Question

The circuit shown in figure consists of a battery of emf $E$ , resistances $R_1$ and $R_2$ inductances $L$ and a two way switch $ABC$. Initially $A$ is connected to $B$ for a long time and at $t=0$ $A$ is connected to $C$. Then,

• A. The current through inductor when the switch is in position $B$ is$\frac{E}{R_1}$
• B. The current through inductor when the switch is in position $B$ is$\frac{E}{R_2}$
• C. The total heat produced in $R_2$ after the switch is thrown to $C$ is$\frac{L{E}^2}{2R_2^2}$
• D. The total heat produced in $R_2$ after the switch is thrown to $C$ is $\frac{L{E}^2}{2R_1^2}$

Answer 1

Answer: (A), (D)

The total heat produced in $R_2$ after the switch is thrown to $C$ is $\frac{L{E}^2}{2R_1^2}$

When the switch is in position $B$ for long time the current through the inductor(acts a short circuit) is by ohms law
$I=\frac{V}{R}=\frac{E}{R_1}$
when switch is in position $B,$ the energy stored in inductor is
$U=\frac{1}{2}L{I}^2=\frac{1}{2}L{\left(\frac{E}{R_1}\right)}^2$
When switch is connected to $C$, total energy stored in inductor is dissipated through resistance $R_2$
$a,d$ are correct