JEE MainPhysicsToughElectromagnetic induction

[IMAGE] As shown in the figure, a battery of emf $\varepsilon$ 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

Show answer and solution

q = \int_{0}^{T_{C}}idt

= \frac{\varepsilon}{R}{\lbrack t - \frac{e^{- t/T_{C}}}{\frac{- 1}{T_{C}}}\rbrack}_{0}^{T_{C}}

;

= \frac{\varepsilon}{R}\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\rbrack

= \frac{\varepsilon}{R} \times \frac{1}{e} \times \frac{L}{R}

= \frac{\varepsilon L}{R^{2}e}

Source: Grass-G / P-23 Physcis Paper 31 Dec FST.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.