Question

In the circuit shown switch $S$ is connected to position $2$ for a long time and then joined to position $1.$ The total heat produced in a resistance $R_1$is

• A. $\frac{L{E}^2}{2{R_2}^2}$
• B. $\frac{L{E}^2}{2{R_1}^2}$
• C. $\frac{L{E}^2}{2R_1{R}_2}$
• D. $\frac{L{E}^2{(R_1+R_2)}^2}{2{{R_1}^2{R}_2}^2}$

Answer 1

The correct option is A $\frac{L{E}^2}{2{R_2}^2}$

At $t=0,2$ is opened and $1$ is closed.

$i\left(u\right)=\frac{E}{R_2}$

$\frac{Ldi\left(t\right)}{dt}\ne R_i\left(i\right(t\left)\right)=0$

$\frac{di\left(t\right)}{dt}=\frac{-R_1}{L}i\left(t\right)$

$\ln i\left(t\right)={\left[\frac{-R_{1}t}{L}\right]}_0^t$

$\left[\ln i\left(t\right)\right]=\frac{-R_{1}t}{L}\frac{i\left(t\right)}{E/R^2}=e^{\frac{-R_{1}t}{L}}i\left(t\right)=\frac{E}{R^2}{e}^{\frac{-R_{1}t}{L}}$

Head produced four twice $dt$ in $R_1$

$=i^2{R}_{1}dt$

$=\frac{{R_{1}E}^2}{{R_1}^2}{e}^{\frac{-2R_{1}t}{L}}dt=\frac{-L}{2R_1}\frac{{R_{1}E}^2}{{R_2}^2}{\left[e^{\frac{-2R_{1}t}{L}}\right]}_0^{\infty }=\frac{{LE}^2}{{{2R}_2}^2}$