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(B) $1$. When $T_{1}$ and $T_{2}$ are connected,the circuit consists of the $6 \, V$ battery,the $R = 6 \, \Omega$ resistor,and the inductor $L$ in se

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$3$

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(B) $1$. When $T_{1}$ and $T_{2}$ are connected,the circuit consists of the $6 \, V$ battery,the $R = 6 \, \Omega$ resistor,and the inductor $L$ in series.$2$. At steady state,the inductor acts as a short circuit (zero resistance). The steady-state current $I$ flowing through the inductor is given by $I = \frac{V}{R} = \frac{6 \, V}{6 \, \Omega} = 1 \, A$.$3$. When $T_{1}$ is disconnected from $T_{2}$ and immediately connected to $T_{3}$,the inductor $L$ is now in series with the $r = 3 \, \Omega$ resistor. Because the current through an inductor cannot change instantaneously,the current $I = 1 \, A$ continues to flow through the new circuit loop.$4$. The potential drop $V_{r}$ across the $r = 3 \, \Omega$ resistor immediately after the switch is flipped is $V_{r} = I 	imes r = 1 \, A 	imes 3 \, \Omega = 3 \, V$.

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