1. Find the current flowing between terminals A and B of the circuit shown below.
a) 1
b) 2
c) 3
d) 4
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: The voltage drop across 5Ω resistor in the circuit is the product of current and resistance => V = 5×7.16 = 35.8 ≅ 36V.