Question

The battery in the circuit shown in Fig. Q26.14 has no internal resistance. After you close the switch S, will the brightness of bulb${\mathit{B}}_{\mathbf{1}}$increase, decrease, or stay the same?

Short answer

The brightness of $B_1$ will increase.

Step-by-step solution

Determining the brightness of bulb B1

The two bulbs are in series while switch S is open, thus they have the same current but not the same voltage. The voltage drop in the battery is caused by their series combination.

When S is closed, the current flows only through$B_1$and the switch S, while$B_2$is cut off because the switch path creates a short circuit with no resistance and draws the current. As it only flows in$B_1$, the current will increase as the total resistance in the circuit decreases.

Hence, the brightness of $B_1$will increase.