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(B) Let the resistance of each identical lamp be $R$ and the battery voltage be $V$.Initially,when switch $S$ is open,lamps $X$ and $Y$ are in series.

Vedclass · Accepted Answer

$X$ increases,$Y$ decreases

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(B) Let the resistance of each identical lamp be $R$ and the battery voltage be $V$.Initially,when switch $S$ is open,lamps $X$ and $Y$ are in series. The total resistance is $2R$,and the current is $I = V / (2R)$. The power dissipated by $X$ and $Y$ is $P = I^2 R = (V^2 / 4R^2) * R = V^2 / (4R)$.When switch $S$ is closed,lamp $Z$ is connected in parallel with lamp $Y$. The equivalent resistance of the parallel combination of $Y$ and $Z$ is $R_p = (R * R) / (R + R) = R/2$.The total resistance of the circuit becomes $R_{total} = R + R/2 = 3R/2$.The total current from the battery is $I' = V / (3R/2) = 2V / (3R)$.Since the current through $X$ is now $I'$,and $I' = 2V / (3R) > V / (2R) = I$,the brightness of lamp $X$ increases.The voltage across the parallel combination of $Y$ and $Z$ is $V_p = I' * R_p = (2V / 3R) * (R/2) = V/3$.Since the voltage across $Y$ was initially $V/2$ (when $S$ was open) and is now $V/3$ (when $S$ is closed),the voltage across $Y$ decreases,so the brightness of lamp $Y$ decreases.

If $X, Y$ and $Z$ in the figure are identical lamps,which of the following changes to the brightness of the lamps occur when switch $S$ is closed?

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