Question

The circuit shown has $3$ identical light bulbs $A,B,C$ and $2$ identical batteries $E_1$, $E_2$. When the switch is open, $A$ and $B$ glow with equal brightness, When the switch is closed:

• A. $A$ and $B$ will maintain their brightness and $C$ will be dimmer than $A$ and $B$.
• B. $A$ and $B$ will become dimmer and $C$ will be brighter than $A$ and $B$.
• C. $A$ and $B$ will maintain their brightness and $C$ will not glow.
• D. $A,B$ and $C$ will be equally bright.

Answer 1

The correct option is C $A$ and $B$ will maintain their brightness and $C$ will not glow.

Answer is C.
If a bulb is connected to a battery, then in the bulb the direction of current is such that it flows from negative terminal of battery to its positive terminal. Whereas inside a battery electrons flows from positive terminal to negative terminal. Also, the current will flow only in a closed loop.
It is given that bulbs A and B glow with equal brightness, when the voltage is supplied with both the batteries E1 and E2.
As the 3 bulbs are identical, the bulb connected to the switch also requires the same voltage for it to glow. But here, when the switch is closed, it forms a closed loop only with one battery. Therefore, it does not glow as the voltage is not sufficient.
Hence, when the switch is closed, A and B will maintain their brightness and C will not glow.