MCQOPTIONS
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MCQOPTIONS
This section includes 13 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. |
In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The value of c in the complementary function of ‘i’ is? |
| A. | c = -0.98cos(π/2-78.6o) |
| B. | c = -0.98cos(π/2+78.6o) |
| C. | c = 0.98cos(π/2+78.6o) |
| D. | c = 0.98cos(π/2-78.6o) |
| Answer» B. c = -0.98cos(π/2+78.6o) | |
| 2. |
The current flowing through the circuit at t = 0 in the circuit shown below is? |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 0 |
| Answer» E. | |
| 3. |
In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complete solution of ‘i’ is? |
| A. | i = ce-200t + 0.98cos(1000t-π/2-78.6o) |
| B. | i = ce-200t + 0.98cos(1000t+π/2-78.6o) |
| C. | i = ce-200t + 0.98cos(1000t+π/2+78.6o) |
| D. | i = ce-200t + 0.98cos(1000t-π/2+78.6o) |
| Answer» C. i = ce-200t + 0.98cos(1000t+π/2+78.6o) | |
| 4. |
In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The particular integral of the solution of ‘ip’ is? |
| A. | ip = 0.98cos(1000t+π/2-78.6o) |
| B. | ip = 0.98cos(1000t-π/2-78.6o) |
| C. | ip = 0.98cos(1000t-π/2+78.6o) |
| D. | ip = 0.98cos(1000t+π/2+78.6o) |
| Answer» B. ip = 0.98cos(1000t-π/2-78.6o) | |
| 5. |
In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complementary function of the solution of ‘i’ is? |
| A. | ic = ce-100t |
| B. | ic = ce100t |
| C. | ic = ce-200t |
| D. | ic = ce200t |
| Answer» D. ic = ce200t | |
| 6. |
THE_COMPLETE_SOLUTION_OF_‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚Àւ§I‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖ¬•_IN_THE_QUESTION_5_IS??$# |
| A. | i = [-0.98 cos⁡(π/2-78.6<sup>o</sup>)] exp⁡(-200t)+0.98cos⁡(1000t+π/2-78.6<sup>o</sup>) |
| B. | i = [-0.98 cos⁡(π/2-78.6<sup>o</sup>)] exp⁡(-200t)-0.98cos⁡(1000t+π/2-78.6<sup>o</sup>) |
| C. | i = [0.98 cos⁡(π/2-78.6<sup>o</sup>)] exp⁡(-200t)-0.98cos⁡(1000t+π/2-78.6<sup>o</sup>) |
| D. | i = [0.98 cos⁡(π/2-78.6<sup>o</sup>)] exp⁡(-200t)+0.98cos⁡(1000t+π/2-78.6<sup>o</sup>) |
| Answer» B. i = [-0.98 cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(‚âà√¨‚àö√ë/2-78.6<sup>o</sup>)] exp‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(-200t)-0.98cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(1000t+‚âà√¨‚àö√ë/2-78.6<sup>o</sup>) | |
| 7. |
THE_VALUE_OF_C_IN_THE_COMPLEMENTARY_FUNCTION_OF_‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚Àւ§I‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖ¬•_IN_THE_QUESTION_5_IS??$# |
| A. | c = -0.98cos⁡(π/2-78.6<sup>o</sup>) |
| B. | c = -0.98cos⁡(π/2+78.6<sup>o</sup>) |
| C. | c = 0.98cos⁡(π/2+78.6<sup>o</sup>) |
| D. | c = 0.98cos⁡(π/2-78.6<sup>o</sup>) |
| Answer» B. c = -0.98cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(‚âà√¨‚àö√ë/2+78.6<sup>o</sup>) | |
| 8. |
The current flowing through the circuit at t = 0 in the circuit shown in the question 5 is? |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 0 |
| Answer» E. | |
| 9. |
The particular integral of the solution of ‘i’ from the information provided in the question 5.$ |
| A. | i<sub>p</sub> = 0.98cos⁡(1000t+π/2-78.6<sup>o</sup>) |
| B. | i<sub>p</sub> = 0.98cos⁡(1000t-π/2-78.6<sup>o</sup>) |
| C. | i<sub>p</sub> = 0.98cos⁡(1000t-π/2+78.6<sup>o</sup>) |
| D. | i<sub>p</sub> = 0.98cos⁡(1000t+π/2+78.6<sup>o</sup>) |
| Answer» B. i<sub>p</sub> = 0.98cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(1000t-‚âà√¨‚àö√ë/2-78.6<sup>o</sup>) | |
| 10. |
The complete solution of the current in the sinusoidal response of R-L circuit is? |
| A. | i = e<sup>-t(R/L)</sup>[V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ-tan<sup>-1</sup>)⁡(ωL/R))]+V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt+θ-tan<sup>-1</sup>)⁡(ωL/R)) |
| B. | i = e<sup>-t(R/L)</sup>[-V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ-tan<sup>-1</sup>)(ωL/R))]-V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt+θ-tan<sup>-1</sup>)⁡(ωL/R)) |
| C. | i = e<sup>-t(R/L)</sup>[V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ-tan<sup>-1</sup>)⁡(ωL/R))]-V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt+θ-tan<sup>-1</sup>)⁡(ωL/R)) |
| D. | i = e<sup>-t(R/L)</sup>[-V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ-tan<sup>-1</sup>)⁡(ωL/R))]+V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt+θ-tan<sup>-1</sup>)⁡(ωL/R)) |
| Answer» E. | |
| 11. |
The value of ‘c’ in complementary function of ‘i’ is?$ |
| A. | c = -V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ+tan<sup>-1</sup>(ωL/R)) |
| B. | c = -V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ-tan<sup>-1</sup>(ωL/R)) |
| C. | c = V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ+tan<sup>-1</sup>(ωL/R)) |
| D. | c = V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(θ-tan<sup>-1</sup>(ωL/R)) |
| Answer» C. c = V/‚Äö√Ñ√∂‚àö‚Ć‚àö‚àÇ(R<sup>2</sup>+(‚âà√¨‚àö¬¢L)<sup>2</sup>) cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(‚âà√≠‚Äö√†√®+tan<sup>-1</sup>(‚âà√¨‚àö¬¢L/R)) | |
| 12. |
The particular current obtained from the solution of i in the sinusoidal response of R-L circuit is? |
| A. | i<sub>p</sub> = V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt+θ+tan<sup>-1</sup>(ωL/R)) |
| B. | i<sub>p</sub> = V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt+θ-tan<sup>-1</sup>(ωL/R)) |
| C. | i<sub>p</sub> = V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt-θ+tan<sup>-1</sup>(ωL/R)) |
| D. | i<sub>p</sub> = V/√(R<sup>2</sup>+(ωL)<sup>2</sup>) cos⁡(ωt-θ+tan<sup>-1</sup>(ωL/R)) |
| Answer» C. i<sub>p</sub> = V/‚Äö√Ñ√∂‚àö‚Ć‚àö‚àÇ(R<sup>2</sup>+(‚âà√¨‚àö¬¢L)<sup>2</sup>) cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(‚âà√¨‚àö¬¢t-‚âà√≠‚Äö√†√®+tan<sup>-1</sup>(‚âà√¨‚àö¬¢L/R)) | |
| 13. |
In the sinusoidal response of R-L circuit, the complementary function of the solution of i is? |
| A. | i<sub>c</sub> = ce<sup>-t(R/L)</sup> |
| B. | i<sub>c</sub> = ce<sup>t(RL)</sup> |
| C. | i<sub>c</sub> = ce<sup>-t(RL)</sup> |
| D. | i<sub>c</sub> = ce<sup>t(R/L)</sup> |
| Answer» B. i<sub>c</sub> = ce<sup>t(RL)</sup> | |