# Network Theory Questions and Answers – Norton’s Theorem

«
»

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Norton’s Theorem”.

1. Norton’s current is equal to the current passing through the ___________ circuited ___________ terminals.
a) short, input
b) short, output
c) open, output
d) open, input

Explanation: Norton’s current is equal to the current passing through short circuited output terminals not the current through open circuited output terminals.

2. The expression of Norton’s current (IN) in the circuit shown below is? a) V/Z1
b) V/Z2
c) V(Z2/(Z1+Z2))
d) VZ1/(Z1+Z2)

Explanation: The Norton’s equivalent form of any complex impedance circuit consists of an equivalent current source and an equivalent impedance. The expression of Norton’s current is IN = V/Z1.

3. The expression of equivalent impedance (ZN) in the circuit shown below is? a) (Z1+Z2)/Z1
b) (Z1+Z2)/Z2
c) Z1Z2/(Z1+Z2)
d) Z1+Z2

Explanation: The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance. The Norton’s equivalent impedance is ZN = Z1Z2/(Z1+Z2).
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. Determine Norton’s equivalent current in the circuit shown below. a) 5∠53.13⁰
b) 4∠53.13⁰
c) 4∠53.13⁰
d) 5∠-53.13⁰

Explanation: The current through the terminals a and b is the Norton’s equivalent current. The Norton’s equivalent current is I = (25∠0o)/(3+j4) = 5∠-53.13⁰A.

5. The Norton’s equivalent impedance in the circuit shown below. a) 4.53∠9.92⁰
b) 4.53∠-9.92⁰
c) 5.53∠9.92⁰
d) 5.53∠-9.92⁰

Explanation: The Norton’s equivalent impedance is Z = (3+j4)(4-j5)/((3+j4)+(4-j5))=4.53∠9.92oΩ. The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance.

6. Determine the Norton’s impedance seen from terminals ‘ab’. a) 6∠90⁰
b) 7∠90⁰
c) 6∠-90⁰
d) 7∠-90⁰

Explanation: The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance. The Norton’s impedance is Zab=(j3)(-j2)/((j3)-(j2))=6∠-90oΩ.

7. Find the Norton’s current passing through ‘ab’ in the circuit shown below. a) 4.16∠126.8⁰
b) 5.16∠126.8⁰
c) 5.16∠-126.8⁰
d) 4.16∠-126.8⁰

Explanation: The Norton’s current is equal to the current passing through the short circuit between the points a and b. The Norton’s current is I=(10∠0o)/(3∠90o)+(5∠90o)/(2∠-90o) =4.16∠-126.8oA.

8. Find the load current in the circuit shown below. a) 3.19∠166.61⁰
b) 3.19∠-166.61⁰
c) 4.19∠166.61⁰
d) 4.19∠-166.61⁰

Explanation: The load current in the circuit is given by IL = I×(6∠-90o)/(5+6∠-90o) = 3.19∠-166.61oA.

9. Determine Norton’s equivalent impedance in the circuit shown below. a) (5+j6) Ω
b) (5-j6) Ω
c) (6+j7) Ω
d) (6-j7) Ω

Explanation: The impedance seen from the terminals when the source is reduced to zero is Z = (5+j6) Ω.

10. Find the Norton’s current in the circuit shown below. a) 40∠30⁰
b) 40∠-30⁰
c) 30∠30⁰
d) 30∠-30⁰

Explanation: The Norton’s current is equal to the current passing through the short circuit between the points a and b. So the current passing through the short circuited terminals ‘a’ and ‘b’ is I = 30∠30⁰A.

Sanfoundry Global Education & Learning Series – Network Theory.

To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers. 