Electric Circuits II - philadelphia.edu.jo · Dr. Firas Obeidat – Philadelphia University 2 1 •Nodal Analysis 2 •Mesh Analysis 3 •Superposition Theorem 4 •Source Transformation

Electric Circuits II Sinusoidal Steady State Analysis

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Dr. Firas Obeidat

Dr. Firas Obeidat – Philadelphia University

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1 • Nodal Analysis

2 • Mesh Analysis

3 • Superposition Theorem

4 • Source Transformation

5 • Thevenin and Norton Equivalent Circuits

Table of Contents


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Sinusoidal steady state analysis

Steps to Analyze AC Circuits: 1. Transform the circuit to the phasor or frequency domain.

2. Solve the problem using circuit techniques (nodal analysis, mesh

analysis, superposition, etc.).

3. Transform the resulting phasor to the time domain.

Frequency domain analysis of an ac circuit via phasors is

much easier than analysis of the circuit in the time domain


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Nodal Analysis

The basis of nodal analysis is Kirchhoff’s current law. Since KCL is valid for

phasors, AC circuits can be analyzed by nodal analysis.

Example: Find the time-domain node voltages v1(t) and v2(t) in the circuit

shown in the figure

Apply KCL on node 1

Apply KCL on node 2

From the above equations, we can find that V1=1−j2 V and V2=−2+j4 V

The time domain

solutions are obtained

by expressing V1 and V2

in polar form:

The time domain

expression is


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Nodal Analysis

Example: Compute V1 and V2 in the circuit

Nodes 1 and 2 form a supernode. Applying KCL

at the supernode gives

But a voltage source is connected between nodes 1

and 2, so that

Substitute the above equation in the first equation

gives


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Mesh Analysis

Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis.

Example: Determine IO current in the circuit using mesh analysis.

Applying KVL to mesh 1, we obtain

For mesh 2

For mesh 3, I3=5 A, Substituting this in the above

two equations, we get


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Mesh Analysis

Example: solve for Vo in the circuit using mesh analysis

meshes 3 and 4 form a supermesh due to the current

source between the meshes. For mesh 1, KVL gives

For mesh 2, KVL gives

For supermesh, KVL gives

Due to the current source between meshes 3 and 4, at

node A

(1)

(2)

(3)

(4)

Substitute equation (2) in equation (1) gives

(5)

Substitute equation (4) in equation (3) gives

(6)


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Mesh Analysis

From equation (5) and equation (6), we obtain the matrix equation

We obtain the following determinants

Current I1 is obtained as

The required voltage Vo is


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Superposition Theorem The superposition theorem applies to ac circuits the same way it applies to dc

circuits. The theorem becomes important if the circuit has sources operating at

different frequencies.

Example: Use the superposition theorem to find Io

in the circuit.

Let

Where Io′ and Io″ are due to the voltage and current

sources, respectively. To find Io′ consider the

circuit in fig(a). If we let Z be the parallel

combination of –j2 and 8+j10, then

The current Io′ is

(1)

(2)


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Superposition Theorem

To get Io″, consider the circuit in fig(b). For mesh 1

For mesh 2

For mesh 3

(3)

(4)

(5)

From (4) & (5)

(6) Expressing I1 in terms of I2 gives

Substituting eq(5) and eq(6) into eq(3), we get

(7)

From eq(2) and eq(7), we get


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Superposition Theorem Example: Find vo of the circuit using the superposition theorem.

Since the circuit operates at three different

frequencies for the dc voltage source), one

way to obtain a solution is to use

superposition,

Where v1 is due to the 5-V dc voltage source, v2 is due to the voltage source, and v3 is

due to the current source. To findv1 we set to zero all sources except the 5-V dc source.

We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a

short circuit to dc. There is an alternative way of looking at this. Since ω=0, jωL=0,

1/ωj=∞.

(1)

(2)

To find v2 we set to zero both the 5-V source and the 2sin5t

current source and transform the circuit to the frequency

domain.

From fig(a)


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The equivalent circuit is now as shown in fig(b).

Let

(3)

By voltage division

In time domain


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To obtain v3 we set the voltage sources to zero and transform what is left to the

frequency domain.

(4)

The equivalent circuit is now as shown in

fig(c). Let

By current division

In time domain

From eq (1), eq (2), eq (3) and eq (4), we get


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Source Transformation Source transformation in the frequency domain

involves transforming a voltage source in series with an

impedance to a current source in parallel with an

impedance, or vice versa.

Example: Calculate Vx in the circuit using the method

of source transformation

Transform the voltage source to a current source as in fig

(a)

The parallel combination of 5 Ω resistance and 3+j4

impedance gives

Converting the current source to a voltage source yields

the circuit in fig (b), where

By voltage division


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Thevenin and Norton Equivalent Circuits

Thevenin’s and Norton’s theorems are applied to

ac circuits in the same way as they are to dc

circuits. The only additional effort arises from

the need to manipulate complex numbers.

A linear circuit is replaced by a voltage source

in series with an impedance

In Norton equivalent circuit, a linear circuit

is replaced by a current source in parallel

with an impedance.

Thevenin’s and Norton’s equivalent circuits

are related as


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Example: Obtain the Thevenin equivalent at terminals a-

b in the circuit.

To find ZTh, set the voltage source to zero. As shown in

fig(a), the 8Ω resistance is in parallel with the –j6

reactance, and the resistance 4Ω is in parallel with the j12

reactance. so that their combination gives

The Thevenin impedance is the series combination of Z1

and Z2 that is,

To find VTh consider the circuit in fig(b). Currents are

obtained as

Applying KVL around loop bcdeab in fig(b) gives


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Example: Find the Thevenin equivalent circuit as seen

from terminals a-b.

To find VTh, apply KCL at node 1 in fig(a)

Applying KVL to the middle loop fig(a)


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To obtain ZTh, remove the independent source.

Due to the presence of the dependent current

source, connect a 3-A current source to terminals

a-b as in fig(b). At the node, KCL gives

Applying KVL to the outer loop in fig(b) gives

The Thevenin impedance is

Example: Obtain Io current using Norton’s theorem.

To find ZTh, set the sources to zero as shown

in fig(a). the 8-j2 and 10+j4 impedances are short

circuited, so that


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Thevenin and Norton Equivalent Circuits To get IN we short-circuit terminals a-b as in fig(b) and apply mesh

analysis. Meshes 2 and 3 form a supermesh because of the current source

linking them. For mesh 1

For the supermesh

At node a, due to the current source between meshes 2 and 3,

(1)

(2)

(3)

Adding eqs. (1) and (2) gives

from eqs. (1)

The Norton current is

Figure (c) shows the Norton equivalent circuit along with the impedance at terminals a-b. By

current division

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Electric Circuits II - philadelphia.edu.jo · Dr. Firas Obeidat – Philadelphia University 2 1 •Nodal Analysis 2 •Mesh Analysis 3 •Superposition Theorem 4 •Source Transformation

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