Question

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. Instantly
B. `t=T/2`
C. `t=2T`
D. `t=oo`

Answer 1

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.