Assume all lamps have resistance $R$ and voltage source is $V$. In steady state, inductors behave as short circuits (wires with zero resistance).

• Current through Lamp B ($I_B$) = $V/R$.
• Current through Lamp C ($I_C$) = $V/R$.
• Inductor $L_2$ is in series with C, so current in $L_2$ is $I_{L2} = I_C = V/R$.
• Inductor $L_1$ feeds both the B-branch and the C-branch. So, current in $L_1$ is $I_{L1} = I_B + I_C = V/R + V/R = 2V/R$.
• Current through Lamp A ($I_A$) = $V/R$ (connected directly across source).

When S is opened, the source is disconnected. However, inductors resist sudden changes in current. Therefore, immediately after $t=0$:

• $I_{L1}$ must remain $2V/R$.
• $I_{L2}$ must remain $V/R$.

Lamp C:
It is in series with $L_2$. Since $I_{L2}$ is unchanged, current through C remains $V/R$.
$\rightarrow$ Brightness of C is unchanged.

Lamp B:
At the node between $L_1$, Lamp B, and $L_2$, Kirchhoff’s Current Law applies.
Current entering from $L_1$ = $2V/R$.
Current leaving to $L_2$ branch = $V/R$.
Remaining current must go through Lamp B: $I_B = I_{L1} – I_{L2} = 2V/R – V/R = V/R$.
$\rightarrow$ Brightness of B is unchanged.

Lamp A:
The entire current $I_{L1}$ ($2V/R$) circulating in the loop must return. With the switch open, the only path for the current from the left side of $L_1$ to complete the circuit is through Lamp A.
Current through A becomes equal to $I_{L1} = 2V/R$.
Previous current was $V/R$. New current is $2V/R$.
$\rightarrow$ Brightness of A increases.