InterviewSolution
Saved Bookmarks
| 1. |
In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. 0 V, 0.47 AB. 1.68 V, 0.47 AC. 0 V, 1.4 AD. 5.6 V, 1.4 A |
|
Answer» Correct Answer - C::D Z = `sqrt ((R)^(2) + (X_L - X_C)^(2))` R = `10Omega`, `X_L` = `omega`L = `2000 xx 5 xx 10^(-3)` = `10Omega` `X_C` = `1/omega`C = `1 / 2000 xx 50 xx 10^(-6)` = 10` Omega` i.e., Z = 10`Omega` Maximum current, `i_o` = `V_o`/Z = 20 / 10 = 2A Hence, `i_rms` = 2/`sqrt 2`= 1.4A and `V_rms` = `4 xx 141` = 5.64V |
|