Question 639: Electrostatics & Capacitance

Question

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

Accepted Answer

Initially, the energy stored in 2 $$\mu $$F capacitor is
$${U_i} = {1 \over 2}C{V^2} = {1 \over 2}\left( {2 	imes {{10}^{ - 6}}} ight){V^2} = {V^2} 	imes {10^{ - 6}}J$$
Initially, the charge stored in 2 $$\mu $$F capacitor is
Q
i
= CV = (2 ÃÂ 10
Ã¢ÂÂ6
)V = 2V ÃÂ 10
Ã¢ÂÂ6
coulomb. When switch S is turned to position 2, the charge flows and both the capacitors share charges till a common potential V
C
is reached.
$${V_C} = {{{m{total }}\,{\mathop{m charge}
olimits} } \over {total\,capacitance}}$$
$$ = {{2V 	imes {{10}^{ - 6}}} \over {\left( {2 + 8} ight) 	imes {{10}^{ - 6}}}} = {V \over 5}volt$$
Finally, the energy stored in both the capacitors
$${U_f} = {1 \over 2}\left[ {\left( {2 + 8} ight) 	imes {{10}^{ - 6}}} ight]\left( {{V \over 5}} ight)$$
$$ = {{{V^2}} \over 5} 	imes {10^{ - 6}}J$$
% loss of energy, $$\Delta U = {{{U_i} - {U_f}} \over {{U_i}}} 	imes 100\% $$
$$ = {{\left( {{V^2} - {V^2}/5} ight) 	imes {{10}^{ - 6}}} \over {{V^2} 	imes {{10}^{ - 6}}}} 	imes 100\%  = 80\% $$

Answer

75%

Answer

80%

Answer

0%

Answer

20%

Electrostatics & Capacitance

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

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