In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is:

Question

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is: (A) (3/8)(Q2/C) (B) (4/3)(Q2/C) (C) (1/8)(Q2/C) (D) (5/8)(Q2/C)

Tardigrade Admin · Accepted Answer

Vi = (1/2)CE2 Vf = ((CE)2/2×4c) = (1/2) (CE2/4) Δ E =(1/2) CE2 ×(3/4) = (3/8) CE2

Q. In the figure shown, after the switch $^{'} S^{'}$ is turned from position $^{'} A^{'}$ to position $^{'} B^{'}$ , the energy dissipated in the circuit in terms of capacitance $^{'} C^{'}$ and total charge $^{'} Q^{'}$ is:

$\frac{3}{8} \frac{Q ^{2}}{C}$

$\frac{4}{3} \frac{Q ^{2}}{C}$

$\frac{1}{8} \frac{Q ^{2}}{C}$

$\frac{5}{8} \frac{Q ^{2}}{C}$

$V_{i} = \frac{1}{2} C E^{2}$
$V_{f} = \frac{( CE ) ^{2}}{2 \times 4 c} = \frac{1}{2} \frac{C E ^{2}}{4}$
$Δ E = \frac{1}{2} C E^{2} \times \frac{3}{4} = \frac{3}{8} C E^{2}$

Q. In the figure shown, after the switch ′S′ is turned from position ′A′ to position ′B′, the energy dissipated in the circuit in terms of capacitance ′C′ and total charge ′Q′ is:

83​CQ2​

34​CQ2​

81​CQ2​

85​CQ2​