above term into the solution;
Switch is turned “on” at: t = 0 sec
Switch
R
^ – t R/L t R/L (R/L) sinwt - w coswt
I(t) = I e
e
---------------------------------- + e – t R/L I(0)
(R/L)2 + w 2
+
+
^ (R/L) sinwt - w coswt
I(t) = I ----------------------------------- + e – t R/L I(0)
(R/L)2 + w 2
I(0)=I0
V(t)
L
_
^
I
I(t) = ------------------- ( sinwt - (wL/R) coswt ) + e – t R/L I(0)
(wL/R)2 + 1
Steady-State Term
I(t)
^
V( t ) = V sin ( wt +  )
Transient Term
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 76
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Solution of the resulting First Order Ordinary Differential Equation
Numerical Example
Switch is turned “on” at: t = 0 sec
Steady-State Term
R
I(t)
+
I(0)=I0
V(t)
L
_
^
I
I(t) = ------------------- ( sinwt - (wL/R) coswt ) + e – t R/L I(0)
(wL/R)2 + 1
Switch
+
Now assume that the parameters of the circuit on
the RHS are as follows;
V(t) = 312 sin wt Volts
R = 1 Ohms
L = 10 mHenry
^ sin ( wt +  )
V( t ) = V
Transient Term
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 77
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Solution of the resulting First Order Ordinary Differential Equation
Numerical Example
I tr (t)
25
20
^^
I
I(t) = ------------------- ( sinwt - (wL/R) coswt ) + e – t R/L I(0)
(wL/R)2 + 1
15
10
5
Transient Term
Steady-State Term
I ss (t)
0
250
200
400
150
100
300
50
200
0
0.5
1
1.5
2
2.5
3
3.5
4
0.5
1
1.5
2
2.5
3
3.5
4
1.5
2
2.5
3
3.5
4
I ss (t) + I tr (t)
100
-50
-100
0
-150
-100
-200
-200
-250
-300
0.5
1
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 78
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Solution for DC Voltage - Two Simple Rules
I(0) = 0
Rule - 1
I(0) = (1 / L)  V(t) dt = I0 = 0
-∞
L=0.002 H
VDC
_
0
+
I(t)
+
An inductor with no initial current acts as
an open circuit to a DC voltage source
initially
R=1
(OC)
R
Then, the initial value of current will be;
I(0) = 0
I(0) = 0
+
VDC
OC
I(0) = 0
The current waveform will then be;
I(t) = I(∞) + [ I(0) - I(∞) ] e -t/
 = L / R = Time Constant: The time required for
the inductor to reach 63 % of full current
= 2 mH / 1  = 0.002 / 1 = 0.002 Sec
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 79
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Solution for DC Voltage - Two Simple Rules
I(∞) = V / R
Rule - 2
R=1
(SC)
I(t)
L=0.002 H
VDC
_
V(t) = L d/dt I(t) = L d/dt (ct) = 0
+
+
An inductor acts as a short circuit to a
DC current (or voltage) finally
Then, the final value of current will be;
R
I(∞) = V / R
VDC
I(0) = 0
The current waveform will then be;
I(t) = I(∞) + [ I(0) - I(∞) ] e -t/
I(∞)
SC
_
I(∞) = V / R
+
+
 = L / R = Time Constant: The time required for
the inductor to reach 63 % of full current
= 2 mH / 1  = 0.002 / 1 = 0.002 Sec
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 80
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Solution for DC Voltage - Two Simple Rules
R = 1 Ohms
The current waveform will then be;
I(t) = I(∞) + [ I(0) - I(∞) ] e -t/
I(t)
+
+
L = 2mH
V= 24 Volts
I(0) = 0
I(t) = V/R - V/R e -t/
= V/R (1- e -t/ )
I(t) (Amps)
or
_
I(∞) = V / R
24.0
20.0
16.0
12.0
 = L / R = Time Constant: The time required for
8.0
the inductor to reach 63 % of full current
= 2 mH / 1  = 0.002 / 1 = 0.002 Sec
4.0
0
% 63 of peak value (24 V)
0

0.002
0.004
0.006
0.008
0.01
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 81
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Example - 6
R = 1 Ohms
Problem
I(t) (Amps)
Inductor will be SC at the end, hence;
I(∞) = V / R = 24 / 1 = 24 Amps
24.0
20.0
16.0
I(∞) = 24 A
12.0
The current waveform will then be;
8.0
I(t) = I(∞) + [ I(0) - I(∞) ] e –t / 
4.0
I(t) = 24 + (6 - 24) e-t / 0.002 = 24 - 18 e-t / 0.002 Amps
L = 2mH
V= 24 Volts
_
Inductor has 6 Amp initial current;
I(0) = I0 = 6 Amp
I(t)
+
Find the current waveform in the 2 mH
inductor with 6 Amps initial current connected
to a 24 Volt DC voltage source through a wire
with 1 Ohm resistance as shown on the RHS
+
I(0) = 6 A
0
0
0.002
0.004
0.006
0.008
0.01
t (sec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 82
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Example - 7
+
V= 24 V
Solve the circuit shown on the RHS for
current waveform flowing in the inductor
R2 = 5 Ohms
_
R1 = 10 Ohms
A
A
+
L = 0.4 H
V= 24 V R2 = 5 Ohms
B
Kill the voltage source, and find Req.
R eq = 10 // 5
= 10 x 5 /(10 + 5)
= 10 / 3 Ohms
L = 0.4 H
I(t)
B
Solution
First take out the branch containing inductor,
and find the Thevenin Equivalent circuit of the
part shown on the LHS seen from the
terminals A and B
A
+
Problem
R1 = 10 Ohms
B
A
R1 = 10 Ohms
A
R2 = 5 Ohms
L = 0.4 H
B
B
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 83
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Example – 7 (Continued)
R1 = 10 Ohms
Solution (Continued)
Open circuit terminals A – B and find VAB
+
VA-B (t)
V= 24 V R2 = 5 Ohms
V = 24 V x R2 / ( R1 + R2 )
= 24 x 5 / 15 = 24 / 3
=8V
• Form the resulting Thevenin equivalent
Circuit,
• Connect the inductance to the resulting
Thevenin equivalent circuit,
• Solve the resulting circuit by using the
straightforward method described in
Example 6
A
A
B
+
R eq= 10 // 5 = 10 x 5 /(10+5)
= 10/3 Ohms
B
A
A
L = 0.4 H
V= 24 / 3 = 8 V
B
B
+
R eq= 10 // 5 = 10 x 5 /(10+5)
= 10/3 Ohms
A
L = 0.4 H
V= 24 / 3 = 8 V
B
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 84
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R-L-C Circuits
Problem
_
R
Writing down KVL for the circuit;
+
L
V(0)=VC0
C
+
V(t)
_
dV(t)/dt = R dI(t)/dt + Ld2I(t)/dt2 +(1/C)I(t)
or
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
C
V(t)
Solution
V(t) = R I(t) + VL(t) + VC(t)
= R I(t) + L dI(t)/dt + (1/C)  I(t)dt + VC(0)
Differentiating both sides wrt time once;
I(t)
+
+
Solve the following circuit for current waveform,
which consists of a resistance, an inductance
and a capacitance connected in series
L I(0)=IL0
R
A second order ordinary differential equation
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 85
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Initial Conditions
Differential Equation
R
L
C
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
+
+
Initial Conditions
_
IL(0) = IL0
VC(0) = VC0
2
L IL(0) = IL0
R
I(t)
+
+
Please note that a differential equation needs initial
conditions in number equal to its order, i.e. two here
1
V(t)
Vc(0)=VC0
C
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 86
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Initial Conditions
Differential Equation
R
L
C
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
+
+
Initial Conditions
_
The voltage initial condition VC(0) may also be
written as,
VC(0) = V(0) – VL(0) – VR(0)
= V(0) – L d/dt IL(0) – R IL(0)
+
I(t)
+
or
V(t)
2
_
d/dt IL(0) = IL’(0) = (1/L) [ V(0) – VC(0) – R IL(0) ]
L IL(0) = IL0
R
Vc(0)=VC0
C
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 87
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AC Circuits
Initial Conditions
Differential Equation
R
L
C
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
+
+
Initial Conditions
_
Thus the initial conditions in terms of the
solution variable I(t) may now be written as,
2
+
I(t)
+
Please note that a differential equation needs initial
conditions in number equal to its order, i.e. two here
L IL(0) = IL0
R
1
V(t)
_
IL(0) = IL0
IL’(0) = (1/L) [ V(0) – VC(0) – R IL(0) ]
Vc(0)=VC0
C
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 88
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Example
R-L-C Circuit
Initial Conditions
Solve the R-L-C circuit with the given parameters
shown on the RHS for current I(t)
VC(0) = VC0 = 87 Volts
IL(0) = IL0 = 0
R=2
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
I(t)
+
1 / (LC) = 1/(1x 2.494 x 10-3) = 401
C = 2.494 mF
+
2/1=2
L=1H
V(t) = 100 e- 4 t
d2I(t)/dt2 + (2/1) dI(t)/dt + 401 I(t) = (1/1) dV(t)/dt
d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = d/dt (100 e-4t ) = - 400 e-4t
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 89
_
Writing down KVL for the circuit shown on the
RHS the following ODE is obtained
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Solution
R=2
R-L-C Circuit
I(t)
+
C = 2.494 mF
+
First, obtain the homogeneous equation by
setting the RHS source function to zero
L=1H
V(t) = 100 e- 4 t
d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = 0
_
Then, solve the characteristic equation
s2 + 2 s + 401 = 0
c = 401
s1, s2 = ( - b
= -1

±
b=2
±
a =1
j 20
b2 - 4 x a x c ) / (2 a)
Eigenvalues of the differential equation
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 90
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Solution
R=2
L=1H
R-L-C Circuit
k1(cos θ - j sin θ)
C = 2.494 mF
V(t) = 100 e- 4 t
_
I(t) = k1 e s1 t + k2 e s2 t
= k1 e (-1 – j20 ) t + k2 e (-1 + j20 ) t
= k1 e - t x e – j20 t + k2 e - t x e
= e - t (k1 e – j20 t + k2 e j20 t )
+
+
Then, the homogeneous solution becomes
I(t)
j20 t
k1(cos θ + j sin θ)
= e - t [ k1 (cos 20 t – j sin 20 t )
+ k2 (cos 20 t + j sin 20 t ) ]
Taken from the Reference: Calculus and
Analytic Geometry, Thomas,Addison
Wesley, Third Ed. 1965, pp. 867
Euler’s Identity
e jθ = cos θ + j sin θ
θ = 20 t
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 91
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Euler’s Identity
Definition
Graphical Representaion
e jθ = cos θ + j sin θ
│e jθ │= │ cos θ + j sin θ │
= √ cos2 θ + sin2 θ
=1
^
sin θ
θ
z
x
cos θ
y
x2 + y 2 = z 2
z = √ x2 + y2
EE
EE 209
209 Fundamentals
Fundamentals of
of Electrical
Electrical and
and Electronics
Electronics Engineering,
Engineering, Prof.
Prof. Dr.
Dr. O.
O. SEVAİOĞLU,
SEVAİOĞLU, Page
Page 92
92
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Solution
R=2
R-L-C Circuit
I(t)
+
B
V(t) = 100 e- 4 t
_
I(t) = e [ k1 (cos 20t – j sin 20t )
+
k2 (cos 20t + j sin 20t ) ]
= e - t [ (k1 + k2 ) cos 20t + (k2 – k1 ) sin 20t ]
-t
C = 2.494 mF
+
Rearranging the terms
A
L=1H
Unknown coefficients to determined
Hence, the homogeneous solution (decaying
sinusoidal term) becomes;
I(t) = e - t ( A cos 20t + B sin 20t )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 93
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Solution
R=2
Nonhomogeneous Solution
(Transient Term)
I(t)
+
C = 2.494 mF
+
V(t) = 100 e- 4 t
Definition of the Nonhomogeneous Solution
General form of the nonhomogeneous solution
(transient term) may be expressed as
In(t) = c e – 4 t
where, In(t) is the nonhomogeneous solution,
c is an unknown coefficient to be determined
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 94
_
Now the nonhomogeneous solution (transient
term) is to be determined
L=1H
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Solution
R=2
Nonhomogeneous Solution
(Transient Term)
I(t)
+
C = 2.494 mF
V(t) = 100 e- 4 t
In(t) = c e – 4 t
_
to the given differential equation;
d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = - 400 e-4t
and solve it for the unknown coefficient c
d2(ce – 4 t)/dt2 + 2 d(ce – 4 t)/dt + 401 ce – 4 t = -400 e-4t
16c e – 4 t + 2c(-4 e – 4 t ) + 401 c e – 4 t = - 400 e-4 t
+
Substitute the nonhomogeneous solution;
L=1H
These terms cancel
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 95
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Solution
Nonhomogeneous Solution
(Transient Term)
These terms cancel
d2(ce – 4 t)/dt2 + 2 d(ce – 4 t)/dt + 401 ce – 4 t = -400 e-4t
16c e – 4 t + 2c(-4 e – 4 t ) + 401 c e – 4 t = - 400 e-4 t
R=2
16 c - 8 c + 401 c = - 400
L=1H
I(t)
C = 2.494 mF
+
+
V(t) = 100 e- 4 t
Thus, the nonhomogeneous solution becomes;
In(t) = - 0.97799 e – 4 t
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 96
_
409 c = - 400 or c = - 400 / 409 = - 0.97799
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Solution
R=2
Complete Solution
I(t)
+
C = 2.494 mF
+
Complete solution is the summation of the
homogeneous (decaying sinusoidal) and
nonhomogeneous solutions (transient term)
L=1H
V(t) = 100 e- 4 t
_
Transient Term
Decaying Sinosoidal Term
I(t) = -0.97799 e – 4 t + e - t ( A cos 20t + B sin 20t )
Unknown coefficients to determined
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 97
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AC Circuits
Solution
R=2
Determination of the Unknown
Coefficients
I(t)
+
C = 2.494 mF
+
I(t) = - 0.97799 e – 4 t + e - t ( A cos 2t + B sin 2t )
L=1H
V(t) = 100 e- 4 t
_
Transient Term
Decaying Sinosoidal Term
The above solution must satisfy the given
initial conditions;
IL (0) = IL0 = 0
VC(0) = VC0 = 87 Volts
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 98
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Solution
Determination of the Unknown
Coefficients
IL (0) = IL0 = 0
VC(0) = VC0 = 87 Volts
Substitute the given initial conditions into the
complete solution equation and solve for the
unknown coefficients A and B;
I(t) = - 0.97799 e – 4 t + e - t ( A cos 2t + B sin 2t )
Transient Term
L=1H
I(t)
+
V(t) = 100 e- 4 t
C = 2.494 mF
+
Decaying Sinosoidal Term
R=2
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 99
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Solution
Determination of the Unknown
Coefficients
IL (0) = IL0 = 0
VC(0) = VC0 = 87 Volts
Substitute the given initial conditions into the
complete solution equation and solve for the
unknown coefficients A and B;
IL(t) = - 0.97799 e – 4 t + e - t ( A cos 20t + B sin 20t )
1
1
1
R=2
L=1H
0
I L(t)
+
V(t) = 100
e- 4 t
C = 2.494 mF
+
IL(0) = - 0.97799 e 0 + e 0 ( A cos 0 + B sin 0 )
= - 0.97799 + A = 0
_
A = 0.97799
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 100
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Solution
Determination of the Unknown
Coefficients
IL (0) = IL0 = 0
VC(0) = VC0 = 87 Volts
IL(t) = - 0.97799 e – 4 t + e - t ( A cos 20t + B sin 20t )
L=1H
I L(t)
87 V
0V
+
V(t) = 100
e- 4 t
C = 2.494 mF
+
V(0) = 100 e- 4 t
= 100 e- 4 x 0
= 100 V
R=2
_
d/dt IL(0) = IL’(0) = (1/L) [ V(0) – VC(0) – R IL(0) ]
= (1/1) (100 – 87 – 2 x 0 )
= 13 Amp/sec
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 101
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Solution
R=2
Determination of the Unknown
Coefficients
e-t
d/dt IL(t) = 0.97799 x 4
(A cos 20t +B sin 20t )
+ e - t ( -20 A sin 20 t + 20 B cos 20t )
I L(t)
+
V(t) = 100 e- 4 t
C = 2.494 mF
+
e–4t
L=1H
_
d/dt IL(0) = 0.97799 x 4
+
e0
1
e0
1
-
e0
0
1
1
0
(A cos 0 + B sin 0 )
1
(-20 A sin 0 + 20 B cos 0 ) =13 Amp/sec
= 0.97799 x 4 – A + 20 B = 13 Amp/sec
= 0.97799 x 4 – 0.97799 + 20 B = 13 Amp/sec
B = (13 - 3 x 0.9799) / 20 = 0.5033
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 102
AC Circuits
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Solution Terms
General form of the Solution
IL(t) = - 0.9799 e – 4 t + e - t ( 0.97799 cos 20t + 0.5033 sin 20t )
Transient Term
Decaying sinosoidal Term
R=2
L=1H
I(t)
C = 2.494 mF
+
+
V(t) = 100 e- 4 t
_
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 103
AC Circuits
METU
Solution Terms
Transient Term
IL(t) = - 0.9799 e – 4 t + e - t ( 0.97799 cos 20t + 0.5033 sin 20t )
Transient Term
0.00
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
R=2
-0.20
-0.60
I(t)
+
C = 2.494 mF
+
-0.40
L=1H
V(t) = 100 e- 4 t
_
-0.80
-1.00
-1.20
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 104
AC Circuits
METU
Solution Terms
Sinusoidal Terms
IL(t) = - 0.9799 e – 4 t + e - t ( 0.97799 cos 20t + 0.5033 sin 20t )
R=2
L=1H
Sinosoidal Terms
1.20
I(t)
C = 2.494 mF
+
+
0.80
V(t) = 100 e- 4 t
0.40
_
0.00
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.40
0.9779 cos 20t
-0.80
0.5033 sin 20t
-1.20
0.9779 cos 20t + 0.5033 sin 20t
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 105
AC Circuits
METU
Solution Terms
Exponentially Decaying Sinusoidal Term
IL(t) = - 0.9799 e – 4 t + e - t ( 0.97799 cos 20t + 0.5033 sin 20t )
1.20
e
0.80
R=2
-t
L=1H
I(t)
+
0
0.5
1
1.5
2
2.5
3
3.5
4
C = 2.494 mF
+
0.40
V(t) = 100 e- 4 t
_
-0.40
-0.80
-1.20
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 106
AC Circuits
METU
Solution Terms
Exponentially Decaying Sinusoidal Term
IL(t) = - 0.9799 e – 4 t + e - t ( 0.97799 cos 20t + 0.5033 sin 20t )
1.20
R=2
e - t x ( 0.97799 cos 20t + 0.5033 sin 20t )
L=1H
0.80
I(t)
+
0
0.5
1
1.5
2
2.5
3
3.5
4
C = 2.494 mF
+
0.40
V(t) = 100 e- 4 t
_
-0.40
-0.80
-1.20
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 107
AC Circuits
METU
Solution Terms
Overall Solution
IL(t) = - 0.9799 e – 4 t + e - t ( 0.97799 cos 20t + 0.5033 sin 20t )
R=2
0.80
0.40
I(t)
1.0
1.5
2.0
2.5
3.0
3.5
4.0
+
C = 2.494 mF
+
0.5
0.0
L=1H
V(t) = 100 e- 4 t
-0.40
_
-0.80
-1.20
-1.60
Homework:
Solve the same problem for the
case that the voltage source has
a sinusoidal waveform
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 108
AC Circuits
METU
RMS Value
Definition: RMS (Root Mean Square)
IDC
The above definition is based on the principle of
equating the heating effects of the AC and DC
currents calculated in both cases
First, calculate the power dissipated (heating
effect) in the resistance R in the DC circuit shown
on the RHS
VR
PDC
+
+
VDC
VR
R
_
RMS value of an AC current is the value of DC
current that would dissipate the same amount of
power on a resistance R
= R x IDC
= VR x IDC
= R x IDC2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 109
AC Circuits
METU
RMS Value
RMS (Root Mean Square)
I(t) = Imax sin wt
2.5
Now, calculate the power dissipated (the
heating effect) in the resistance R in the
AC circuit shown on the RHS
2.0
VR (t) = I(t) x R
1.5
PAC (t) = V(t) x I(t)
= R I(t)2 = R ((Vmax / R) sin wt )2
= R (I max sin wt )2
1.0
-0.5
/2

2
3/2
Angle (Radians)
+
VR(t)
R
_
V(t) = Vmax sin wt
0
0
V(t) +
Imax
0.5
I(t)
I(t) = (Vmax / R) sin wt
I 2(t) = (Imax sin wt )2
-1.0
-1.5
-2.0
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 110
AC Circuits
METU
RMS Value
Definition
Now let us equate the power dissipations
in the above cases
IDC
VR = R x IDC
PDC = VR x IDC
= R x IDC2
+
+
_
VR (t) = R x I(t)
PAC (t) = V(t) x I(t)
= R I(t)2 = R ((Vmax / R) sin wt )2
= R (I max sin wt )2
R
VR
VDC
I(t) = (Vmax / R) sin wt
Imax
I(t)
+
V(t) +
V(t)
R
_
V(t) = Vmax sin wt
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 111
AC Circuits
METU
RMS Value
Definition
VR
PDC
= R x IDC
= VR x IDC
= R x IDC2
60
45
40
R x IDC2
20
0
VR (t) = R x I(t)
PAC (t) = V(t) x I(t)
= R I(t)2 = R ((Vmax / R) sin wt )2
= R (I max sin wt )2
PDC
100
PAC(t)
/2

2
3/2
Angle (Radians)
80
60
40
20
0
0
/2

2
3/2
Angle (Radians)
R (I max sin wt )2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 112
AC Circuits
METU
RMS Value
Definition
60
In principle, the DC power waveform is NOT
equal to the AC power waveform
They may however, be equal in terms of their
averages, i.e. the mean of DC power transferred
to the load over a period may be equated to that
of AC power transferred to the load within a
period
Average(PAC )
PDC
45
40
R x IDC2
20
0
100
PAC(t)
/2

2
3/2
Angle (Radians)
80
60
45
40
20
0
0
/2

3/2
2
Angle (Radians)
R (I max sin wt )2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 113
AC Circuits
METU
RMS Value
Definition
60
Hence, equating the averages of the two terms;
Average( PDC ) = Average(PAC )
T
45
40
R x IDC2
20
T
PDC avg = (1/T)  R IDC2 dt = (1/T) R IDC2 t | = R IDC 2
0
PDC
0
0
100
PAC(t)
/2

2
3/2
Angle (Radians)
80
Average(PAC )
60
45
40
20
0
0
/2

2
3/2
Angle (Radians)
R (I max sin wt )2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 114
AC Circuits
METU
RMS Value
Definition
Please note that average value of an AC
waveform is calculated as
Area under the Curve
Average = -------------------------------Period
T
= (1/T)  PAC(t) dt
0
T
P(t) avg = (1/T)  PAC(t) dt
0
T
= (1/T)  R
I(t)2
dt
100
PAC(t)
80
60
0
45
40
20
0
0
/2

2
3/2
Angle (Radians)
T
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 115
AC Circuits
METU
RMS Value
PDC
Definition
60
Hence, equating the average of the two terms;
45
40
R IDC 2 = (1/T) R  I(t)2 dt
0
0
T
IDC 2 = (1/T)  I(t)2 dt
PAC(t)
/2

3/2
100
2
Angle (Radians)
80
0
Irms = √ ( 1/T )  I(t)2 dt
R x IDC2
20
T
Irms
60
45
40
20
0
0
/2

3/2
2
Angle (Radians)
R (I max sin wt )2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 116
AC Circuits
METU
Example
PAC(t)
Problem
100
Find the RMS value of the sinusoidal current
waveform;
I(t) = Imax sinwt
Irms
80
60
45
40
20
0
shown on the RHS
0
/2

3/2
2
Angle (Radians)
Solution
R (I max sin wt )2
Irms = √ ( 1/T )  I(t)2 dt
= √ ( 1/T )  [ Imax sin wt ] 2 dt
= Imax √ ( 1/T )  sin2 wt dt
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 117
AC Circuits
METU
RMS Value
PAC(t)
Solution
I RMS = Imax √ ( 1/T ) 
sin2
100
80
wt dt
60
45
40
Now use;
sin2
20
cos2
wt = 1 –
wt
= 1 – (1 + cos 2wt) / 2
= 1 – ½ – ½ cos 2wt
= ½ – ½ cos 2wt
0
0
/2

3/2
Angle (Radians)
R (I max sin wt )2
I RMS = Imax √ ( 1/T )  (½ –2½ cos 2wt ) dt
I RMS = Imax √ ( 1/T ) [ ½ dt –  ½ cos 2wt dt ]
Taken from the Reference: Calculus and
Analytic Geometry, Thomas, Addison Wesley,
Third Ed. 1965, pp. 348
=0
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 118
AC Circuits
METU
RMS Value
PAC(t)
Solution
100
80
I RMS = Imax √ ( 1/T ) [  ½ dt –  ½ cos 2wt dt ]
=0
I RMS = Imax √ ( 1/T ) ½  dt = Imax√ ½ ( 1 / T ) T
IDC = √ ½ I max
= I max / √ 2 = I rms = Imax x 0.7071
60
45
40
20
0
0
/2

3/2
2
Angle (Radians)
R (I max sin wt )2
Rule: RMS value of a sinusoidal waveform is 1/ √ 2 of its peak value
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 119
AC Circuits
METU
RMS Value of a Sinusoidal Waveform
Voltage (Volts)
Problem
Calculate the RMS value of the
sinusoidal voltage waveform shown on
the RHS
Vrms = Vmax / √ 2
= V max x 0.7071
= 312 x 0.7071 = 220 Volts
312
220
Vrms = 220 Volts
Angle (Radians)
0
/2

3/2
- 312
RMS value of the domestic voltage in Turkey
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 120
2
AC Circuits
METU
Example - 8
Problem
Calculate the RMS value of the rectangular
voltage waveform shown on the RHS
I(t)
V(t)
Voltage (Volts)
Vrms = √ ( 1/T )  V(t)2 dt
0
0
0.4
0.5
0.6
32
24
16
0.3
dt +  ( -
0.3
-4
Voltage2 (Volts)
√ (1/0.3) [ 
42
0.2
Time (msec)
_
Vrms = √ ( 1/T )  V(t)2 dt
0.1
0.1
-2
VR(t)
= 4 Volts
2
+
+
=
4
4)2
dt ]
8
V2(t) = 16
0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Time (msec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 121
METU
AC Circuits
Any Questions Please ?
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 122