The bulb which glows with maximum intensity in the given circuit is
- $4 \Omega$ bulb
- $2 \Omega$ bulb
- $3 \Omega$ bulb
- $6 \Omega$ bulb
Solution
As, resistance of portion $B$ is higher. $\therefore$ A greater potential drop across $B$ occurs. Now, in section $B$, Power, $P=\frac{V^2}{R}$ So, least resistance gives a higher power output. $\therefore 4 \Omega$ bulb glows with maximum intensity.
Asked in: AP EAMCET 2018 (23 Apr Shift 2)