Question 92: Electromagnetic Induction

Question

A 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 :

Accepted Answer

$$i = {i_0}\left( {1 - {e^{ - t/{t_c}}}} \right)$$
q = $$\int\limits_0^{{t_c}} {idt} $$
= $$\int\limits_0^{{t_c}} {{ \in  \over R}\left( {1 - {e^{ - t/{t_c}}}} \right)dt} $$
= $${ \in  \over R}\left[ {t - {{{e^{ - t/{t_c}}}} \over {\left( { - {1 \over {{t_c}}}} \right)}}} \right]_0^{{t_c}}$$
= $${ \in  \over R}\left[ {\left( {{t_c} + {t_c}{e^{ - 1}}} \right) - \left( {0 + {t_c}} \right)} \right]$$
= $${ \in  \over R}{t_c}.{1 \over e}$$
= $${ \in  \over R}.{L \over R}.{1 \over e}$$
= $${{ \in L} \over {e{R^2}}}$$

Answer

$${{\varepsilon L} \over {e{R^2}}}$$

Answer

$${{\varepsilon L} \over {{R^2}}}$$

Answer

$${{\varepsilon L} \over {{R^2}}}\left( {1 - {1 \over e}} \right)$$

Answer

$${{\varepsilon R} \over {e{L^2}}}$$

Electromagnetic Induction

A 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 :

Choose the correct answer: