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Handout 11: AC circuit
AC generator
Figure 1 compares the voltage across the directcurrent (DC) generator and that across the alternatingcurrent (AC) generator. For DC generator, the voltage is
constant. For AC type, the voltage takes a periodic form,
varying in size and sign. The standard form of the AC
voltage is a sinusoidal wave. Assume that the amplitude
is 𝑉0 and angular frequency is 𝜔. The voltage 𝑉 is given
by
𝑉 = 𝑉0 cos 𝜔𝑡
or
𝑉 = 𝑉0 sin 𝜔𝑡.
Figure 1: Comparison between DC voltage
and AC voltage
The function will be cosine or sine, depending on the
initial condition. However, both forms are contained in a
single expression
𝑉 = 𝑉0 𝑒 𝑖𝜔𝑡 .
The cosine form is the real part of the above equation
while the sine form is the imaginary part.
We can write down similar expression for AC
current, oscillating with amplitude 𝐼0 and angular
frequency 𝜔:
𝐼 = 𝐼0 𝑒 𝑖𝜔𝑡 .
Complex Impedance
An AC generator is connected, one at a time, to a
resistor, an inductor and a capacitor. The generator
produces current 𝐼 = 𝐼0 𝑒 𝑖𝜔𝑡 .
Resistor Consider the circuit in Fig. 2. The
voltage across the resistor
𝑉 = 𝐼𝑅 =
𝐼0 𝑅 𝑒 𝑖𝜔𝑡 .
The voltage varies the same angular frequency as
current and there is no phase difference (Fig. 4(b)). We
define complex impedance for the resistor
𝑍𝑅 =
𝑉
= 𝑅.
𝐼
The impedance is the resistance itself which is real.
Figure 2: Voltage and current across
resistor
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Inductor Consider the circuit in Fig. 3(a). The
voltage across the inductor is
𝑉 = 𝐿
𝑑𝐼
= 𝑖𝜔𝐿𝐼0 𝑒 𝑖𝜔𝑡 = 𝜔𝐿𝐼0 𝑒 (𝑖𝜔𝑡 +𝜋
𝑑𝑡
2)
The voltage varies with the same frequency as current
but with the phase 𝜋 2 ahead of the current. We define a
complex impedance for the inductor
(a)
𝑉
𝑍𝐿 =
= 𝑖𝜔𝐿.
𝐼
(b)
Figure 3: Voltage and current across (a)
inductor and (b) capacitor
The unit of the impedance is ohm.
Capacitor Consider the circuit in Fig. 3(b). The
charge stored in the capacitor varies as 𝑄 = 𝐼𝑑𝑡 =
𝐼0 𝑖𝜔 𝑒 𝑖𝜔𝑡 . The voltage across the capacitor is
𝑉 =
𝑄
𝐼0 𝑖𝜔𝑡
𝐼0 (𝑖𝜔𝑡 −𝜋
=
𝑒
=
𝑒
𝐶
𝑖𝜔𝐶
𝜔𝐶
2)
.
The voltage varies with the same frequency as current
but with the phase 𝜋 2 lag behind current. We define a
complex impedance for the capacitor
𝑍𝐶 =
𝑉
1
=
.
𝐼
𝑖𝜔𝐶
Series circuit
Consider a series circuit in Fig. 4 which consists
of a resistor, an inductor and a capacitor. The total
impedance of the circuit 𝑍 = 𝑍𝑅 + 𝑍𝐿 + 𝑍𝐶 is given by.
𝑍 = 𝑅 + 𝑖𝜔𝐿 +
1
1
= 𝑅 + 𝑖 𝜔𝐿 −
𝑖𝜔𝐶
𝜔𝐶
Note the the total impedance varies with angular
frequency 𝜔. Assume that the AC generator produces a
varying voltage 𝑉 = 𝑉0 𝑒 𝑖𝜔𝑡 . The current in the circuit is
given by
𝑉
𝑉0 𝑒 𝑖𝜔𝑡
𝐼 =
=
1
𝑍
𝑅 + 𝑖 𝜔𝐿 − 𝜔𝐶
The above equation can be simplified by writing 𝑍 in the
polar form
𝑍 = 𝑍 𝑒 𝑖𝜃 ,
Figure 4: Series circuit
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where
𝑍 =
1
𝑅 2 + 𝜔𝐿 −
𝜔𝐶
1
𝜔𝐿 − 𝜔𝐶
2
,
𝜃 = tan
−1
𝑅
.
Hence, the expression for the current in the circuit takes
the form
Figure 5: Variation of 𝑍 and phase angle 𝜃
with angular frequency 𝜔.
𝑉0 𝑒 𝑖𝜔𝑡
𝑉0 𝑒 𝑖(𝜔𝑡 −𝜃)
𝐼 =
=
.
𝑍
𝑍 𝑒 𝑖𝜃
It can be seen the amplitude of the current 𝐼 = 𝑉0 𝑍
and the phase of current differs from that of voltage by
𝜃.
Note the magnitude of the impedance is
minimum when the imaginary part of 𝑍 is zero, i.e.,
𝜔 = 1 𝐿𝐶 . This is known as the “resonance angular
frequency”𝜔0 . The variation of total impedance with 𝜔 is
shown in Fig. 5. At resonance, 𝑍 = 𝑅 and 𝑍 is
minimum here. At low frequency, 𝜃 = − 𝜋 2 and at high
frequency, 𝜃 = 𝜋 2. Because 𝑍 is minimum at the
resonance, the amplitude of the current 𝐼 becomes
maximum here (Fig. 6).
Figure 6: Variation of current
amplitude with 𝜔
Parallel circuit
Consider a parallel circuit in Fig. 7. The total
impedance 𝑍 satisfies
1
1
1
1
1
1
1
1
=
+ +
= +
+ 𝑖𝜔𝐶 = + 𝑖 𝜔𝐶 −
𝑍
𝑍𝑅 𝑍𝐿 𝑍𝐶
𝑅 𝑖𝜔𝐿
𝑅
𝜔𝐿
Assume that the AC generator produces voltage
𝑉 = 𝑉0 𝑒 𝑖𝜔𝑡 . The total current is the circuit is given by
𝑉
1
1
𝐼 =
= 𝑉0 + 𝑖 𝜔𝐶 −
𝑍
𝑅
𝜔𝐿
Figure 7: Parallel circuit
𝑒 𝑖𝜔𝑡 .
The above can be simplified by writing 1 𝑍 as
1
=
𝑍
1 𝑖𝜃
𝑒 ,
𝑍
where
1
=
𝑍
1
𝑅
2
1
+ 𝜔𝐶 −
𝜔𝐿
2
,
𝜃 = tan−1
1
𝜔𝐶 − 𝜔𝐿
1 𝑅
.
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Hence,
𝐼 = 𝑉0
𝑒 𝑖𝜃 𝑖𝜔𝑡
𝑉0 𝑒 𝑖(𝜔𝑡 +𝜃)
𝑒
=
.
𝑍
𝑍
The amplitude of the current is 𝐼 = 𝑉0 𝑍 and the
phase of current differs from that of voltage by 𝜃.
At a particular angular frequency 𝜔 = 1 𝐿𝐶 , the
reciprocal 1 𝑍 is minimum. Therefore, the current
amplitude is also minimum at this frequency.
Example 1 A series RLC circuit with 𝐿 = 160 mH,
𝐶 = 100 µF, and 𝑅 = 40.0 Ω is connected to a
sinusoidal voltage 𝑉 volts. At time 𝑡 seconds, the voltage
𝑉 𝑡 = 40𝑒 𝑖𝜔𝑡 , with 𝜔 = 200 rad s-1.
a) What is the impedance of the circuit?
b) Find the amplitude of the current in the circuit.
c) What is the phase difference between the voltage
and the current of the circuit? Which one leads?
Example 2 An inductor with inductance 𝐿 is connected
in series to a resistor with resistance 𝑅. The input
voltage takes the form 𝑉𝑖𝑛 = 𝑉0 𝑒 𝑖𝜔𝑡 . The output voltage
is taken across the resistor.
a) Show that
𝑉out
=
𝑉0
.
𝜔 2 𝐿2
1+ 2
𝑅
b) Sketch 𝑉out against 𝜔. Explain why this circuit is
called a “low-pass filter”.
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*Example 3 Consider the bridge circuit shown below.
When the circuit is balanced, show that
𝜔 =
1
𝑅2 𝑅4 𝐶2 𝐶4
.
Power
Let 𝑉 be the voltage across a device and 𝐼 be the
current passing through it. The electrical power P is
given by
𝑃 = 𝐼𝑉.
(∗)
The sign of 𝑃 varies. We are interested in the average
power 𝑃 = 𝐼𝑉 . The complex product of 𝐼 and 𝑉
cannot be used to find average because it is nonlinear.
To find the average, we must first take the real part,
multiply and find the average.
The voltage is given by 𝑉 = 𝑉0 𝑒 𝑖𝜔𝑡 . The real part
of the voltage Re 𝑉 = 𝑉0 cos 𝜔𝑡. The impedance of the
device takes the form 𝑍 = 𝑍 𝑒 𝑖𝜃 . Hence the current
through the device is 𝐼 = 𝑉 𝑍 = 𝑉0 𝑍 𝑒 𝑖(𝜔𝑡 −𝜃) with
amplitude 𝐼0 = 𝑉0 𝑍 . The real part of the current
Re 𝐼 = 𝑉0 𝑍 cos 𝜔𝑡 − 𝜃 . Hence, the power becomes
𝑃 = 𝑉02 𝑍 cos 𝜔𝑡 cos 𝜔𝑡 − 𝜃 . To find 𝑃 , we need to
find the average cos 𝜔𝑡 cos 𝜔𝑡 − 𝜃
= 1 2 cos 𝜃.
Then, the time average of power is given by
𝑃 =
𝑉02
cos 𝜃.
2𝑍
For a pure capacitance or pure inductance, 𝑍 is
pure imaginary, so 𝜃 = ±𝜋/2, so the average power is
zero. That means that no energy is dissipated in those
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circuit elements. They store energy but they do not
dissipate it. For a pure resistance 𝑍 = 𝑅 is real, so 𝜃 = 0,
so the average power is 𝑃 = 𝑉02 2𝑅 .This may not
immediately look like the usual relation for DC circuits,
𝑃 = 𝑉 2 𝑅 , but it is in fact equivalent, since the average
value of 𝑉 2 is just 𝑉02 2.
By introducing the root mean square voltage
𝑉rms =
𝑉2 =
𝑉0
,
2
the average power in equation (∗) can be written as
2
𝑃 = 𝑉𝑟𝑚𝑠
𝑍 cos 𝜃. Similarly, we define root mean
square current 𝐼𝑟𝑚𝑠 = 𝐼 2 = 𝐼0 2, where 𝐼0 is the
amplitude of the current. The rms voltage and current
are the quantities usually referred to for AC circuits,
rather than the amplitudes themselves which are a
factor of 2 larger.
Example 4 A light bulb uses average power of 70 W
when connected to a 60-Hz power source with a peak
voltage 170 V. Calculate the resistance of the light bulb.
Example 5 Consider a series RLC circuit with 𝑅 = 25 Ω,
𝐿 = 6.0 mH and 𝐶 = 25 𝜇F. The circuit is connected to a
600-Hz AC source with the root-mean-square voltage
𝑉rms = 10 V. Calculate the average power delivered to
the circuit.
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Supplementary note: complex number
Complex number
A complex number 𝑧 takes the form 𝑧 = 𝑎 + 𝑖𝑏
where 𝑎 and 𝑏 are real numbers and 𝑖 = −1. The value
𝑎 is called the “real part” and 𝑏 is called the “imaginary
part” of the complex number 𝑧.
Let 𝑧1 = 𝑎1 + 𝑖𝑏1 and 𝑧2 = 𝑎2 + 𝑖𝑏2 are complex
numbers.
Zero complex number 𝑧 = 0 means that 𝑎 = 0
and 𝑏 = 0.
Equal complex numbes If 𝑧1 = 𝑧2 , this implies
that 𝑎1 = 𝑎2 and 𝑏1 = 𝑏2 .
Complex conjugate The complex conjugate of 𝑧
is denoted by 𝑧 ∗ = 𝑎 − 𝑖𝑏.
Addition/Subtraction
𝑧1 + 𝑧2 = 𝑎1 + 𝑎2 + 𝑖 𝑏1 + 𝑏2
𝑧1 − 𝑧2 = 𝑎1 − 𝑎2 + 𝑖 𝑏1 − 𝑏2
For example, 3 − 2𝑖 − 1 + 𝑖 = 2 − 3𝑖.
Complex modulus A complex modulus 𝑧 is
defined by the equation 𝑧 2 = 𝑧𝑧 ∗ = 𝑎 + 𝑖𝑏 𝑎 − 𝑖𝑏 =
𝑎2 + 𝑏 2 . Therefore, the complex modulus
𝑧
𝑎2 + 𝑏 2 .
=
Some useful properties of complex modulus are
𝑧1 𝑧2
=
𝑧1 𝑧2
𝑧1
𝑧2
=
𝑧1
.
𝑧2
Division
Multiply and divide by complex
conjugate of the denominator. For example,
3−𝑖
3 − 𝑖 1 − 2𝑖
1 − 7𝑖
1 7
=
×
=
= − 𝑖.
1 + 2𝑖
1 + 2𝑖 1 − 2𝑖
5
5 5
Polar form Figure 1 shows a complex number
𝑧 = 𝑎 + 𝑖𝑏 plotted on a complex plane. The horizontal
axis is real part. The vertical axis is the imaginary part.
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The vector makes angle 𝜃 to the real axis. One can write
𝑎 = 𝑟 cos 𝜃 and 𝑏 = 𝑟 sin 𝜃. Hence,
𝑧 = 𝑎 + 𝑖𝑏 = 𝑟 cos 𝜃 + 𝑖𝑟 sin 𝜃 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 .
The above is known as “polar form” of complex
number. From Fig. 1, 𝑟 = 𝑎2 + 𝑏 2 = 𝑧 . By using
Euler’s formula
𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃,
one can express 𝑧 in the form
𝑧 =
𝑧 𝑒 𝑖𝜃 ,
where 𝑧 = 𝑎2 + 𝑏 2 and 𝜃 = tan−1 𝑏 𝑎 . Note that
𝑒 𝑖𝜃 = 1.
Example 1 Evaluate
a) 2 −3 + 𝑖
b) 1 − 2𝑖 2
c) 1 + 𝑖 (−2 + 𝑖)
Example 2 Express each of the followings in the form
𝑧 = 𝑧 𝑒 𝑖𝜃 .
a) 𝑧 = 2 + 2𝑖
b) 𝑧 = −1 + 𝑖 3
c) 𝑧 = 1 + 𝑖 𝑒 𝑖𝜋 /2
Figure 1: Complex number z plotted
on a complex plane called Argand
diagram