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This section includes 16 Mcqs, each offering curated multiple-choice questions to sharpen your Network Theory knowledge and support exam preparation. Choose a topic below to get started.
1. |
Find the Norton’s current in the circuit shown below. |
A. | 40∠30⁰ |
B. | 40∠-30⁰ |
C. | 30∠30⁰ |
D. | 30∠-30⁰ |
Answer» D. 30∠-30⁰ | |
2. |
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 |
Answer» B. (5-j6) Ω | |
3. |
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 |
Answer» C. 4.19∠166.61⁰ | |
4. |
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 |
Answer» E. | |
5. |
Determine the Norton’s impedance seen from terminals ‘ab’. |
A. | 6∠90⁰ |
B. | 7∠90⁰ |
C. | 6∠-90⁰ |
D. | 7∠-90⁰ |
Answer» D. 7∠-90⁰ | |
6. |
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 |
Answer» B. 4.53∠-9.92⁰ | |
7. |
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 |
Answer» E. | |
8. |
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 |
Answer» D. Z1+Z2 | |
9. |
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) |
Answer» B. V/Z2 | |
10. |
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 |
Answer» C. open, output | |
11. |
Find the Norton’s current in the circuit shown in the question 9.$# |
A. | 40∠30⁰ |
B. | 40∠-30⁰ |
C. | 30∠30⁰ |
D. | 30∠-30⁰ |
Answer» D. 30‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ-30‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ª | |
12. |
Find the load current in the circuit shown in the question 6? |
A. | 3.19∠166.61⁰ |
B. | 3.19∠-166.61⁰ |
C. | 4.19∠166.61⁰ |
D. | 4.19∠-166.61⁰ |
Answer» C. 4.19‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ166.61‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ª | |
13. |
Find the Norton’s current passing through ‘ab’ in the circuit shown in the question 6.$ |
A. | 4.16∠126.8⁰ |
B. | 5.16∠126.8⁰ |
C. | 5.16∠-126.8⁰ |
D. | 4.16∠-126.8⁰ |
Answer» E. | |
14. |
The Norton’s equivalent impedance in the circuit shown in the question 4.$ |
A. | 4.53∠9.92⁰ |
B. | 4.53∠-9.92⁰ |
C. | 5.53∠9.92⁰ |
D. | 5.53∠-9.92⁰ |
Answer» B. 4.53‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ-9.92‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ª | |
15. |
The expression of equivalent impedance (ZN) in the circuit shown in the question 2 is? |
A. | (Z<sub>1</sub>+Z<sub>2</sub>)/Z<sub>1</sub> |
B. | (Z<sub>1</sub>+Z<sub>2</sub>)/Z<sub>2</sub> |
C. | Z<sub>1</sub>Z<sub>2</sub>/(Z<sub>1</sub>+Z<sub>2</sub>) |
D. | Z<sub>1</sub>+Z<sub>2</sub> |
Answer» D. Z<sub>1</sub>+Z<sub>2</sub> | |
16. |
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 |
Answer» C. open, output | |