Question 13 Solution

State 1 (Both Open): Components $R, L, C$ in series. $$ Z_1 = \sqrt{R^2 + (X_L – X_C)^2} $$ State 2 (Switch A Closed): Inductor $L$ shorted. Components $R, C$ in series. $$ Z_2 = \sqrt{R^2 + X_C^2} $$ Given that brightness (current) is unchanged: $|Z_1| = |Z_2|$. $$ (X_L – X_C)^2 = X_C^2 \implies X_L – X_C = X_C \implies X_L = 2X_C $$

Switch B shorts the capacitor. Components $R, L$ in series. $$ Z_3 = \sqrt{R^2 + X_L^2} $$ Substitute $X_L = 2X_C$: $$ Z_3 = \sqrt{R^2 + (2X_C)^2} = \sqrt{R^2 + 4X_C^2} $$

Compare $Z_3$ with $Z_2$: $$ Z_2 = \sqrt{R^2 + X_C^2} $$ $$ Z_3 = \sqrt{R^2 + 4X_C^2} $$ Clearly, $Z_3 > Z_2$.
Since impedance increases, current decreases. Therefore, the brightness of the lamp decreases.