SATHEE: Electrostatics 5 Question 23

Electrostatics 5 Question 23

25. A $2 μ F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 is

(2011)

(a) $0$

(b) $20$

(d) $80$

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Answer:

Correct Answer: 25. (d)

Solution:

$q_{i} = C_{i} V = 2 V = q$

This charge will remain constant after switch is shifted from position 1 to position 2 .

$\begin{aligned} U_{i} & = \frac{1}{2} \frac{q^{2}}{C_{i}} = \frac{q^{2}}{2 \times 2} = \frac{q^{2}}{4} \\ U_{f} & = \frac{1}{2} \frac{q^{2}}{C_{f}} = \frac{q^{2}}{2 \times 10} = \frac{q^{2}}{20} \end{aligned}$

$∴$ Energy dissipated $= U_{i} - U_{f} = \frac{q^{2}}{5}$

This energy dissipated $= \frac{q^{2}}{5}$ is $80$ of the initial stored energy $= \frac{q^{2}}{4}$ .

$∴$ Correct option is (d).

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Electrostatics 5 Question 23

25. A 2μF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is

Answer:

Solution:

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25. A $2 μ F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 is