As shown in the figure, a battery of emf ∈ is connected to an...

As shown in the figure, a battery of emf

\in

is connected to an inductor

L

and resistance

R

in series. The switch is closed at

t = 0 .

The total charge that flows from the battery, between

t = 0

and

t = {t}_{c}

(

{t}_{c}

is the time constant of the circuit) is:

Options

A $\dfrac{\in L}{e{R}^{2}}$
B $\dfrac{\in L}{{R}^{2}}\left(1 - \dfrac{1}{e}\right)$
C $\dfrac{\in L}{{R}^{2}}$
D $\dfrac{\in R}{e{L}^{2}}$

Correct answer

A. $\dfrac{\in L}{e{R}^{2}}$

Step-by-step solution

$q = \mathop{∫}_{0}^{{T}_{c}}idt$
$= \dfrac{\in}{R}{\left[t - \dfrac{{e}^{- \dfrac{t}{{T}_{c}}}}{- \dfrac{1}{{T}_{c}}}\right]}_{0}^{{T}_{c}} = \dfrac{\in}{R}\left[{T}_{c} + {T}_{c}{e}^{- 1} - {T}_{c}\right]$
$= \dfrac{\in}{R} \times \dfrac{1}{e} \times \dfrac{L}{R}$

$q = \dfrac{\in L}{{R}^{2}e}$

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