JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 21)
A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected
in parallel with another uncharged capacitor of capacitance $${C \over 2}$$. The energy loss in the process
after the charge is distributed between the two capacitors is :
$${1 \over 2}CV_0^2$$
$${1 \over 4}CV_0^2$$
$${1 \over 3}CV_0^2$$
$${1 \over 6}CV_0^2$$
Explanation
Heat loss = $${1 \over 2}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)V_0^2$$
= $${1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2$$
= $${1 \over 6}CV_0^2$$
= $${1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2$$
= $${1 \over 6}CV_0^2$$
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