Question

In the circuit shown in the figure switch S is closed at time t = 0. Find the charge which passes through the battery in one time constant.

Answer 1

We have,

An inductor and a resistance connected in a circuit.

The current flowing in LR circuit at a given time t is

$i=i_0(1-e^{-t/T})$

$i_0=\frac{E}{R}$ and $T=\frac{L}{R}$

Charge flowing in time $t=T$

$q={\int }_0^{T}idt$

$={\int }_0^r{i}_0(1-e^{-t}r)dt$

$=\frac{i_{0}T}{e}$

$=\frac{\frac{E}{R}\times \frac{L}{R}}{e}$

$=\frac{EL}{R^{2}e}$