# EE2010 Fundamentals of Electric Circuits

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EE2010 Fundamentals of Electric Circuits
Lecture 09 Network Theorems

Introduction A large complex circuits Simplify circuit analysis
Circuit Theorems ‧Thevenin’s theorem ‧ Norton theorem ‧ Superposition ‧ max. power transfer

linear circuit A linear circuit is one whose output is linearly related (or directly proportional) to its input

Superposition principle
The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. Turn off, killed, inactive source: independent voltage source: 0 V (short circuit) independent current source: 0 A (open circuit) Dependent sources are left intact.

Superposition principle
Steps to apply superposition principle: Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to the independent sources. Turn off voltages sources = short voltage sources; make it equal to zero voltage Turn off current sources = open current sources; make it equal to zero current Superposition involves more work but simpler circuits. Superposition is not applicable to the effect on power.

Superposition principle
Removing the effect of ideal sources Voltage source is replaced by a S/C Current source is replaced by a O/C Removing the effect of practical sources

Superposition principle
Dependent Source (a) Dependent Voltage Source A voltage source whose parameters are controlled by voltage/current else where in the system v = ρix CDVS (Current Dependent Voltage source) v = µVx VDVS (Voltage Dependent Voltage source) (b) Dependent Current Source A current source whose parameters are controlled by voltage/current else where in the system v = βix CDCS (Current Dependent Current source) v = αVx VDCS (Voltage Dependent Current source) For Superposition, All dependent sources must be left intact!! You can’t apply O/C and S/C on dependent sources

Example -1 Using the superposition theorem, determine current I1 for the network in Fig. Solution: Since two sources are present, there are two networks to be analyzed. First let us determine the effects of the voltage source by setting the current source to zero amperes as shown in Fig.

Example -1 Since I1’ and I1’’ have the same defined direction, the total current is defined by The voltage source is in parallel with the current source and load resistor R1, so the voltage across each must be 30 V. The result is that I1 must be determined solely by

Example -2 Using the superposition theorem, determine the current through the 12Ω resistor in Fig. Considering the effects of the 54 V source requires replacing the 48 V source by a short-circuit equivalent as shown in Fig. The result is that the 12Ω and 4Ω resistors are in parallel.

Example -2 The total resistance seen by the source is
and the source current is Using the current divider rule results in the contribution to I2 due to the 54 V source:

Example -2 The total resistance seen by the 48 V source is
Applying the current divider rule results in

Example -3 Use the superposition theorem to find v in the circuit in Fig.

Example -3 Since there are two sources, let Voltage division to get
Current division, to get Hence And we find

Example -4 a. Using the superposition theorem, determine the current through resistor R2 for the network in Fig. b. Demonstrate that the superposition theorem is not applicable to power levels.

Example -4 Solutions: (a) Simple series circuit with a current equal to Parallel combination of resistors R1 and R2. Applying the current divider rule The total solution for current I2 is the sum of the currents established by the two sources.

(b) The power delivered to the 6 Ω resistor is
Example -4 (b) The power delivered to the 6 Ω resistor is Using the total resultant current

Example -5 Using the principle of superposition, find the current l2 through the 12 k resistor in Fig.

Example -5 Solution: Considering the effect of the 6 mA current source
Current divider rule:

Example -5 Considering the effect of the 9 V voltage source