Network Theory Multiple Choice Questions on “Norton’s Theorem”.
1. Find the current flowing between terminals A and B of the circuit shown below.
A. 1
B. 2
C. 3
D. 4
Answer: D
Clarification: The magnitude of the current in Norton’s equivalent circuit is equal to the current passing through the short circuited terminals that are I=20/5=4A.
2. Find the equivalent resistance between terminals A and B of the circuit shown below.
A. 0.33
B. 3.33
C. 33.3
D. 333
Answer: B
Clarification: Norton’s resistance is equal to the parallel combination of both the 5Ω and 10Ω resistors that is R = (5×10)/15 = 3.33Ω.
3. Find the current through 6Ω resistor in the circuit shown below.
A. 1
B. 1.43
C. 2
D. 2.43
Answer: B
Clarification: The current passing through the 6Ω resistor and the voltage across it due to Norton’s equivalent circuit is I = 4×3.33/(6+3.33) = 1.43A.
4. Find the voltage drop across 6Ω resistor in the circuit shown below.
A. 6.58
B. 7.58
C. 8.58
D. 9.58
Answer: C
Clarification: The voltage across the 6Ω resistor is V = 1.43×6 = 8.58V. So the current and voltage have the same values both in the original circuit and Norton’s equivalent circuit.
5. Find the current flowing between terminals A and B in the following circuit.
A. 1
B. 2
C. 3
D. 4
Answer: D
Clarification: Short circuiting terminals A and B, 20-10(I1)=0, I1=2A. 10-5(I2), I2=2A. Current flowing through terminals A and B = 2+2 = 4A.
6. Find the equivalent resistance between terminals A and B in the following circuit.
A. 3
B. 3.03
C. 3.33
D. 3.63
Answer: C
Clarification: The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 5Ω resistor => R = ((10×5))/(10+5) = 3.33Ω.
7. Find the current flowing between terminals A and B obtained in the equivalent Nortan’s circuit.
A. 8
B. 9
C. 10
D. 11
Answer: D
Clarification: To solve for Norton’s current we have to find the current passing through the terminals A and B. Short circuiting the terminals a and b, I=100/((6×10)/(6+10)+(15×8)/(15+8))=11.16 ≅ 11A.
8. Find the equivalent resistance between terminals A and B obtained in the equivalent Nortan’s circuit.
A. 8
B. 9
C. 10
D. 11
Answer: B
Clarification: The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 6Ω resistor and parallel combination of the 15Ω resistor and the 8Ω resistor => R=(10×6)/(10+6)+(15×8)/(15+8)=8.96≅9Ω.
9. Find the current through 5Ω resistor in the circuit shown below.
A. 7
B. 8
C. 9
D. 10
Answer: A
Clarification: To solve for Norton’s current we have to find the current passing through the terminals A and B. Short circuiting the terminals a and b I=11.16×8.96/(5+8.96) = 7.16A.
10. Find the voltage drop across 5Ω resistor in the circuit shown below.
A. 33
B. 34
C. 35
D. 36
Answer: D
Clarification: The voltage drop across 5Ω resistor in the circuit is the product of current and resistance => V = 5×7.16 = 35.8 ≅ 36V.