Question 210: AC fundamentals

Question

A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then
the time at which the energy stored in the inductor reaches $$\left( {{1 \over n}} \right)$$  times of its maximum value, is :

Accepted Answer

P.E. in inductor, $$U = {1 \over 2}L{I^2}$$
$$U \propto {I^2}$$
$${U \over {{U_0}}} = {\left( {{I \over {{I_0}}}} \right)^2}$$
$${1 \over n} = {\left( {{I \over {{I_0}}}} \right)^2}$$
$$I = {{{I_0}} \over {\sqrt n }}$$
We know, $$I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
$${{{I_0}} \over {\sqrt n }} = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
$$ \Rightarrow $$ $${e^{ - {{Rt} \over L}}}$$ = 1 - $${1 \over {\sqrt n }}$$
taking ln & solving we get,
$$t = {L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n  - 1}}} \right)$$

Answer

$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n  + 1}}} \right)$$

Answer

$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n  - 1}}} \right)$$

Answer

$${L \over R}\ln \left( {{{\sqrt n  + 1} \over {\sqrt n  - 1}}} \right)$$

Answer

$${L \over R}\ln \left( {{{\sqrt n  - 1} \over {\sqrt n }}} \right)$$

AC fundamentals

A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches $\left( {{1 \over n}} \right)$ times of its maximum value, is :

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