Electric Circuits II Sinusoidal Steady State Analysis Dr. Firas Obeidat 1
Table of Contents 1 2 3 4 5 Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin and Norton Equivalent Circuits 2
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
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 v 1 (t) and v 2 (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 V 1 =1 j2 V and V 2 = 2+j4 V The time domain solutions are obtained by expressing V 1 and V 2 in polar form: The time domain expression is 4
Nodal Analysis Example: Compute V 1 and V 2 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
Mesh Analysis Kirchhoff s voltage law (KVL) forms the basis of mesh analysis. Example: Determine I O current in the circuit using mesh analysis. Applying KVL to mesh 1, we obtain For mesh 2 For mesh 3, I 3 =5 A, Substituting this in the above two equations, we get 6
Mesh Analysis Example: solve for V o 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 Substitute equation (4) in equation (3) gives (5) (6) 7
Mesh Analysis From equation (5) and equation (6), we obtain the matrix equation We obtain the following determinants Current I 1 is obtained as The required voltage V o is 8
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 I o in the circuit. Let (1) Where I o and I o are due to the voltage and current sources, respectively. To find I o consider the circuit in fig(a). If we let Z be the parallel combination of j2 and 8+j10, then The current I o is (2) 9
Superposition Theorem To get I o, consider the circuit in fig(b). For mesh 1 For mesh 2 For mesh 3 From (4) & (5) Expressing I 1 in terms of I 2 gives Substituting eq(5) and eq(6) into eq(3), we get (3) (4) (5) (6) From eq(2) and eq(7), we get (7) 10
Superposition Theorem Example: Find v 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 v 1 is due to the 5-V dc voltage source, v 2 is due to the voltage source, and v 3 is due to the current source. To findv 1 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 v 2 we set to zero both the 5-V source and the 2sin5t current source and transform the circuit to the frequency domain. 11
Superposition Theorem The equivalent circuit is now as shown in fig(b). Let By voltage division In time domain (3) 12
Superposition Theorem To obtain v 3 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 From eq (1), eq (2), eq (3) and eq (4), we get (4) 13
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 V x 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
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
Thevenin and Norton Equivalent Circuits Example: Obtain the Thevenin equivalent at terminals a- b in the circuit. To find Z Th, 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 Z 1 and Z 2 that is, To find V Th consider the circuit in fig(b). Currents are obtained as Applying KVL around loop bcdeab in fig(b) gives 16
Thevenin and Norton Equivalent Circuits Example: Find the Thevenin equivalent circuit as seen from terminals a-b. To find V Th, apply KCL at node 1 in fig(a) Applying KVL to the middle loop fig(a) 17
Thevenin and Norton Equivalent Circuits To obtain Z Th, 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 I o current using Norton s theorem. To find Z Th, set the sources to zero as shown in fig(a). the 8-j2 and 10+j4 impedances are short circuited, so that 18
Thevenin and Norton Equivalent Circuits To get I N 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, (3) (1) (2) 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
20