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A current of 1 A flows through a coil, when a 100 V d.c. is applied to it. But a current of 0.5 A flows through the same coil, when a 100V a.c. of frequency 50 Hz is applied. The resistance and inductance of the coil are given by (take `pi^(2)=10`)A. `100 Omega , sqrt(0.2) H`B. `50 Omega , sqrt(0.3) H`C. `100 Omega , sqrt(0.3) H`D. `100 Omega , sqrt(0.2) H` |
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Answer» Correct Answer - C `R = 100/1 = 100 Omega for d.c.` for a.c., `I = V/(sqrt(R^(2)+L^(2)omega^(2)))` `:. R^(2)+L^(2)omega^(2) = (V^2)/(I^2) = (100 xx 100)/((1)/(2) xx (1)/(2)) = 4 xx 10^(4)` `:. 4 xx 10^(4) = (100)^(2) + 4pi^(2)L^(2)f^(2) = 10^(4)(1+10L^(2))` `3 = 10 L^(2) :. L = sqrt(0.3)H` (Thus `R = 100 Omega and L = sqrt(0.3)H)`. |
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