In the circuit shown in figure, the time constants of the two branches 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 wires are negligible)

See the diagram in the problem
The (L - R) and (C - R) sections having the same time constant are in parallel.
The current through the R - L branch will be unaffected by the parallel R-C branch

$i_L\left(t\right)=\frac{E}{R}\left[1-e^{\frac{-Rt}{L}}\right] .....\left(i\right)$

Similarly the current in the RC branch is given by

$i_C\left(t\right)=\frac{E}{R}{e}^{\frac{-t}{RC}} .....\left(ii\right)$

Hence the current through the battery is

$i\left(t\right)=i_L\left(t\right)+i_C\left(t\right)=\frac{E}{R}\left[1-e^{-\frac{R}{C}t}+e^{-\frac{t}{RC}}\right] ....\left(iii\right)$

Time constant of RL branch is ${\tau }_L=\frac{L}{R}$ and the time constant of RC branch is ${\tau }_C=RC$

If these are equal, ${\tau }_L={\tau }_C=T$, equation (iii) becomes

$i\left(t\right)=\frac{E}{R}\left[1-e^{-\frac{t}{T}}+e^{-\frac{t}{T}}\right]=\frac{E}{R}$ for all values of t >0. Hence the result.