Capacitors

JEE Main level, numerical answer, medium

137 of 180

Question 137 of 180

A 5μF5\mu F capacitor is charged fully by a 220V220V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5μF2.5\mu F capacitor. If the energy change during the charge redistribution is X100J\dfrac{X}{100}J then value of XX to the nearest integer is :

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Answer: 4

C1=5μFV1=220Volt{C}_{1}=5\mu F{V}_{1}=220Volt
C2=2.5μFV2=0{C}_{2}=2.5\mu F{V}_{2}=0
Heat loss;
ΔH=U1−Ut=12C1C2C1+C2(v1−v2)2\Delta H={U}_{1}-{U}_{t}=\dfrac{1}{2}\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}{\left({v}_{1}-{v}_{2}\right)}^{2}
=12×5×2.5(5+2.5)(220−0)2μJ=\dfrac{1}{2}\times \dfrac{5\times 2.5}{(5+2.5)}(220-0{)}^{2}\mu J
=52×3×22×22×100×10−6J=\dfrac{5}{2\times 3}\times 22\times 22\times 100\times {10}^{-6}J
=5×11×223×10−4J=55×223×10−4J=\dfrac{5\times 11\times 22}{3}\times {10}^{-4}J=\dfrac{55\times 22}{3}\times {10}^{-4}J
=12103×10−4J=12103×10−3J=4×10−2=\dfrac{1210}{3}\times {10}^{-4}J=\dfrac{1210}{3}\times {10}^{-3}J=4\times {10}^{-2}
According to questions
x100=4×10−2\dfrac{x}{100}=4\times {10}^{-2}
so, x=4x=4

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