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In the circuit shown in figure switch `S` is closed at time `t=0`. The charge which passes through the battery in one time constant is A. `(eR^(2)E)/(L)`B. `E = ((L)/(R ))`C. `(E L)/(eR^(2))`D. `(e L)/(E R)` |
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Answer» Correct Answer - C The current at time `t` is given by `i = i_(0) (1-e^(-t//tau))` Here `i_(0) = (E)/(R )` and `tau = (L)/(R )` `:. q = int_(0)^(tau)i d t = int_(0)^(t)i_(0)(1-e^(-t//tau))dt` `= (i_(0) tau)/(e) = (((E)/(R ))((L)/(R )))/(e) = (EL)/(eR^(2))` |
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