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A 1 mu F capacitor C is connected to a battery of 10 V through a resistance 1 M Omega. The voltage across C after 1 sec is approximately |
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Answer» Solution :`C=1 mu F = 10^(-6)F, R = 1M Omega = 10^(6)Omega` `V_(0)=10 V` Time constant of circuit, `tau_(c )=RC=10^(6)xx10^(-6)=1s` Voltage across C in time `t, V=V_(0)(1-e^(-t//tau_(c )))` Fort = 1 s, and `tau_(c )=1s` `therefore V=10(1-(1)/(e ))=6.3 V` |
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