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In the circuit shown in fig, the time constant of the two braches are equal(=T). Then if the key S is closed at the instant t=0, the time in which the current in the circuit through the battery will rise to its final value of `(E//R)` will be, (assume that the internal resistance of the battery and of the connecting wire are negligible) A. InstantlyB. `t=T/2`C. `t=2T`D. `t=oo` |
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Answer» Correct Answer - A The (L-R) and (C-R) sections having the same time constant are in parallel. The current through the R-L branch will be unaffacted by the parallel R-C branch `i_(L)(t)=E/R [1-r^((-Rt)/(L))]` ...(i) Similarly the current in the RC branch is given by `i_(C)(t)=E/R e^((-t)/(RC))`...(ii) Hence the current through the battery is `i(t)=i_(L)(t)+i_(C)(t)=E/R[1-e^(-(R)/(C)t)+e^(-(t)/(RC))]`....(iii) Time cosntant of RL branch is `tau_(L)=L/R` and the time constant of RC branch is `tau(C)=RC`. If these are equal , `tau_(L)=tau_(C)=T`, equation (iii) becomes `i(t)=E/R[1-e^(-(R)/(C)t)+e^(-(t)/(RC))]=E/R` for all value of tgto=0. Hence the result. |
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