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

Choose the correct answer: