Alternating Current (AC) Circuits
Dr Miguel Cavero
August 21, 2014
AC Circuits
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Alternating Current And Voltages
The alternating current and voltage can be expressed as
I = I0 cos ωt
V
= V0 cos (ωt + α)
The voltages across the three electrical components (R, L and C) are
VR = V0R cos ωt
VC
= V0C cos (ωt − π/2)
VL = V0L cos (ωt + π/2)
AC Circuits
AC Circuits
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Opposition To Current
Each component provides an opposition to current flow:
V0
I0
V0
I0
V0
I0
= R
=
1
= XC
ωC
= ωL = XL
resistance
capacitative reactance
inductive reactance
The impedance Z is the combined opposition to current in an AC
circuit.
q
Z = R2 + (XL − XC )2
Since R, XC and XL all have units of ohms (Ω), the units of impedance
is also the ohm (Ω).
AC Circuits
AC Circuits
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Capacitor In An AC Circuit
For an uncharged capacitor and a resistor in a DC circuit, initially
charges in the circuit can move freely so the current is large.
As the charge accumulates on the capacitor, the voltage across it
increases, opposing the current in the circuit.
After some time interval, that depends on the time constant τ = RC,
the current in the circuit approaches zero.
For a capacitor in an AC circuit, the alternating current has the same
effect as the capacitor continuously charging and discharging as the
current changes direction.
AC Circuits
AC Circuits
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Capacitor In An AC Circuit
Initially the current is large, as charge moves freely to accumulate on
the capacitor (point a).
As charge accumulates on the plates of the capacitor, the voltage
across it increases (c to d) while the current in the capacitor decreases
(a to b).
At point d, the voltage across the capacitor is a maximum and there is
no more current in the capacitor (point b).
AC Circuits
AC Circuits
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Capacitor In An AC Circuit
The current then flows in the opposite direction (b to e) as if the
capacitor is discharging, so the voltage across it decreases (d to f ).
When the voltage across the capacitor is zero (point f ), the current is
at its maximum as charge is free to move in the circuit.
The remainder of the cycle is the same as the first half, but for current
moving in the opposite direction.
AC Circuits
AC Circuits
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Inductor In An AC Circuit
At point a, the rate of change of the current (the slope of the current
curve) is at a maximum. The peak voltage across the inductor is
therefore also a maximum.
As the current increases (a to b), the slope of the curve decreases,
until it reaches a minimum at point b. The voltage is also a minimum at
the corresponding point d.
AC Circuits
AC Circuits
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Inductor In An AC Circuit
As the current decreases (b to e) to voltage increases, but in the
opposite direction that is was before. The voltage is negative (d to f )
since the induced (back) emf always opposes the change in the
current.
AC Circuits
AC Circuits
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Power Dissipation
The power P dissipated in a resistor R is given by
P = I 2R
where I is the instantaneous current.
The power dissipated in the resistor depends on the square of the
instantaneous current.
For an alternating current with a peak value of I0 , the power dissipated
in a resistor is not the same as the power dissipated in a resistor with a
direct current of the same value as I0 .
An alternating current is only at the peak value I0 for one instant in
time in its cycle.
The rms current Irms is the average value that is used - the direct
current that dissipates the same amount of energy in a resistor as the
energy dissipated in the resistor by the actual alternating current.
AC Circuits
Root-Mean-Square Current/Voltage
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Example
Consider a resistor R = 10 Ω, with a direct current of I = 5 A through it.
The power dissipated in the resistor is
P = I 2 R = 52 × 10 = 250 W
What alternating current through that resistor leads to the same
amount of power dissipated in it?
The average power hP i is given by
hP i = I 2 R
= (I0 cos ωt)2 R
= I02 (cos ωt)2 R
= I02 R (cos ωt)2
1 2
=
I R
2 0
AC Circuits
Root-Mean-Square Current/Voltage
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Root-Mean-Square
2 I R
(square the current, then take the mean)
1 2
=
I R
2 0 I0
I0
√
√
=
R (then take the square-root)
2
2
= (Irms ) (Irms ) R
hP i =
2
= Irms
R
The rms current is related to the peak current by
I0
Irms = √
2
AC Circuits
Root-Mean-Square Current/Voltage
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Example
Consider a resistor R = 10 Ω, with a direct current of I = 5 A through it.
The power dissipated in the resistor is
P = I 2 R = 52 × 10 = 250 W
An alternating current, which has a peak value of 7.0711 A, passes
through a resistor of 10 Ω. What is the average power dissipated in the
resistor?
D
E 1
1
hP i = (I0 cos ωt)2 R = I02 R = (7.0711)2 × 10 = 250 W
2
2
√
√
An alternating current with an rms value of Irms = I0 / 2 = 7.0711
A
2
leads to an average power dissipated in a 10 Ω resistor as the power
disippated through the same resistor when a direct current of 5 Ω
passes through it.
AC Circuits
Root-Mean-Square Current/Voltage
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Rms Voltage
The average value for the voltage in an AC circuit is also given in terms
of the peak voltage.
V0
Vrms = √
2
The average power dissipated in a resistor is
2
hP i = Vrms Irms = Irms
R=
2
Vrms
R
The rms current/voltage is the effective value of the current/voltage in
an AC circuit.
√
The mains supply is 230 V rms. The peak voltage is 2V0 = 325 V.
AC Circuits
Root-Mean-Square Current/Voltage
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Power Factor
Consider an AC circuit that has resistance as well as reactance (the
current and voltage are not in phase).
The instantaneous power P is
P
= V I = (V0 cos(ωt + α)) × (I0 cos ωt)
= V0 I0 [cos ωt cos(ωt + α)]
1
2
= V0 I0 cos ωt cos α + sin α sin ωt
2
(α is the phase angle between the voltage and the current.)
The average power can be shown to be
hP i = Vrms Irms cos α
Here, cos α is called the power factor.
AC Circuits
Root-Mean-Square Current/Voltage
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RLC AC Circuits
Consider a simple circuit containing a resistor R, inductor L and a
capacitor C all in series with an alternating source of emf.
What is the voltage across the combination (i.e. across all three
components)?
The total voltage at some point in time must equal the source voltage
at that instant.
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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RLC AC Circuits
What is the voltage across the resistor and inductor?
The voltage across R is not in phase with the voltage across L.
VRL 6= VR + VL
The voltage across the combination is
VRL = V0RL cos(ωt + α)
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Phasor Diagrams
Phasor diagrams are used in order to show the relationship between
the voltages across components and their phases.
The voltage across a component, the phasor, is represented by a
rotating vector. The diagram showing these is called a phasor diagram.
The length of the phasor is the value of the voltage, while α is the
phase angle.
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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RLC AC Circuits
The phasor diagram shows the relationship between the peak voltages
across each component and the peak voltage across the combination.
q
RL
=
V0
(V0R )2 + (V0L )2
p
=
(I0 R)2 + (I0 XL )2
q
= I0 R2 + XL2
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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RLC AC Circuits
What is the peak voltage across the three components?
The phasor VC is subtracted from the phasor VL , since they lie along
the same line but point in opposite directions. The phasor VL − VC is
always perpendicular to the phasor VR .
p
V0RLC = I0 R2 + (XL − XC )2
The ratio of voltage across a component/combination to the current is
the opposition to current
V 0 = I0 Z
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Impedance
The impedance Z in an AC circuit is the analogue to the resistance R
in a DC circuit.
Note that the impedance is a function of R, L and C, as well as the
angular frequency ω:
s
p
1 2
2
2
2
Z = R + (XL − XC ) = R + ωL −
ωC
The phase angle for the circuit is given by
tan α =
AC Circuits
VL − VC
XL − XC
=
VR
R
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1591.5 Hz.
(a) Calculate the peak voltage.
(b) Find the reactances XL and XC and the impedance Z.
(c) Find the peak current in the circuit.
(d) Determine the phase angle α.
(e) Calculate the peak voltage across each element in the circuit.
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1591.5 Hz.
(a) Calculate the peak voltage.
Given the rms voltage, the peak voltage V0 is
√
√
V0 = 2Vrms = 2 × 35.36 = 50.01 V
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1592 Hz.
(b) Find the reactances XL and XC and the impedance Z.
The frequency given is the cyclic frequency. The angular
frequency is
ω = 2πf = 2π × 1592 = 1.000 × 104 rad s−1
The reactances are
XL = ωL = 10000 × 60.0 × 10−3
XC
AC Circuits
= 600 Ω
1
1
=
=
ωC
10000 × 0.500 × 10−6
= 200 Ω
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1592 Hz.
(b) Find the reactances XL and XC and the impedance Z.
The impedance Z is then
p
Z =
R2 + (XL − XC )2
p
=
(300)2 + (600 − 200)2
= 500 Ω
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1591.5 Hz.
(c) Find the peak current in the circuit.
The Ohm’s Law equivalent for an AC circuit is V0 = I0 Z, therefore
I0 =
AC Circuits
V0
50.01
=
= 0.100 A
Z
500
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1591.5 Hz.
(d) Determine the phase angle α.
The phase angle is given by the peak voltages or the reactances
and resistance.
tan α =
VL − VC
XL − XC
400
=
=
VR
R
300
The phase angle α is therefore 53◦ .
Which reactance, XL or XC , is bigger?
Since XL > XC , the voltage leads the current by 53◦ .
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 1
A series RLC circuit has a resistor R = 300 Ω, an inductor L = 60.0 mH
and a capacitor C = 0.500 µF connected to an alternating voltage
source of 35.36 V rms at 1591.5 Hz.
(e) Calculate the peak voltage across each element in the circuit.
V0R = I0 R = 0.100 × 300
V0L = I0 XL = 0.100 × 600
V0C
= I0 XC = 0.100 × 200
Note that the peak voltage V0 of the source is not the sum of the
peak voltages of each component.
V0 is the vector sum of the phasors V0R , V0L and V0C .
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 2
A series RLC circuit has resistance R = 250 Ω, inductance
L = 0.600 H, capacitance C = 3.50 µF, frequency f = 60.0 Hz and a
peak voltage V0 = 150 V.
(a) Find the impedance of the circuit.
(b) Find the maximum current in the circuit.
(c) Calculate the phase angle.
(d) Determine the peak voltages across each component.
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 2
A series RLC circuit has resistance R = 250 Ω, inductance
L = 0.600 H, capacitance C = 3.50 µF, frequency f = 60.0 Hz and a
peak voltage V0 = 150 V.
(a) Find the impedance of the circuit.
The angular frequency is
ω = 2πf = 2π × 60.0 = 377 rad s−1
The reactances are
XL = ωL = 377 × 0.600 = 226 Ω
1
1
XC =
=
= 758 Ω
ωC
377 × 3.50 × 10−6
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 2
A series RLC circuit has resistance R = 250 Ω, inductance
L = 0.600 H, capacitance C = 3.50 µF, frequency f = 60.0 Hz and a
peak voltage V0 = 150 V.
(a) Find the impedance of the circuit.
The impedance Z is therefore
p
R2 + (XL − XC )2
Z =
p
=
(250)2 + (226 − 758)2
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 2
A series RLC circuit has resistance R = 250 Ω, inductance
L = 0.600 H, capacitance C = 3.50 µF, frequency f = 60.0 Hz and a
peak voltage V0 = 150 V.
(b) Find the maximum current in the circuit.
The maxiumum/peak current is related to the peak voltage and the
impedance.
V0
150
I0 =
=
Z
Z
where Z is the value from part (a).
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 2
A series RLC circuit has resistance R = 250 Ω, inductance
L = 0.600 H, capacitance C = 3.50 µF, frequency f = 60.0 Hz and a
peak voltage V0 = 150 V.
(c) Calculate the phase angle.
The phase angle α between the current and the voltage is given by
tan α =
XL − XC
226 − 758
=
R
250
What is the significance of the minus sign?
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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Example 2
A series RLC circuit has resistance R = 250 Ω, inductance
L = 0.600 H, capacitance C = 3.50 µF, frequency f = 60.0 Hz and a
peak voltage V0 = 150 V.
(d) Determine the peak voltages across each component.
Given the value of the peak current I0 from part (b), the peak
voltages across each component are
V0R = I0 R = I0 × 250
V0L = I0 XL = I0 × 226
V0C
= I0 XC = I0 × 758
What is the sum V0R + V0L + V0C compared to the peak voltage
V0 = 150 V?
AC Circuits
Combinations Of Resistors, Inductors And
Capacitors
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