JEE Main 2024 — 5 April, Shift 1 — Question 58

In parallel combination : Potential difference is same across all Energy

$=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\right) \mathrm{V}^{2}$

$=\frac{1}{2}(25+30+45) \times(100)^{2} \times 10^{-6}=0.5=\mathrm{E}$

In series combination: Charge is same on all.

$\frac{1}{\mathrm{C}_{\text {equ }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45}$

$\frac{1}{\mathrm{C}_{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}_{\text {equ }}=\frac{450}{43}$

Energy $=\frac{Q^{2}}{2 C_{1}}+\frac{Q^{2}}{2 C_{2}}+\frac{Q^{2}}{2 C_{3}}$

$=\frac{\mathrm{Q}^{2}}{2}\left[\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right]$ $\frac{\left(\mathrm{V} \times \mathrm{C}_{\text {equ }}\right)^{2}}{2} \times \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{\mathrm{V}^{2} \mathrm{C}_{\text {equ }}}{2}$ $\frac{(100)^{2}}{2} \times \frac{450}{43} \times 10^{-6}$

$\Rightarrow \frac{4.5}{86}=\frac{9}{x} E=\frac{9}{x} \times 0.5 \Rightarrow x=86$