The circuit involves two ideal cells connected to a 1 µF capacitor via a key K. Initially the key K is in position 1 and the capacitor is charged fully by 2V cell. The key is then 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.
Column I | Column II | ||
(A) | The net charge crossing the 4 volt cell in µC is | (P) | 2 |
(B) | The magnitude of work done by 4 Volt cell in µJ is | (Q) | 6 |
(C) | The gain in potential energy of capacitor in µJ is | (R) | 8 |
(D) | The net heat produced in circuit in µJ is | (S) | 16 |
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(P) - (R) - (Q) - (P)
Sol. The initial charge on capacitor = CV i = 2 × 1 µC = 2 µC
The final charge on capacitor = CV f = 4 × 1 µC = 4 µC
∴ Net charge crossing the cell of emf 4V is
q f – q i = 4 – 2 = 2 µC
The magnitude of work done by cell of emf 4V is
W = (q f – q i ) 4 = 8 µJ
The gain in potential energy of capacitor is
Δ U = 
= 
1 × [4 2 – 2 2 ] µJ = 6 µJ
Net heat produced in circuit is
Δ H = 
W – Δ U = 8 – 6 = 2 µJ
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