Verified Solution • 2025–26 Session moderate

QJEEM20Ph1700: JEE Main 2020 Physics - Class 12 JEEM20Ph1700 Solution: A 5 F capacitor is charged fully by a 220 V supply. It is then…

A 5 μ\mu F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μ\mu F capacitor. If the energy change during the charge redistribution is X100J{X \over {100}}J then value of X to the nearest integer is _____.

Step-by-Step Verified Solution

ui = 12\frac{1}{2} ×\times 5 ×\times 10-6 ×\times 220

Final common potential

= 220×5+0×2.55+2.5\frac{220\times 5+0\times 2.5}{5+2.5} = 220 ×\times 23\frac{2}{3}

uf = 12\frac{1}{2} ×\times (5 + 2.5) ×\times 10-6 ×\times (220×23)2\left( 220\times \frac{2}{3} \right)^{2}

Δ\Delta u = ui - uf

⇒\Rightarrow Δ\Delta u = –403.33 × 10–4

⇒\Rightarrow –403.33 × 10–4 = X100J{X \over {100}}J

⇒\Rightarrow X = -4.03

Value of X is approximate 4

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