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When an inductor L and a resistor R in series are connected across a 12 V, 50 Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by pi/3. radian. Calculate the value of R. |
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Answer» Solution :Here, `V_(rms) = 12 V, v = 50 Hz, I_(rms) = 0.5 A`, and phase difference `phi = pi/3` `therefore` IMPEDANCE `Z = (V_(rms)/I_(rms)) = (12 V)/(0.5 A) = 24 OMEGA` and `tan phi = tan pi/3 = X_(L)/R` or `sqrt(3) = X_(L)/R rArr X_(L) = sqrt(3) R` As `Z = sqrt(R^(2) + X_(L)^(2))`, hence `24 = sqrt(R^(2) + (sqrt(3)R)^(2)) = 2R` `rArr R = 24/2 = 12 Omega` |
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