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At time t=0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them beocme 4V? `[Take : In 5 = 1.6, In3 = 1.1]`. . |
Answer» Here, supply voltage, `V_(s) = 10 vol t`. Effective resistance, `R = (2xx2)/(2+2) = 1 M Omega = 10^(6) Omega` Effective capacitanace, `C = 2+2 = 4 muF = 4xx10^(-6)F` While charging the capacitor with voltage applied, voltage across the capacitor is given by `V_(c) = V_(s) [1-e^(-t//RC )]=10 [1-e^(-t//10^(6))xx4xx10^(-6)]` `4 = 10 [1-e^(-t//4)]` `1-e^(-t//4) = (4)/(10) = 0.4` `e^(-t//4) = 1 - 0.4 = 0.6` `- (t)/(4) log_(e) e = log_(e) 0.6 = log_(e) ((3)/(5)) = log_(e) 3 - log_(e) 5` `- (t)/(4) = 1.1 - 1.6 = -0.5` `t = 4xx0.5 = 2s` |
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