Module
In
Electrical Circuits 2
Module 5 Resistance (R), Inductance (L), and Capacitance (C) Circuits
Module 6 Energy in an Inductive Circuit
Module 7 The Series Inductance-Capacitance (L-C) Circuit.
Module 8 The Series Resistance-Inductance-Capacitance (R-L-C) Circuit
Resistance (R), Inductance (L), and Capacitance (C) Circuits
Definition of Terms: Resistance (R) – is the opposition of current flow in a conductor. Its unit is ohms (Ω).
- It is the ratio of voltage to current for constant voltage and current.
Conductance (G)– a measure of how well the material will conduct electricity. It is the reciprocal of resistance.
Inductance (L) – is a property, which resists changes in current. Its unit is Henry (H). – reciprocal of inductance.
Capacitance (C) – it is the ability to store electrical charge. Its unit is Farad (F).
Elastance – a measure on how “susceptible” an element is to the passage of current thru it. And it is the reciprocal of
capacitance.
Inductive Reactance (XL) – is the opposition to current flow, which results in the continual interchange of energy between
the source and the magnetic field of the inductor.
- it is also the property of an inductor that makes the current to lag behind the voltage by 90 elec deg. (For purely
inductive load only)
Capacitive Reactance (XC) – is the opposition to the flow of charge, which results to the continual interchange of energy
between the source and the electric field of the capacitor.
- it is the property of capacitor that makes the current to lead the voltage by 90 elec deg. (For purely capacitive
circuit only.)
Impedance (Z) – is the geometric sum of the IR and IXL voltage drops (for R-L circuit) and IR and IXC drops (for R-C circuits.
- it is measured in Siemens (S) or Mho.
Admittance (Y) – a measure of how easily a network will “admit” the passage of current thru the system. And it is the
reciprocal of impedance.
Impedance diagram – a vector display that clearly depicts the magnitude of the impedance of the resistive, reactive, and
capacitive components of a network, and the magnitude and angle of the total impedance of the system.
Phasor – a radius vector that has a constant magnitude at a fixed angle from the positive real axis and that represents a
sinusoidal voltage or current in the vector domain.
Phasor Diagram – a vector display that provides at a glance the magnitude and phase relationships among the various
voltages and current of a network.
Basic Types of Circuit:
The study of circuits involves three basic types of units. These are the following:
1.
The Resistance (R) – circuit a. Circuit Diagram:
2. Wave Diagrams
3. Phasor Diagram
The instantaneous power (p) in the resistance circuit:
The instantaneous power (p) in the purely resistive circuit can be expressed by the formula…
p=exi
Substituting e and i; p = (Emsinwt) x (Imsinwt) = EmImsin2wt
p = (Emsinwt) x (Imsinwt) = EmImsin2wt
2
2
But: Sin wt = ½ - ½ cos wt
By proper substitution, the equation may be rewritten in the ff. form:
p = EmIm ( ½ - ½ cos2wt)
= (EmIm/2) – [(EmIm/2)cos2wt]
Note:
“The average power per cycle represented by the cos2wt term is zero because the positive and negative halves of
each cycle of a cosine (or sine) function cancel each other.
This means therefore that the constant term, represented by EmIm / 2, is the average power delivered to the
resistor.
That is;
Pave = EmIm / 2 = EI watts (Formula 1)
Example
An incandescent lamp load, generally considered to be made up of resistors, takes 4.8 kw from a 120-volt ac
source. Calculate (a) the total current, (b) the instantaneous value of power, (c) the resistance of the load.
Solution:
I = P/E =4,800/120 = 40 amps
Pmax = ( √𝟐 𝑬) (√𝟐 𝑻) = 2EI = 2 x 4,800 = 9,600 watts
R = E/I = 120/40 = 3 ohms
2. The Inductance (L) – circuit
a. Circuit Diagram
b. Wave Diagram
c. Phasor Diagram
When a sinusoidal voltage is impressed across a pure inductance, the current wave will also be sinusoidal.
However, unlike the pure resistance circuit in which e and i are in phase, the current will lag behind the voltage by 90 elec
deg.
Current Growth in Inductive Circuits, it was shown that a direct current builds up in an R-L circuit in accordance
with the equation;
E = iR + L di/dt
When applied to ac circuits, where R is zero and impressed emf is a sinusoidal function, it becomes;
e = Em sin wt = L di/dt
Rearranging the equation becomes;
di = (Em/L) sin wt dt
By integration;
i = (Em/wL) sin(wt - 2/‫)ח‬
or;
i = Im sin(wt - 2/‫)ח‬
Where:
Im = Em/wL ; XL = 2‫ח‬fL = wL
The instantaneous power (p):
p = e x I = (Em sinwt) x (Im coswt)
p = - EI sin2wt
Where:
Pave = 0 for a purely inductive circuit.
Pmax = EI
Example:
An inductance of 0.106 Henry is connected to a 120-volt 60-cycle source. Calculate (a) the inductive reactance,
(b) the current in the circuit, (c) the average power taken by 27 the inductor, (d) the maximum power delivered to the
inductor or returned to the source. Write the equations for (e) the current, and (f) the power.
Solution:
a. XL = 2πfL = 2π x 60 x 0.106 = 40 amps
b. I = E/XL = 120/40 = 30 3 amps
c. Pave = 0
d. Pmax = EI = 120 x 3 = 360
e. i = Em/wL sin(wt – π/2) 3 √𝟐 sin (377t – π/2)
f.
p = -EI sin 2wt = - 120 x 3 sin 2 x 377t = -360 sin 754t
Energy in an Inductive Circuit:
To determine the energy in joules(watt-sec) that is stored in the inductor, it is important to understand that this
storage takes place in continually changing increments of p dt as the current rises from zero to a maximum. This is between
2/‫ ח‬to ‫ ח‬radians.
From the instantaneous power equation;
that is,
p = - EI sin 2wt
It follows that the differential energy delivered to the inductor in time dt is…
dW = p dt
Substituting the instantaneous power p and by integrating this equation;
W = LI2 joules (Formula)
Where:
W = energy stored in the inductor, Joules
L = inductance of the circuit, Henry
I = current in Amp
Example:
When a 117-volt 60-cycle source is connected to a pure inductor, the current is 3.9 amp. (a) Write the equation
for the current in the circuit. Determine (b) the energy stored 28 in the inductor bet wt1 = 2/‫ ח‬and wt2 = ‫; )ח‬c) the stored
energy bet wt1 = 4/‫ ח‬and wt2 = 34/‫; )ח‬d) the energy stored in the magnetic field bet wt1 = 4/‫ ח‬and wt2 = ‫ח‬.
Solution:
a. i = 3.9 √𝟐 sin (377t – π/2)
𝟏𝟏𝟕
b. L = 𝟑𝟕𝟕 𝒙 𝟑.𝟗 = 0.0795 henry
W = 0.0795 x (3.9)2 = 1.21 joules
𝟎.𝟎𝟕𝟗𝟓 𝒙 (𝟑.𝟗)𝟐
𝟎.𝟎𝟕𝟗𝟓 𝒙 (𝟑.𝟗)𝟐
+
=0
𝟐
𝟐
𝟎.𝟎𝟕𝟗𝟓 𝒙 (𝟑.𝟗)𝟐
+ 0.0795 x (3.9)2 = 0.605
𝟐
c. W = d. W =
joule
The Capacitance (C) – Circuit
a. Circuit Diagram
b. Wave
Diagram
c. Phasor Diagram
When a sinusoidal voltage is impressed across a pure capacitance, the current wave will also be sinusoidal.
However, unlike the pure resistance circuit in which e and 29 i are in phase, the current will lead the voltage by 90 elec
deg.
Charging Current in an RC Circuits, and in accordance with Kirchhoff’s Law,
𝒕 𝒊𝒅𝒕
E = iR + ∫𝒐
𝑪
When applied to ac circuits, where R is zero and impressed emf is a sinusoidal function, it becomes;
𝒕
e = Em sin wt = 1/C ∫𝒐 𝒊𝒅𝒕
Rearranging the equation becomes;
q = CEm sin wt
Getting the derivatives;
dq/dt = wCEm coswt
or
i = wCEm sin (wt + 2/‫)ח‬
Where:
Im = wCEm ; XL = 2‫ח‬fL = wL
The instantaneous power (p):
p = e x i = (Em sinwt) x wCEm sin (wt + 2/‫)ח‬
p = EI sin 2wt
Where:
Pave = 0 for a purely capacitive circuit.
Pmax = EI
Example:
A 127-μf capacitor is connected to a 125-volt 50-cycle source. Calculate (a) the capacitive reactance, (b) the
current in the circuit, (c) the average power taken by the 30 capacitor, (d) the maximum power delivered to the
capacitor or returned to the source. Write the equations for (e) the current and (f) the power.
Solution:
𝟏𝟎𝟔
𝟏𝟎𝟔
a. Xc = 𝟐𝝅𝒇𝒄 = 𝟐𝝅 𝒙 𝟓𝟎 𝒙 𝟏𝟐𝟕 = 25 ohms
𝑬
𝟏𝟐𝟓
b. I = 𝑿𝒄 = 𝟐𝟓 = 5 amps
c. Pave = 0
d. Pm = EI = 125 x 5 625 watts
𝝅
𝝅
e. i= wCEm sin (wt + 𝟐 ) = 5√𝟐 sin (314t + 𝟐 )
f. p = EI sin 2wt = 125 x 5 sin 628t = 625 sin 628t
Energy in a capacitive Circuit:
To determine the energy in joules (watt-sec) that is stored in the capacitor, it is important to understand that this
storage takes place in continually changing increments of p dt as the current rises from zero to a maximum. This is between
0 to 2/‫ ח‬radians.
From the instantaneous power equation; that is,
p = EI sin2wt
It follows that the differential energy delivered to the capacitor in time dt is…
dW = p dt
Substituting the instantaneous power p and by integrating this equation;
W = CE2 joules (Formula)
Where:
W = energy stored in the capacitor, Joules
C = capacitance of the circuit, Henry
I = current in Amp
Example:
The current in a circuit is 1.96 amp when a capacitor is connected to a 250-volt 50-cycle source. (a) Write the
current 31 and power equations. (b) Determine the energy stored in the capacitor during the positive half of the power
wave.
Solution:
𝝅
a. i = 1.96 √𝟐 sin (314t + 𝟐 )
p = 250 x 1.96 sin ( 2 x 314)t = 490 sin 628t
𝟏.𝟗𝟔 𝒙 𝟏𝟎𝟔
b. C = 𝟑𝟏𝟒 𝒙 𝟐𝟓𝟎 = 25𝝁𝒇
W= 25 x 10-6 x ( 250) 2 = 1.56 joules
The Series Inductance-Capacitance (L-C) Circuit
Circuit Diagram
Formula:
If XL > XC (The series L-C behave like an inductance)
a. Xeq = XL - XC
If XC > XL (The series L-C behave like a capacitor)
c. a. Xeq = XC - XL
Note: “an increase in L or C will result in an over-all increase in the inductive reactance of such a circuit; a decrease in
L or C will, on the other hand, result in an overall increase in the capacitive reactance of a similar circuit.”
Example:
1. A series circuit consisting of a 0.0795-henry inductor and a 177-F capacitor is connected to a 120-volt 60-cycle
source. Calculate (a) the equivalent reactance of the 32 circuit, (b) the circuit current, indicating whether the latter lags or
leads.
Solution:
a. XL =2π x 60 x 0.0795 = 30 ohms
𝟏𝟎𝟔
XC =
= 15 ohms
𝟐𝛑 𝐱 𝟔𝟎 𝐱 𝟏𝟕𝟕
Xeq = 30 -15 = 15 ohms (inductive)
b. I =
𝟏𝟐𝟎
𝟏𝟓
= 8 amp
2. A series circuit consisting of a 0.0795-henry inductor and a 177-F capacitor is connected to a 120-volt variable
frequency source. At what frequency will the circuit take a lagging current of 4 amp?
Solution:
Xeq. =
𝟏𝟐𝟎
𝟒
= 30 ohms
Also
Xeq = XL – XC
XL = 2πf x 0.0795 = 0.5
𝟏𝟎𝟔
𝟏
XC = 𝟐𝛑𝐟 x 𝟏𝟕𝟕 =
30 = 0.5f -
𝟗𝟎𝟎
𝒇
𝟗𝟎𝟎
𝒇
For which
f=
𝟔𝟎±√𝟑,𝟔𝟎𝟎+𝟕,𝟐𝟎𝟎 𝟔𝟎 ±𝟏𝟎𝟒
= 𝟐
𝟐
= 82 cycles
The Series Resistance-Inductance (R-L) Circuit.
Circuit diagram
Phasor Diagram
Working Formulas:
𝑧 = √𝑅 2 + 𝑥𝐿2
E=IxZ;
Cos  = p.f. = ER / E = IR / IZ = R/Z = P / EI
e = Em sinα ; i = Im sin (α-)
P = EI cos 
Example
A load of 18.4-kw operating at a power factor of 0.8 lagging is connected to a 460-volt 60-cycle source. Calculate
(a) the load current; (b) the power-factor angle; (c) the 33 equivalent impedance, resistance, and reactance of the load.
(d) Write the equations for the voltage and current.
Solution:
𝑃
18,000
a. I = 𝐸 cos 𝜃 = 460 𝑥 0.8 = 50 amp
b. Cos- 1 0.8 = 36. 8 elec deg.
460
c. 𝑍 = 50 = 9.2 ohms
R = Z cos 𝜃 = 9.2 x 0.8 = 7.36 ohms
XL = √9.22 − 7.362 = 5.52 ohms
d. 𝑒 = 460√2 sin 2𝜋 𝑥 60𝑡 = 650 sin 377𝑡
36.8
𝑖 = 50 √2 sin ( 377𝑡 − 180 𝜋) = 70.7 sin(377𝑡 − 0.205𝜋)
The Series Resistance-Capacitance (R-C) Circuit.
a. Circuit Diagram
b. Phasor Diagram
Working Formulas:
E = I x Z ; Z =  R 2 + XC 2
Cos  = p.f. = ER / E = IR / IZ = R/Z = P / EI
e = EM sinα ; i = IM sin (α+)
P = EI cos 
Example:
A 125-volt 25-cycle source is connected to a series circuit consisting of a 30-ohm resistor and a 159-µF capacitor.
Calculate the ff: impedance, current, power factor, power.
Solution:
106
)2
2𝜋 𝑥 25 𝑥 159
𝑧 = √𝑅 2 + 𝑋𝐿2 = √(30)2 + (
= √302 + 402 = 50 𝑜ℎ𝑚𝑠
𝐸
𝑍
=
𝑃𝐹 =
𝑅
𝑍
𝐼=
125
50
=
= 2.5 𝑎𝑚𝑝
30
50
= 0.6 𝑙𝑒𝑎𝑑𝑖𝑛𝑔
𝑃 = 𝐸𝐼 cos 𝜃 = 125 𝑥 2.5 𝑥 0.6 = 187.5 𝑤𝑎𝑡𝑡𝑠
𝑃 = 𝐼 2 𝑅 = (2.5)2 𝑥 30 = 187.5 𝑤𝑎𝑡𝑡𝑠
A series R-C circuit takes a current whose equation is I = 0.85 sin (754t + п/4) when connected to a source of emf
having the equation e = 340 sin 754t. Calculate (a) the values of Z, R and XC,; (b) the capacitance of the capacitor; (c) the
circuit power factor and power.
Solution:
a. 𝐼 =
𝑃
𝐸𝑅
b. 𝑃𝐹 =
=
4,800
115
= 41.7 𝑎𝑚𝑝
𝑃
4,800
𝐸𝐼 230 𝑥 41.7
= 0.5 = cos 60°
c. 𝑍 =
𝐸 230
𝐼 41.7
5.51 𝑜ℎ𝑚𝑠
d. 𝑅 =
𝐸𝑅
𝐼
115
41.7
=
= 2.75 𝑜ℎ𝑚𝑠
e. 𝑋𝑐 = √𝑍 2 + 𝑅 2 = √(5.51)2 − (2.75)2 = 4.77 ohms
𝐶=
f.
106
2𝜋𝑓𝑥𝑐
=
106
377 𝑥 4.77
= 555𝜇𝑓
𝑒 = 230 √2 sin(2𝜋 𝑥 60𝑡) = 325 sin 377𝑡
60
𝜋
𝑖 = 41.7 √2 sin (2𝜋 𝑥 60𝑡 + 180 𝜋) = 59 sin(377𝑡 + 3 )
The Series Resistance-Inductance-Capacitance (R-L-C) Circuit
a. Circuit Diagram
b. Phasor Diagram
Working Formulas:
At XL > XC ; The circuit acts like an R-L circuit and the current is lagging.
𝐸 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
At XL < XC ; The circuit acts like an R-C circuit and the current is leading.
𝐸 = √𝑅 2 + (𝑋𝑐 − 𝑋𝐿 )2
E = I x Z ; P = EI cos 
Note:
At XL = XC, the circuit is said to produce resonance. And the circuit is like a purely resistive circuit.
That is;
Z = R and XL = XC
Example:
A series circuit consisting of an 80-ohm resistor, a 0.3-henry inductor, and a 50-µF capacitor is connected to a 120volt 60-cycle source. Calculate the ff: (a) equivalent impedance of the circuit, (b) current, (c) voltage drops across the
several units, (d) power and power factor.
Solution:
a. 𝑋𝐿 = 2𝜋 𝑥 60 𝑥 0.3 = 113 𝑜ℎ𝑚𝑠
𝑋𝑐 =
106
2𝜋 𝑥 60 𝑥 50
= 53 𝑜ℎ𝑚𝑠
𝑍 = √(80)2 + ( 113 − 53)2 = 100 𝑜ℎ𝑚𝑠
120
b. 𝐼 =
= 1.2 𝑎𝑚𝑝
100
c. 𝐸𝑅 = 1.2 𝑥 80 = 96 𝑣𝑜𝑙𝑡𝑠
𝐸𝐿 = 1.2 𝑥 113 = 135.6 𝑣𝑜𝑙𝑡𝑠
𝐸𝑐 = 1.2 𝑥 53 = 63.6 𝑣𝑜𝑙𝑡𝑠
d. 𝑃 = (1.2)2 𝑥 80 = 115. 2 𝑤𝑎𝑡𝑡𝑠
𝑃𝐹 =
115.2
120 𝑥 1.2
= 0.8 𝑙𝑎𝑔𝑔𝑖𝑛𝑔
The Impedance-Coil Resistance (RL-R) Circuit
a. Circuit diagram
b. Phasor Diagram
Working Formulas:
Cos  = (E2 – ER 2 – EZ 2) / 2EREZ
Rcoil = (EZ cos ) / I
Lcoil = (EZ sin ) / (I x 2Πf)
Example:
A 115-volt 60-cycle source is connected to a series circuit of a fixed resistor and an impedance coil. If the resistorand coil-voltage drops are 55.4 and 80 volts, respectively, under which condition the current is 1.69 amp, calculate the
resistance and inductance of the impedance coil.
Solution:
𝑐𝑜𝑠 𝜃 =
(115)2 − (55.4)2 − (80)2
2 𝑥 55.4 𝑥 80
=
3,753
8,8865
= 0.423
𝜃 = 65°
sin 𝜃 = 0.906
𝑅𝑐𝑜𝑖𝑙 =
𝐿=
80 𝑥 0.423
1.69
80 𝑥 0.906
1.69 𝑥 377
= 20 ohms
= 0.114 ℎ𝑒𝑛𝑟𝑦
Volt-amperes and Reactive Volt-amperes
The Power Triangle Diagram:
a. Inductive load
b. Capacitive load
Power Formulas:
a. True, Real or Active Power (P)
b. P = EI cos  = I2R watts
b. Reactive or Idle Power (Q)
Q = EI sin  = I2X vars or reactive volt-amperes
Reactive Factor (RF) = sin  = Q / S
c. Apparent Power (S)
S = EI = I2Z Volt-amperes
Example:
A load of 250 KVA, operating at a power factor of 0.86 lagging, is connected to a 2,300-volt a-c source. Calculate
(a) power, (b) current, (c) reactive kilovolt-amperes, (d) reactive factor.
Solution:
a. 𝑃𝑜𝑤𝑒𝑟 = 𝑘𝑣𝑎 𝑥 𝑃𝐹 = 250 𝑥 0.86 = 215 𝑘𝑤
1,000
250 𝑥 1,000
b. 𝐼 = 𝑘𝑣𝑎 𝑥 𝐸 =
108.7 𝑎𝑚𝑝
2,300
c. 𝜃 = 𝑐𝑜𝑠 −1 0.86 = 30°41′
sin 30°41′ = 0
𝑅 − 𝑘𝑣𝑎 = 𝑘𝑣𝑎 sin 𝜃 = 250 𝑥 0.51 = 127.5 𝑘𝑣𝑎𝑟
d. 𝑅𝐹 = √1 − (𝑃𝐹)2 = √1 − (0.86)2 = 0.51
Supplementary Problems:
1.
A 46-kv 0.8-lagging power factor load is connected to the end of a short transmission line where the voltage is
230. If the line resistance and reactance are 0.06 and 0.08 ohm, respectively, calculate the voltage at the sending
end.
2. A telephone receiver has an impedance of 306 ohms at 800 cps and a resistance of 60 ohms. For what value of
capacitance of a capacitor in series with the coil will the power factor be unity at 1,000 cps?
3. Two relays each with 20-ohm resistance and 0.16-henry inductance are connected in series. What is the equivalent
impedance?
a. 20 + j102.2 ohms
b. 20 + j120.63 ohms
c. 20 + j95.32 ohms
d. 10 + j25.32 ohms
4. A 50-µF and 100-µF capacitors are connected in series and across a 100 sin (wt + 30o) voltage. Write the equation
of the current.
a. 1.26 sin (wt + 120o) c. 5.65 sin (wt + 120o)
b. 1.26 sin (wt + 90o) d. 5.65 sin (wt + 90o)s
5.
If e = 100sin(wt+30o) – 50 cos3wt + 25 sin(5wt+150o) and i = 20 sin(wt+40o) + 10 sin(3wt + 30o) – 5 sin(5wt-50o).
Calculate the power in watts.
a. 1177 c. 1043
b. 937.6
d. 1224 38
6. An impedance draws a current I = 10 cos (wt – 300) Amp from a voltage v = 220 sin wt volts. What is the power?
(EE B.P. Oct. ’97)
a. 2200 W
b. 1100 W
c. 190.5 W
d. 1320 W
7. An impedance draws a current i = 10 cos (wt – 300) Amp from a voltage v = 220 sin (wt + 300) Volts. What is the
impedance of the circuit?
a. 15.6 – j15.6 ohms
c. 19.1 – j11 ohms
b. 15.6 + j15.6 ohms
d. 11 + j19.1 ohms
8. A series circuit has an applied voltage of V = 220 sin (wt + 300) and draws a current I = 10 sin (wt – 30o). What is the
average power and power factor of the circuit?
a. 1,905 W, 86.6% lagging
c. 2,200 W, 100 %
b. 1,905 W, 86.6% lagging
d. 1,100 W, 50% lagging
9.A circuit has resistance of 20 ohms and a reactance of 30 ohms. What is the power factor of the circuit?
a. 0
b. 0.55
c. 0.832 d. 0.99
10. A two-element series circuit with R = 15 ohms, L = 20 mH has an impedance of 30 ohms and an unknown angle. What
is the frequency in HZ?
a. 2.0
b. 21
c. 2,067
d. 207
11. A capacitor in series with a 200 ohms resistor draws a current of 0.3 ampere from 120 volts, 60 hz source. What is the
value of the capacitor in microfarad?
a. 8.7 b. 9.7
c. 6.7
d. 7.7
12. A 10-ohm inductive resistor is connected in series with an unknown capacitance. At 60 hz the impedance of the circuit
is 10 + j11.72. At 30 hz the impedance of the circuit is 10 – j5. What is the value of L in millihenrys?
a. 50
b. 500
c. 100 d. 250
13. A resistor R and capacitor C are connected in series across a 100 V, 60 cycle source. The reading of an ammeter
connected in the circuit is 2A and the reading of a voltmeter connected across the capacitor is 80 V. Calculate the values
of R and C.
a. 66 Ohms and 30 μF
c. 30 Ohms and 66 μF
b. 30 Ohm and 60 μF
d. 36 Ohms and 60 μF