Question

For the circuit shown, a shorting wire of negligible resistance is added to the circuit between points $A$ and $B$. When this shorting wire is added, bulb $3$ goes out. Which bulb(s) in the circuit brighten? All bulbs are identical.

• A. Only bulb 2
• B. Only bulb 4
• C. Only bulbs 1 and 4
• D. Only bulbs 2 and 4

Answer 1

Answer: (C)

Let the resistance of each bulb be $R .$ Initially:
$R_{ eq }=5 R / 3, $ Finally: $R_{ eq }=3 R / 2$
Equivalent resistance decreases, so current increases in circuit and in '1' also. Hence brightness of '1' increases. It means potential difference across '1' increases, so across '2' potential difference decreases, hence brightness of 2 decreases.
Initially: Potential difference across '4'
$V_{4 i}=\frac{1}{2}\left[\frac{(2 R / 3) \varepsilon}{2 R / 3+R}\right]=\frac{\varepsilon}{5}$
Finally: $V_{4 f}=\frac{(R / 2) \varepsilon}{R / 2+R}=\frac{\varepsilon}{3}$
Since $V_{4 f} > V_{4 i}$, hence brightness of 4 increases.