In the figure both cells A and B are of equal emf. Find R for which potential difference across battery A will be zero, long time after the switch is closed. Internal resistance of batteries A and B `(r_1) and (r_2)` respectively`(r_(1)gtr_(2))`.

Explanation

After a long time, steady state is reached in which impendance due to inductor `(omegaL` for `dc`) is zero and that due to capcitance
`((1)/(omega C))` becomes infinite, so equivalent circuit is shows in Fig
Net external resistance `R_(ext) = ((R )/(2) + R)/(2) = (3)/(4) R`
Net internal resistance `R_("int") = r_(1) + r_(2)`
`:.` Current in circuit `I = (2E)/((3)/(4) R + r_(1) + r_(2))`

The potential difference across the terminals of cell `A` is zero, so
`E - Ir_(1) = 0 rArr E - (2Er_(1))/((3)/(4) R + r_(1) + r_(2)) rArr R = (4)/(3) (r_(1) - r_(2))`.