1.

In the circuits shows in S_(1) and S_(2) are switches. S_(2) remains closed for a long time and S_(1) is opened. Now S_(1) is also closed. Just after S_(1) is closed, find the potential difference (V) across R and di//dt (with sign) in L.

Answer»


Solution :Before closing `S_(1)`, current in the inductor is `I = varepsilon//2R`. Just after closing `S_(1)`, current in inductor will remain same, and other currents are as shows

For LEFT LOOP: `varepsilon = i_(1) R + L(di)//dt)` (i)
For right loop: `varepsilon = i_(2)(2R) + L(di//dt)` (II)
and `i_(1) + i_(2) = i` (iii)
Solve to get `(di)/(dt) = (2varepsilon)/(3L)`
p.d.across `R: V_(R) = i_(1)R = varepsilon - L(di//dt) = varepsilon - L ((2 E)/(3L)) = (varepsilon)/(3)`


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