91Ó°ÊÓ

Three identical bulbs A, B, and C are connected. When switch S is closed, the current flows through the branch with zero resistance, which is the wire, and bulb C turns off. The resistance adds in series. The current is equal to the electromotive force (emf) of the battery divided by the circuit resistance as:

$I=\frac{\mathrm{Î\mu }}{∑R}$

Before closing the switch, the resistance $R_A+R_B+R_C$, so the current is;

$I=\frac{\mathrm{Î\mu }}{R_A+R_B+R_C}$

Asrole="math" localid="1663853209391" $R_A+R_{B}and{R}_C$are the resistance of the bulbs, and when the switch has closed the resistance of the circuit $R_A+R_B$, therefore the current is;

$I^{\text{'}}=\frac{\mathrm{Î\mu }}{R_A+R_B}$

On comparing;

$I^{\text{'}}>I$

Because the brightness of the light is determined by the current flowing through it, the current will increase when the switch is closed.

Hence, the brightness of bulbs A & B increases.