A 2 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]

Question

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]

Smart Achievers · Accepted Answer

(4)

Initially and when switch S is turned to position -2, charge $(q)$ will remain constant.

Initially, energy stored

$U_{i} = \frac{1}{2} \frac{q^{2}}{C_{i}} = \frac{1}{2} \times \frac{q^{2}}{2} = \frac{q^{2}}{4}$

When switch S is turned to position -2, then energy stored

$U_{f} = \frac{1}{2} \frac{q^{2}}{C_{f}} = \frac{1}{2} \times \frac{q^{2}}{(2 + 8)} = \frac{q^{2}}{20}$

$∴$ $Energy dissipated, Δ U = U_{i} - U_{f} = \frac{q^{2}}{4} - \frac{q^{2}}{20} = \frac{q^{2}}{5}$

$∴$ $% of stored energy dissipated = \frac{Δ U}{U_{i}} \times 100 = \frac{\frac{q^{2}}{5}}{\frac{q^{2}}{4}} \times 100 = \frac{4}{5} \times 100 = 80 %$

Solution

Q.
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]

1 0%

2 20%

3 75%

4 80%

Want Us to Email you About Special Offers & Updates?