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In figure the key K is closed at t=0 After a long time the potential difference between A and B is zero the value of R will be $[r_{1} = r_{2} = 1 Ω, E_{1} = 3 V a n d E_{2} = 7 V, C = 2 μ F, L = 4 m H,$ where $r_{1}$ and $r_{2}$ are the internal resistances of cells $E_{1}$ and $E_{2}$ respectively]-

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$\frac{2}{3} Ω$

$\frac{4}{9} Ω$

$\frac{4}{3} Ω$

$4 Ω$

answer is B.

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Detailed Solution

$V_{A} - V_{B} = E_{1} - I r_{1} = 0, E_{1} = I r_{1}$

$E_{1} = (\frac{E_{1} + E_{2}}{2 R + 1 + R + 1}) r_{1}, 3 = \frac{10 \times 1}{3 R + 2}, 9 R + 6 = 10, 9 R = 4, R = \frac{4}{9} Ω$

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