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A series circuit consists of two capacitors, a resistors, and an ideal voltage source. The circuit is initially at steady state. Now a conducting wire is shorted across the capacitor C as shown by dotted line. After shorting the wire, select the correct statement// statements. |
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Answer» Heat developed in the circuit is `2 CV^2//3` Before SHORTING the wire at steady STATE the charges on each capacitors will be EQUAL to  `q_0 = q_1 = q_2 = (2C)/3 V` After shorting the capacitor marked (2) will be useless in circuit. Hence `q'_1 = 2CV = q_f and q'_2 = 0` The charge supplied by battery after shorting  `Deltaq = q_(f) - q_0 = 2CV - (2CV)/3 = (4CV)/3` Hence work done by battery `W = Deltaq V = (4CV^2)/3` Initial potential energy of system. `V_0 = 1/2 (2/3C) V^2 = 1/3 CV^2` Final potential energy of system `U_(f) = 1/2 (2C)V^2 = CV^2` Hence change in PE `DeltaU = U_(f) - U_(i)` or `DeltaU = CV^2 - 1/3 CV^2 = 2/3 CV^2` Work done by battery `W = DeltaU + Heat` Heat developed `H = W-DeltaU` or `H = 4/3CV^2 - 2/3 CV^2 = 2/3CV^2` . |
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