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A L-R combination is connected to a battery of emf 4 volt. If L = 0.1 H and R = 4.0 `Omega`, then the time taken to reach a current of 0.631 ampere is:-A. (1/40) secB. 0.4 secC. 1.6 secD. 0.63 sec |
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Answer» Correct Answer - A Final maximum current `=E/R=4/4=1` amp Instantaneous current , `I=E/R (1-e^(-Rt//L))` Here , I=0.6321 ampere 0.6321 = 1(1–e–Rt/L) 0.6321 = 1–e–Rt/L e–Rt/L = 1–0.6321 = 0.3679 = `1/e` or `"Rt"/L=1` or `t=L/R = 0.1/4=1/40` sec. |
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