Q.Three identical lamps, each having a resistance R, are connected to a battery of emf $\varepsilon$ (of negligible internal resistance): lamps A and B are connected in series with the battery, and lamp C is connected in parallel with a switch S such that closing S short-circuits lamp C. Suddenly the switch S is closed.

Step 1. With switch S open, lamps A, B and C are all in series with the battery, giving total resistance $3R$, so the current is $I_{open}=\dfrac{\varepsilon}{3R}$.

Step 2 (a). Closing S short-circuits lamp C (all current now bypasses C through the zero-resistance switch), leaving only A and B in series, total resistance $2R$: $I_{closed}=\dfrac{\varepsilon}{2R}$ -- a LARGER current than before, since the total resistance in the circuit has decreased.

Step 3 (b). Because C is bypassed entirely once S is closed, no current flows through it at all, so lamp C goes completely dark. Lamps A and B, still carrying the (now larger) current $\varepsilon/2R$ instead of the smaller $\varepsilon/3R$, glow MORE brightly than before.

Step 4 (c). Voltage: with S open, each of the three identical lamps carries the same current through the same resistance R, so each has voltage $V_A=V_B=V_C=\varepsilon/3$. With S closed, A and B (in series, total $2R$) each carry current $\varepsilon/2R$, giving $V_A=V_B=(\varepsilon/2R)R=\varepsilon/2$; C, entirely bypassed, has $V_C=0$.

Step 5 (d). Power: with S open, each lamp dissipates $P=I^2R=(\varepsilon/3R)^2R=\varepsilon^2/9R$, so $P_A=P_B=P_C=\varepsilon^2/9R$. With S closed, A and B each dissipate $P=(\varepsilon/2R)^2R=\varepsilon^2/4R$, while $P_C=0$ (C carries no current). …

Recognise that closing S short-circuits lamp C, reducing the series circuit from 3R to 2R, and recompu …

• Assuming closing the switch removes lamp C's resistance without also recognising the current through it drops to exactly zero. …