Electromagnetic Induction And Alternating Current Ques 54

Answer:

Correct Answer: 54.$R=\frac{4}{3}(r_1-r_2)$

Solution:

Formula:

Growth Of Current in Series R-L Circuit:

After a long time, resistance across an inductor becomes infinite while resistance across a capacitor becomes zero. Hence, net external resistance,

$$ R_{\mathrm{net}}=\frac{\frac{R}{2}+R}{2}=\frac{3 R}{4} $$

Current through the batteries, $i=\frac{2 E}{\frac{3 R}{4}+r_{1}+r_{2}}$

Given that the potential across the terminals of cell $A$ is zero.

$$ \begin{alignedat} & \therefore \quad E-i r_{1}=0 \\ & \text { or } \quad E-\frac{2 E}{3 R / 4 + r_{1}+r_{2}} r_{1}=0 \end{aligned} $$

Solving this equation, we get, $R=\frac{4}{3}\left(r_{1}-r_{2}\right)$