In the given LR circuit, the switch S is closed at time t = 0. Then,

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The ratio of induced emfs in the inductors of inductance L and 2L will be constant

The potential difference ‘ $V_{A} {- V}_{B}$ ’ increases with time

The potential difference ‘ $V_{A} {- V}_{B}$ ’ will be constant.

The ratio of induced emf’s in the inductor of inductance L and 2L will be time varying

answer is A, D.

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Detailed Solution

$\begin{array}{l} i_{1} = \frac{ε}{R} (1 - e^{- R t / L}) \\ i_{2} = \frac{ε}{2 R} (1 - e^{- 2 R t / 2 L}) = \frac{ε}{2 R} (1 - e^{- R t / L}) \\ {(ε_{ind})}_{L} = L \frac{d i_{1}}{d t} = ε (e^{- R t / L}) \\ {(ε_{ind})}_{2 L} = 2 L \frac{d i_{2}}{d t} = ε (e^{- R t / L}) \\ V_{A} - V_{B} = - 2 L \frac{d l_{2}}{d t} + L \frac{d l_{1}}{d t} = 0 \end{array}$