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In the circuit shown in figure, capacitor A has capacitance C_(1)=2muF when filled with a di-electric slab of dielectric constant k=2. Capacitors B and C are air capacitors and have capacitances C_(2)=3muF and C_(3)=6muF, respectively. A is charged by closing switch S_(1) alone (a) Calculate energy supplied by battery during process of charging. Switch S_(1) is now opened and S_(2) is closed. (b) Calculate charge on B and energy stored in the system when electrical equilibrium is attained. Now switch S_(2) is also opened, slab of A is removed. Another di-electric slab of K=2, which can just fill the space in B, is inserted into it and then switch S_(2) alone is closed. (c) Calculate by how many times electric field in B is increased. Calculate also, loss. of energy during redistribution of charge.

Answer»


ANSWER :(a) `0.0648J`, ENERGY stored in capacitor is `1/2CV^(2)`, but energy SUPPLIED by BATTERY is `CV^(2)`; (B) `180muC, 0.0162J`; © `0.75, 0.0054J`


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