Question

In the figure below, the switches $S_1$ and $S_2$ are closed simultaneously at $t=0$ and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current $I$ in the middle wire reaches its maximum magnitude $I_{max}$ at time $t=\tau$. Which of the following statement(s) is (are) true?

• A. $I_{max}=\frac{V}{2R}$
• B. $I_{max}=\frac{V}{4R}$
• C. $t=\frac{L}{R}\ell n2$
• D. $t=\frac{2L}{R}\ell n2$

Answer 1

Answer: (B), (D)

$t=\frac{2L}{R}\ell n2$


Consider current in left part to be $i_1$ and in the right part to be $i_2$
$\therefore i$ through middle branch
$i_{max}=(i_2-i_1{)}_{max}$

$\Delta i=(i_2-i_1)={\frac{V}{R}}^{[1-e^{-\left(\frac{R}{2L}\right)t}]}$$-{\frac{V}{R}}^{[1-e^{-\left(\frac{R}{L}\right)t}]}$

${\frac{V}{R}}^{\left(e^{-\left(\frac{R}{L}\right)t}-e^{-\left(\frac{R}{2L}\right)t}\right]}$
For $(\Delta i{)}_{max}\frac{d\left(\Delta i\right)}{dt}=0$

$\frac{V}{R}[-\frac{R}{L}{e}^{-\left(\frac{R}{L}\right)t}-(-\frac{R}{2L})e^{-\left(\frac{R}{2L}\right)t}]=0$

$e^{-\left(\frac{R}{L}\right)t}=\frac{1}{2}{e}^{-\left(\frac{R}{2L}\right)t}$

$e^{-\left(\frac{R}{2L}\right)t}=\frac{1}{2}$

$\left(\frac{R}{2L}\right)t=\ell n2$

$t=\frac{2L}{R}\ell n2\to$ time when $i$ is maximum.
Henne, option D is correct.

$i_{max}={\frac{V}{R}}^{[e^{-\frac{R}{L}\left(\frac{2L}{R}\ell n2\right)}-e^{-\left(\frac{R}{2L}\right)\left(\frac{2L}{R}\ell n2\right)}]}$

so, $\frac{V}{R}\left|[\frac{1}{4}-\frac{1}{2}]\right|=\frac{1}{4}\frac{V}{R}$
Hence option B is correct.