question_answer

The circuit shown in the figure is in steady state. Find the rate of change of current through L immediately after the switch S is closed.

A) \[\frac{\varepsilon }{3L}\]                 

B) \[\frac{2\varepsilon }{3L}\]

C) \[\frac{3\varepsilon }{5L}\]                

D) \[\frac{\varepsilon }{5L}\]

Correct Answer: B

Solution :

[b] In steady state the current through the inductor is The current through the inductor will not change immediately after the switch is closed. Let          \[{{V}_{AB}}=V\] In left loop      \[R{{i}_{1}}+V=\varepsilon \]              ... (i) In right loop     \[2R{{i}_{2}}+V=\varepsilon \]             ... (ii) From (i) and (ii) \[{{i}_{1}}+{{i}_{2}}=\frac{\varepsilon -V}{R}+\frac{\varepsilon -V}{2R}\] Since  \[{{i}_{1}}+{{i}_{2}}={{i}_{0}}=\frac{\varepsilon }{2R}\] [Because\[{{i}_{0}}=\frac{\varepsilon }{2R}\]] \[\therefore \,\,\,\,\,\frac{\varepsilon -V}{R}+\frac{\varepsilon -V}{2R}=\frac{\varepsilon }{2R}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,V=\frac{2\varepsilon }{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,L=\frac{di}{dt}=\frac{2\varepsilon }{\varepsilon }\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\frac{di}{dt}=\frac{2\varepsilon }{3L}\] At this instant the potential difference across R is \[\varepsilon -\frac{2\varepsilon }{3}=\frac{\varepsilon }{3}\]. This potential difference will eventually become e. It means current is increasing. \[\frac{di}{dt}\] is positive.