MCQOPTIONS
Saved Bookmarks
MCQOPTIONS
This section includes 7 Mcqs, each offering curated multiple-choice questions to sharpen your Network Theory Questions and Answers knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The complete solution of current obtained by substituting the values of c1 and c2 in the following equation is? |
| A. | i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰). |
| B. | i = e-37.5t(0.49cos1290t – 1.3sin1290t) + 0.7cos(500t + 133.5⁰) |
| C. | i = e-37.5t(0.49cos1290t – 1.3sin1290t) – 0.7cos(500t + 133.5⁰) |
| D. | i = e-37.5t(0.49cos1290t + 1.3sin1290t) – 0.7cos(500t + 133.5⁰) |
| E. | i = e-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰) |
| Answer» E. i = e-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰) | |
| 2. |
The value of the c2 in the following equation is? |
| A. | i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰). |
| B. | 2.3 |
| C. | -2.3 |
| D. | 1.3 |
| E. | -1.3 |
| Answer» D. 1.3 | |
| 3. |
The value of the c1 in the following equation is? |
| A. | i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰). |
| B. | -0.5 |
| C. | 0.5 |
| D. | 0.6 |
| E. | -0.6 |
| Answer» C. 0.5 | |
| 4. |
In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complete solution of current. |
| A. | i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰) |
| B. | i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰) |
| C. | i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t – π/4 + 88.5⁰) |
| D. | i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t + π/4 + 88.5⁰) |
| Answer» B. i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰) | |
| 5. |
In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the particular solution. |
| A. | ip = 0.6cos(500t + π/4 + 88.5⁰) |
| B. | ip = 0.6cos(500t + π/4 + 89.5⁰) |
| C. | ip = 0.7cos(500t + π/4 + 89.5⁰) |
| D. | ip = 0.7cos(500t + π/4 + 88.5⁰) |
| Answer» E. | |
| 6. |
In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complementary current. |
| A. | ic = e-37.5t(c1cos1290t + c2sin1290t) |
| B. | ic = e-37.5t(c1cos1290t – c2sin1290t) |
| C. | ic = e37.5t(c1cos1290t – c2sin1290t) |
| D. | ic = e37.5t(c1cos1290t + c2sin1290t) |
| Answer» B. ic = e-37.5t(c1cos1290t – c2sin1290t) | |
| 7. |
In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation. |
| A. | -38.5±j1290 |
| B. | 38.5±j1290 |
| C. | 37.5±j1290 |
| D. | -37.5±j1290 |
| Answer» E. | |