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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. `(EL)/(eR^(2))`D. `(eL)/(ER)` |
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Answer» Correct Answer - C The current at time `T` is given by `i=i_(0)(1-e^(-t//tau))` where `i_(0)=(E)/(R )` and `tau=(L)/(R )` So `dq=I dt=i_(0)(1-e^(-t//tau))dt` `Q=intdQ=in_(0)^(R )i_(0)(1-e^(t//tau))dt=i0{[in_(0)^(R )dt-int_(0)^(R )e^(-t//tau)dt]}` `=(i_(0)tau)/(e)impliesQ=((E)/(R ).(L)/(R))/(e)=(EL)/(eR^(2))impliesQ=(EL)/(ER^(2))` |
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