Question 698: Electrostatics & Capacitance

Question

A $$2$$  $$\mu F$$ 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

Accepted Answer

When switch S is connected to terminal 1, the potential difference across the 2 $$\mu$$F capacitor is V volt. Therefore, energy stored in the system is
$${U_1} = {1 \over 2}{C_1}{V^2} = {1 \over 2} 	imes 2 	imes {V^2}$$
$$ = {V^2}\,\mu J$$
When switch S is turned to terminal 2, the charge will flow from 2 $$\mu$$F capacitor to 8 $$\mu$$F capacitor until their potentials are equalized. The common potential is
$${V^2} = {q \over {{C_1} + {C_2}}} = {{{C_1}V} \over {{C_1} + {C_2}}}$$
$$ = {{2V} \over {(2 + 8)}} = {V \over 5}$$ volt
$$	herefore$$ Energy stored in the system now will be
$${U_2} = {1 \over 2}({C_1} + {C_2})V_2^2$$
$$ = {1 \over 2}(2 + 8) 	imes {\left( {{V \over 5}} ight)^2} = {{{V^2}} \over 5}\mu J$$
$$	herefore$$ Percentage loss of energy is
$${{{U_1} - {U_2}} \over {{U_1}}} 	imes 100 = {{\left( {{V^2} - {{{V^2}} \over 5}} ight)} \over {{V^2}}} 	imes 100 = 80\% $$

Answer

$$0\% $$

Answer

$$20\% $$

Answer

$$75\% $$

Answer

$$80\% $$

Electrostatics & Capacitance

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

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