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Current in resistance is 1 A , then A. `underset(s)(V)` = 5 VB. impedance of network is `5 Omega`C. power factor of given circuit is (0.6) lagging (current is lagging )D. All the above |
Answer» Correct Answer - D `V_S^(2)` = `(3)^(2)` + `(8 - 4)^(2)` , `V_S` = 5 v Now, Z = `V_S`/t = 5/t = 5 `Omega` Also, `V_R` = IR or R = 3/1 = `3 Omega` So, PF = R/Z = 0.6 as `V_L gt V_C` `implies` I lags V, so this a lagging nature. |
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