Electric Circuits II Sinusoidal Steady State Analysis

Dr. Firas Obeidat

1 Table of Contents

• Nodal Analysis 1

• Mesh Analysis 2

• Superposition Theorem 3

• Source Transformation 4

• Thevenin and Norton Equivalent Circuits 5

2 Dr. Firas Obeidat – Philadelphia University 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

3 Dr. Firas Obeidat – Philadelphia University 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 The time domain

by expressing V1 and V2 expression is in polar form: 4 Dr. Firas Obeidat – Philadelphia University 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

5 Dr. Firas Obeidat – Philadelphia University 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

6 Dr. Firas Obeidat – Philadelphia University 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

(1) For mesh 2, KVL gives (2) For supermesh, KVL gives (3) Due to the current source between meshes 3 and 4, at node A (4)

Substitute equation (2) in equation (1) gives (5) Substitute equation (4) in equation (3) gives (6)

7 Dr. Firas Obeidat – Philadelphia University 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

8 Dr. Firas Obeidat – Philadelphia University 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 (1)

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

(2)

9 Dr. Firas Obeidat – Philadelphia University Superposition Theorem

To get Io″, consider the circuit in fig(b). For mesh 1 (3) For mesh 2 (4) For mesh 3 (5) From (4) & (5)

Expressing I1 in terms of I2 gives (6) Substituting eq(5) and eq(6) into eq(3), we get

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

10 Dr. Firas Obeidat – Philadelphia University 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, (1)

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=∞.

From fig(a) (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. 11 Dr. Firas Obeidat – Philadelphia University Superposition Theorem

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

By voltage division

In time domain

(3)

12 Dr. Firas Obeidat – Philadelphia University Superposition Theorem

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

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

By current division

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

13 Dr. Firas Obeidat – Philadelphia University 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

14 Dr. Firas Obeidat – Philadelphia University 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

15 Dr. Firas Obeidat – Philadelphia University Thevenin and Norton Equivalent Circuits

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

16 Dr. Firas Obeidat – Philadelphia University Thevenin and Norton Equivalent Circuits

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)

17 Dr. Firas Obeidat – Philadelphia University Thevenin and Norton Equivalent Circuits

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

18 Dr. Firas Obeidat – Philadelphia University 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 (1) For the supermesh (2) At node a, due to the current source between meshes 2 and 3, (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

19 Dr. Firas Obeidat – Philadelphia University 20