The circuit involves two ideal cells connected to a 1 μ F capacitor via a key K . Initially the key K is in position 1 and the capacitor is charged fully by 2 V cell. The key is pushed to position 2 .

Question

The circuit involves two ideal cells connected to a $1 μ F$ capacitor via a key K . Initially the key K is in position 1 and the capacitor is charged fully by 2 V cell. The key is pushed to position 2 . Column-I gives physical quantities involving the circuit after the key is pushed from position 1. Column-II gives corresponding results. Match the statements in Column-I with the corresponding values in Column-II and select the current options.

List - I	List - II
(P) The net charge crossing the 4 volt cell in $μ C$ is	(1) 2
(Q) The magnitude of work done by 4 Volt cell in $μ J$ is	(2) 6
(R) The gain in potential energy of capacitor in $μ J$ is	(3) 8
(S) The net heat produced in circuit in $μ J$ is	(4) 16

TestHub · Accepted Answer

(Moderate) The initial charge on capacitor $= CV_{i} = 1 \times 2 μ C = 2 μ C$

The final charge on capacitor $= CV_{f} = 1 \times 4 μ C = 4 μ C$

∴ Net charge crossing the cell of emf 4 V is

$q_{f} - q_{i} = 4 - 2 = 2 μ C$

The magnitude of work done by cell of emf 4 V is

$W = (q_{f} - q_{i}) 4 = 8 μ J$ The gain in potential energy of capacitor is $Δ U = \frac{1}{2} C (V_{f}^{2} - V_{i}^{2}) = \frac{1}{2} \times [42 - 22] μ J = 6 μ J$

Net heat produced in circuit is

$Δ H = W - Δ U = 8 - 6 = 2 μ J$

Physics - Capacitor Question with Solution | TestHub

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