This document discusses sinusoidal steady state analysis of AC circuits. It covers sinusoids and their properties, phasor analysis using complex exponentials, and phasor relationships for basic circuit elements like resistors and inductors. Sinusoids are periodic functions that can be used to model AC voltages and currents. Phasor analysis allows solving circuits in the frequency domain by replacing time-varying quantities with phasors, making the equations algebraic. This allows determining voltage-current phase relationships and impedances.

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AC STEADY STATE
ANALYSIS
Chapter 8

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Chapter Outline
 Sinusoids
 Sinusoidal and complex forcing functions
 Phasor diagrams
 Impedance and admittance
 Analysis techniques

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Sinusoids
 Considering the sine wave 𝑥 𝑡 = 𝑋𝑀 sin 𝜔𝑡 where
𝑥 𝑡 could represent v(t) or i(t). 𝑋𝑀is the amplitude,
maximum value, or peak value; 𝜔 is the radian or
angular frequency; and 𝜔𝑡 is the argument of the sine
function. The function repeats itself every 2π radians.
This condition is described mathematically as
 𝑥 𝜔𝑡 + 2𝜋 = 𝑥(𝜔𝑡), or in general for period T, as
 𝑥 𝜔(𝑡 + 𝑇) = 𝑥(𝜔𝑡)
 meaning that the function has the same value at time
t+T as it does at time t.

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Sinusoids
Figure 8.1: Plots of a sine wave as a function of both ωt and t.

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Sinusoids
 The number of cycles per second, called Hertz, is the
frequency f, where
 𝑓 =
1
𝑇
 Since ωT=2π(Fig. 8.1a)
 𝜔 =
2𝜋
𝑇
= 2𝜋𝑓
 let us consider the following general expression for a
sinusoidal function:
 𝑥 𝑡 = 𝑋𝑀 sin(𝜔𝑡 + 𝜃)
 𝜃 is called the phase angle

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Sinusoids
 Because of the presence
of the phase angle, any
point on the waveform
𝑋𝑀 sin(𝜔𝑡 + 𝜃) occurs 𝜃
radians earlier in time
than the corresponding
point on the waveform
𝑋𝑀 sin 𝜔𝑡. Therefore, we
say that 𝑋𝑀 sin 𝜔𝑡 lags
𝑋𝑀 sin(𝜔𝑡 + 𝜃) by 𝜃
radians.
Figure 8.2: Graphical illustration of
𝑋𝑀 𝑠𝑖𝑛(𝜔𝑡 + 𝜃) leading 𝑋𝑀 𝑠𝑖𝑛 𝜔𝑡 by 𝜃
radians.

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Sinusoids
 Adding to the argument integer multiples of either 2π
radians or 360° does not change the original function
 The cosine function could be easily used as well,
since the two waveforms differ only by a phase angle;
that is,
 cos 𝜔𝑡 = sin 𝜔𝑡 +
𝜋
2
and sin 𝜔𝑡 = cos 𝜔𝑡 −
𝜋
2
 Note:
 − cos 𝜔𝑡 = cos(𝜔𝑡 ± 1800
) 𝑎𝑛𝑑
 −sin 𝜔𝑡 = sin(𝜔𝑡 ± 1800)

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Examples
 Determine the frequency and the phase angle
between the two voltages 𝑣1 𝑡 = 12 sin(1000𝑡 +

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Exercise
1. Given that 𝑣 𝑡 = 120 cos 314𝑡 + 𝜋/4 𝑉, determine
the frequency of the voltage in Hertz and the phase
angle in degrees.
2. Three branch currents in a network are known to be
𝑖1 𝑡
= 2 sin 377𝑡 + 450 𝐴 , 𝑖2 cos(377𝑡 + 100 𝑡

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Sinusoidal and Complex
Forcing Functions
Figure 8.3: A simple RL circuit

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Sinusoidal and Complex
Forcing Functions
 KVL equation for the circuit is
 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
+ 𝑅𝑖 𝑡 = 𝑉𝑀 cos 𝜔𝑡
 Since the input forcing function is 𝑉𝑀 cos 𝜔𝑡 , we
assume that the forced response component of the
current i(t) is of the form
 𝑖 𝑡 = 𝐴 cos(𝜔𝑡 + ∅)
 𝑖 𝑡 = 𝐴 cos(𝜔𝑡 + ∅) =
𝐴cos∅ cos 𝜔𝑡 − 𝐴𝑠𝑖𝑛∅ sin 𝜔𝑡 = 𝐴1 cos 𝜔𝑡 + 𝐴2 sin 𝜔𝑡
 Substituting into the DE

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Sinusoidal and Complex
Forcing Functions
 𝐿
𝑑
𝑑𝑡
𝐴1 cos 𝜔𝑡 + 𝐴2 sin 𝜔𝑡 + 𝑅 𝐴1 cos 𝜔𝑡 + 𝐴2 sin 𝜔𝑡 =
𝑉𝑀 cos 𝜔𝑡
Evaluating and equating the coefficients
 −𝐴1𝜔𝐿 + 𝐴2𝑅 = 0
 𝐴1𝑅 + 𝐴2𝜔𝐿 = 𝑉𝑀
Solving the above equations simultaneously and
substituting in 𝑖 𝑡 gives
 𝑖 𝑡 =
𝑅𝑉𝑀
𝑅2+𝜔2𝐿2 cos 𝜔𝑡 +
𝜔𝐿𝑉𝑀
𝑅2+𝜔2𝐿2 sin 𝜔𝑡

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Sinusoidal and Complex
Forcing Functions
Which can be written as
 𝑖 𝑡 = 𝐴 cos(𝜔𝑡 + ∅)
 Where 𝐴𝑐𝑜𝑠 ∅ =
𝑅𝑉𝑀
𝑅2+𝜔2𝐿2 and 𝐴𝑠𝑖𝑛 ∅ = −
𝜔𝐿𝑉𝑀
𝑅2+𝜔2𝐿2
 Thus tan ∅ = −
𝜔𝐿
𝑅
 𝐴𝑐𝑜𝑠 ∅ 2
+ 𝐴𝑠𝑖𝑛 ∅ 2
= 𝐴2
=
𝑉𝑀
2
𝑅2+𝜔2𝐿2
 𝐴 =
𝑉𝑀
𝑅2+𝜔2𝐿2
 Hence the final expression is
 𝑖 𝑡 =
𝑉𝑀
𝑅2+𝜔2𝐿2
cos 𝜔𝑡 − 𝑡𝑎𝑛−1 𝜔𝐿
𝑅

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Sinusoidal and Complex
Forcing Functions
 The preceding analysis indicates that ∅ is zero if L=0
and hencei(t) is in phase with v(t). If R=0, ∅ = −900,
and the current lags the voltage by 90°. If L and R are
both present, the current lags the voltage by some
angle between 0° and 90°.
 This example illustrates an important point: solving
even a simple one-loop circuit containing one resistor
and one inductor is very complicated compared to the
solution of a singleloop circuit containing only two
resistors. It would be more complicated to solve a
more complicated circuit using this procedure.

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Sinusoidal and Complex
Forcing Functions
 Let us determine the current in the RL circuit
examined in Figure 8.3.
 We will apply 𝑉𝑀𝑒𝑗𝜔𝑡
 The forced response will be
 𝑖 𝑡 = 𝐼𝑀𝑒𝑗 𝜔𝑡+∅
 Substituting this into the DE, and differentiating, then
dividing by the common factor will give.
 𝑅𝐼𝑀𝑒𝑗∅
+ 𝑗𝜔𝐿𝐼𝑀𝑒𝑗∅
= 𝑉𝑀
 which is an algebraic equation with complex
coefficients.

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Sinusoidal and Complex
Forcing Functions
 This equation can be written as
 𝐼𝑀𝑒𝑗∅ =
𝑉𝑀
𝑅+𝑗𝜔𝐿
 Converting the right-hand side of the equation to
exponential or polar form produces the equation
 𝐼𝑀𝑒𝑗∅
=
𝑉𝑀
𝑅2+𝜔2𝐿2
𝑒
𝑗 −𝑡𝑎𝑛−1𝜔𝐿
𝑅
 This shows that
 𝐼𝑀 =
𝑉𝑀
𝑅2+𝜔2𝐿2
𝑎𝑛𝑑 ∅ = −𝑡𝑎𝑛−1 𝜔𝐿
𝑅

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Sinusoidal and Complex
Forcing Functions
 However, since our actual forcing function was
𝑉𝑀 cos 𝜔𝑡 rather than 𝑉𝑀𝑒𝑗𝜔𝑡
our actual response is the
real part of the complex response:
 𝑖 𝑡 = 𝐴 cos(𝜔𝑡 + ∅) =
𝑉𝑀
𝑅2+𝜔2𝐿2
cos 𝜔𝑡 − 𝑡𝑎𝑛−1 𝜔𝐿
𝑅
 Similar to the one obtained previously.

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Phasors
 Again, we consider the RL circuit in Figure 8.3. Let us
use phasors to determine the expression for the
current.
 The DE is 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
+ 𝑅𝑖 𝑡 = 𝑉𝑀 cos 𝜔𝑡
 The forcing function can be replaced by a complex
forcing function that is written as 𝑽𝑒𝑗𝜔𝑡 with phasor
𝑽 = 𝑉𝑀∠00. Similarly, the forced response component
of the current i(t) can be replaced by a complex
function 𝑰𝑒𝑗𝜔𝑡
that is written as with phasor 𝑰 = 𝐼𝑀∠∅

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Phasors
 Using the complex forcing function, we find that the
differential equation becomes
 𝐿
𝑑
𝑑𝑡
𝑰𝑒𝑗𝜔𝑡
+ 𝑅𝑰𝑒𝑗𝜔𝑡
= 𝑽𝑒𝑗𝜔𝑡
 The common factor can be eliminated, leaving the
phasors; that is,
 𝑗𝜔𝐿𝑰 + 𝑅𝑰 = 𝑽
 Thus𝑰 =
𝑉
𝑅+𝑗𝜔𝐿
= 𝐼𝑀∠∅ =
𝑉𝑀
𝑅2+𝜔2𝐿2
∠ − 𝑡𝑎𝑛−1 𝜔𝐿
𝑅
 Therefore, 𝑖 𝑡 =
𝑉𝑀
𝑅2+𝜔2𝐿2
cos 𝜔𝑡 − 𝑡𝑎𝑛−1 𝜔𝐿
𝑅

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Phasors
 We define relations between phasors after the 𝑒𝑗𝜔𝑡
term has been eliminated as “phasor, or frequency
domain, analysis.”
 The phasors are then simply transformed back to the
time domain to yield the solution of the original set of
differential equations.
 In addition, we note that the solution of sinusoidal
steady-state circuits would be relatively simple if we
could write the phasor equation directly from the circuit
description.

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Phasors
Time Domain Frequency Domain
𝐴 cos 𝜔𝑡 ± 𝜃 𝐴∠ ± 𝜃
𝐴 sin 𝜔𝑡 ± 𝜃 𝐴∠ ± 𝜃 − 900
Table 8.1: Phasor Representation

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Exercise
1. Convert the following voltage functions to phasors
 𝑣1 𝑡 = 12 cos 377𝑡 − 4250 𝑉
 𝑣2 𝑡 = 18 sin 2513𝑡 + 4.20 𝑉
2. Convert the following Phasors to the time domain if
the frequency is 400 Hz.
 𝑉1 = 10∠200
 𝑉2 = 12∠ −600

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Phasor Relationships for Circuit
Elements
 Resistor
Figure 8.4: Voltage–current relationships for a resistor.

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Resistor
 𝑣 𝑡 = 𝑅𝑖(𝑡)
Applying the complex voltage 𝑉𝑀𝑒𝑗 𝜔𝑡+𝜃𝑣 results in the
complex current 𝐼𝑀𝑒𝑗 𝜔𝑡+𝜃𝑖 and therefore
 𝑉𝑀𝑒𝑗 𝜔𝑡+𝜃𝑣 = 𝑅𝐼𝑀𝑒𝑗 𝜔𝑡+𝜃𝑖
 𝑉𝑀𝑒𝑗𝜃𝑣 = 𝑅𝐼𝑀𝑒𝑗𝜃𝑖
In phasor form
 𝑽 = 𝑅𝑰
 Where𝑽 = 𝑉𝑀𝑒𝑗𝜃𝑣 = 𝑉𝑀∠𝜃𝑣 𝑎𝑛𝑑 𝑰 = 𝐼𝑀𝑒𝑗𝜃𝑖 = 𝐼𝑀∠𝜃𝑖.
 Where 𝜃𝑣 = 𝜃𝑖. Thus current and voltage for this circuit
are in phase.

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Resistor
 Exercise
The current in a 4Ω resistor is known to be 𝑰 =
12∠600𝐴.
Express the voltage across the resistor as a time
function if the frequency of the current is 4 kHz.

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Inductor
Figure 8.5: Voltage–current relationships for an inductor.

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Inductor
 𝑣 𝑡 = 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
 𝑉𝑀𝑒𝑗 𝜔𝑡+𝜃𝑣 = 𝐿
𝑑
𝑑𝑡
𝐼𝑀𝑒𝑗 𝜔𝑡+𝜃𝑖
 𝑉𝑀𝑒𝑗𝜃𝑣 = 𝑗𝜔𝐿𝐼𝑀𝑒𝑗𝜃𝑖
 In phasor notation
 𝑽 = 𝑗𝜔𝐿𝑰
 Since the imaginary operator 𝑗 = 1𝑒𝑗900
= 1∠900
=
−1
 𝑉𝑀𝑒𝑗𝜃𝑣 = 𝜔𝐿𝐼𝑀𝑒𝑗 𝜃𝑖+900

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Inductor
 Therefore, the voltage and current are 90° out of
phase, and in particular the voltage leads the current
by 90° or the current lags the voltage by 90°. The
phasor diagram and the sinusoidal waveforms for the
inductor circuit are shown in Figs. 8.5c and d,
respectively.
Example
 The voltage 𝑣 𝑡 = 12 cos 377𝑡 + 200
𝑉 is applied to a
20-mH inductor as shown in Fig. 8.7a. Find the
resultant current.

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Capacitor
Figure 8.6: Voltage–current relationships for a capacitor.

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Capacitor
 𝑖 𝑡 = 𝐶
𝑑𝑣(𝑡)
𝑑𝑡
 𝐼𝑀𝑒𝑗 𝜔𝑡+𝜃𝑖 = 𝐶
𝑑
𝑑𝑡
𝑉𝑀𝑒𝑗 𝜔𝑡+𝜃𝑣
 Which reduces to
 𝐼𝑀𝑒𝑗𝜃𝑖 = 𝑗𝜔𝐶𝑉𝑀𝑒𝑗𝜃𝑣
 In phasor notation this becomes
 𝐼 = 𝑗𝜔𝐶𝑉
 And 𝐼𝑀𝑒𝑗𝜃𝑖 = 𝜔𝐶𝑉𝑀𝑒𝑗 𝜃𝑣+900
 Note that the voltage and current are 90° out of phase.
In particular, the current leads the voltage by 90°

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Exercise
1. The voltage 𝑣 𝑡 = 100 cos 314𝑡 + 150
𝑉 is
applied to a 100 μF capacitor as shown in Fig. 8.6a.
Find the current.
2. The current in a 150- μF capacitor is 𝐼 =
3.6∠ −1450 𝐴 . If the frequency of the current is 60
Hz, determine the voltage across the capacitor.

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Impedance and Admittance
 Impedance is defined as the ratio of the phasor
voltage V to the phasor current I:
 𝒁 =
𝑽
𝑰
Figure 8.7: General impedance relationship.

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Impedance and Admittance
 𝑍 =
𝑉𝑀∠𝜃𝑣
𝐼𝑀∠𝜃𝑖
=
𝑉𝑀
𝐼𝑀
∠𝜃𝑣 − 𝜃𝑖 = 𝑍∠𝜃𝑧
 Since Z is the ratio of V to I, the units of Z are ohms.
Thus, impedance in an ac circuit is analogous to
resistance in a dc circuit.
 In rectangular form, impedance is expressed as
 𝑍 𝜔 = 𝑅 𝜔 + 𝑗𝑋(𝜔)
 Where 𝑅 𝜔 is the real, or resistive, component and
𝑋 𝜔 is the imaginary, or reactive, component. In
general, we simply refer to R as the resistance and 𝑋 as
the reactance.
 The above equations indicate that𝑍 𝜃𝑧 = 𝑅 + 𝑗𝑋

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Impedance and Admittance
 Thus𝑍 = 𝑅2 + 𝑋2 𝑎𝑛𝑑 𝜃𝑧 = tan−1 𝑋
𝑅
 Where 𝑅 = 𝑍𝑐𝑜𝑠𝜃𝑧 𝑎𝑛𝑑 𝑋 = 𝑍𝑠𝑖𝑛𝜃𝑧 .
 For the individual passive elements the impedance is
as shown in Table 8.2.

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Impedance and Admittance
Passive
Element
Impedance
R 𝑍 = 𝑅
L 𝑍 = 𝑗𝜔𝐿 = 𝑗𝑋𝐿 = 𝜔𝐿∠900
, 𝑋𝐿 = 𝜔𝐿
C
𝑍 =
1
𝑗𝜔𝐶
= 𝑗𝑋𝐶 = −
1
𝜔𝐶
∠900
, 𝑋𝐶 = −
1
𝜔𝐶
Table 8.2: Passive Element Impedance

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Impedance and Admittance
 KCL and KVL are both valid in the frequency domain.
 Impedances can be combined using the same rules
that we established for resistor combinations.
 𝑍𝑠 = 𝑍1 + 𝑍2 + ⋯ 𝑍𝑛 𝑎𝑛𝑑
1
𝑍𝑝
=
1
𝑍1
+
1
𝑍2
+ ⋯
1
𝑍𝑛

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Impedance and Admittance
 Determine the equivalent
impedance of the network
shown in Fig. 8.8 if the
frequency is f=60 Hz.
Then compute the current
i(t) if the voltage source is
𝑣 𝑡 = 50 cos 𝜔𝑡 + 300
𝑉
 Finally, calculate the
equivalent impedance if
the frequency is f=400
Hz.
Figure 8.8: Series ac circuit.

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Impedance and Admittance
 Find the current i(t) in the network if 𝑣 𝑡 =
120 sin 377𝑡 + 600 𝑉

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Impedance and Admittance
 Another quantity that is very useful in the analysis of ac
circuits is the two-terminal input admittance, which is
the reciprocal of impedance; that is,
 𝒀 =
1
𝒁
=
𝑰
𝑽
 The units of Y are siemens, and this quantity is
analogous to conductance in resistive dc circuits. Since
Z is a complex number, Y is also a complex number.
 𝒀 = 𝑌𝑀 𝜽𝒚
 which is written in rectangular form as
 𝑌 = 𝐺 + 𝑗𝐵

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Impedance and Admittance
 From the expression
 𝐺 =
𝑅
𝑅2+𝑋2 𝑎𝑛𝑑 𝐵 = −
𝑋
𝑅2+𝑋2
 and in a similar manner, we can show that
 𝑅 =
𝐺
𝐺2+𝐵2 , 𝑋 =
−𝐵
𝐺2+𝐵2
 The admittance of the individual passive elements are
 𝑌𝑅 =
1
𝑅
= 𝐺 , 𝑌𝐿 =
1
𝑗𝜔𝐿
= −
1
𝜔𝐿
∠900
, 𝑌𝐶 = 𝑗𝜔𝐶 =
𝜔𝐶∠900

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Impedance and Admittance
 The rules for combining admittances are the same as
those for combining conductances;
 𝑌𝑝 = 𝑌1 + 𝑌2 + ⋯ 𝑌𝑛 𝑎𝑛𝑑
1
𝑌𝑠
=
1
𝑌1
+
1
𝑌2
+ ⋯
1
𝑌𝑛

42. © Niwareeba Roland
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Impedance and Admittance
 Calculate the equivalent admittance𝑌𝑝 for the network
in Fig. 8.9 and use it to determine the current I if 𝑉
𝑠 =
60 450 𝑉.
Figure 8.9: Example parallel circuit
𝑌𝑅 =
1
𝑍𝑅
=
1
2
𝑆
𝑌𝐿 =
1
𝑍𝐿
= −
𝑗
4
𝑆
Thus 𝑌𝑝 =
1
2
− 𝑗
1
4
𝑆
𝐻𝑒𝑛𝑐𝑒 𝐼 = 𝑌𝑝𝑉
𝑠
=
1
2
− 𝑗
1
4
60∠400
= 33.5 18.430 𝐴

43. © Niwareeba Roland
43
Phasor Diagrams
 Impedance and admittance are functions of frequency,
and therefore their values change as the frequency
changes.
 These changes in Z and Y have a resultant effect on
the current–voltage relationships in a network. This
impact of changes in frequency on circuit parameters
can be easily seen via a phasor diagram

44. © Niwareeba Roland
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Phasor Diagrams-Example
Figure 8.11: Example parallel circuit

45. © Niwareeba Roland
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Phasor diagrams-example
 At the upper node in the circuit KCL is
 𝐼𝑆 = 𝐼𝑅 + 𝐼𝐿 + 𝐼𝐶 =
𝑉
𝑅
+
𝑉
𝑗𝜔𝐿
+
𝑉
1/𝑗𝜔𝐿
 𝑆𝑖𝑛𝑐𝑒 𝑉 = 𝑉𝑀∠00
, 𝑡ℎ𝑒𝑛 𝐼𝑆 =
𝑉𝑀∠00
𝑅
+
𝑉𝑀∠−900
𝜔𝐿
+ 𝑉𝑀𝜔𝐶∠900
 Note that 𝐼𝑆 is in phase with V when IC = IL or, in other
words, when 𝜔𝐿 =
1
𝜔𝐶
. Hence, the node voltage V is in
phase with the current source when 𝜔 =
1
𝐿𝐶
 This can also be seen from the KCL equation
 𝐼 =
1
𝑅
+ 𝑗 𝜔𝐶 −
1
𝜔𝐿
𝑉

46. © Niwareeba Roland
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Phasor diagrams-Example
Figure 8.12: Phasor
diagrams for the
circuit in Fig. 8.11.

47. © Niwareeba Roland
47
Basic Analysis Using
Kirchhoff’s Laws
 We wish to calculate all the voltages and currents in
the circuit shown in Fig. 8.14a.
Figure 8.14 (a) Example ac circuit, (b) phasor diagram for the currents
(plots are not drawn to scale).

48. © Niwareeba Roland
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Basic Analysis Using
Kirchhoff’s Laws
 Solution
 𝑍𝑒𝑞 = 4 +
(𝑗6)(8−𝑗4)
𝑗6+8−𝑗4
= 8.24 + 𝑗4.94 = 9.61∠30.940
𝛺
 𝐼1 =
𝑉𝑆
𝑍𝑒𝑞
=
24 600
9.61 30.940 = 2.5∠29.060 𝐴
 𝑉1 can be determined using KVL:
 𝑉1 = 𝑉𝑆 − 4𝐼1 = 16.26∠78.430𝐴 (This can also be
determined by voltage division)
 𝐼2 =
𝑉1
𝑗6
= 2.71∠ −11.580
𝐴

49. © Niwareeba Roland
49
Basic Analysis Using
Kirchhoff’s Laws
 𝐼3 =
𝑉1
8−𝑗4
= 1.82∠1050
𝐴
 (𝐼2 𝑎𝑛𝑑 𝐼3 can also be determined by current division.)
 𝑉2 = 𝐼3 −𝑗4 = 7.28∠150
𝑉
 The phasor diagram for the currents 𝐼1, 𝐼2 𝑎𝑛𝑑 𝐼3 and is
shown in Fig. 8.14b and is an illustration of KCL.

50. © Niwareeba Roland
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Analysis Techniques-Nodal
Figure 8.15 Circuits for node analysis

51. © Niwareeba Roland
51
Analysis Techniques-Nodal
 The KCL equation for the supernode that includes the
voltage source is.

𝑉1
1+𝑗
+
𝑉2
1
+
𝑉2
1−𝑗
= 2∠00
 and the associated KVL constraint equation is
 𝑉2 − 𝑉1 = 6∠00
 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑉1 = 𝑉2 − 6. Substituting this into the first
equation and solving gives
 𝑉2 =
5
2
− 𝑗
3
2
𝑉
 Therefore 𝐼0 =
5
2
− 𝑗
3
2
𝐴

52. © Niwareeba Roland
52
Analysis Techniques-Loop

53. © Niwareeba Roland
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Analysis Techniques-Loop
 The three loop equations are
 𝐼1 = −2 00
 𝐼1 1 + 𝑗 + 𝐼2 2 + 𝐼3 1 − 𝑗 = 6∠00
 𝐼2 1 − 𝑗 + 𝐼3 2 − 𝑗 = 0
 Solving the above equations gives
 𝐼3 = −
5
2
+ 𝑗
3
2
𝐴
 And finally 𝐼0 = −𝐼3 =
5
2
− 𝑗
3
2
𝐴

54. © Niwareeba Roland
54
Analysis Techniques-
Superposition

55. © Niwareeba Roland
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Exercise
 Use Thevenin’s and Norton’s analysis techniques to
obtain a similar result.