IIT-JEE 2011 · Paper 1 · Physics Q30

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PhysicsSingle correct+3 / -1Question 30 of 69

A 2 μF capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is

A battery V, a 2 μF capacitor and an uncharged 8 μF capacitor share a common bottom wire. The switch S joins the top of the 2 μF capacitor either to the battery (position 1) or to the 8 μF capacitor (position 2).

20%

75%

80%

Answer and worked solution

Official answer: D

Charge sharing. The 2 μF capacitor (charge $2V$ μC) is connected to the uncharged 8 μF one. Final energy $\dfrac{Q^2}{2(C_1+C_2)}$ is a fraction $\dfrac{C_1}{C_1+C_2}=\frac2{10}$ of the initial $\dfrac{Q^2}{2C_1}$ .

Dissipated 80%.

(D)

Topics:Energy stored in a capacitor Series and parallel combinations

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IIT-JEE 2011 · Paper 1, Physics question 30