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2. Electric Potential and Capacitance
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A body of capacity $4\,\mu \,F$ is charged to $80\,V$ and another body of capacity $6\,\mu \,F$ is charged to $30\,V$. When they are connected the energy lost by $4\,\mu \,F$ capacitor is.......$mJ$
A
$7.8$
B
$4.6$
C
$3.2$
D
$2.5$
Solution
(a) Initial energy of body of capacitance $4 \,µF$ is ${U_i} = \frac{1}{2} \times (4 \times {10^{ – 6}})\,{(80)^2} = 0.0128\,J$
Final potential on this body after connection is $V = \frac{{4 \times 80 + 6 \times 30}}{{4 + 6}} = 50\,V.$ So final energy on it
${U_f} = \frac{1}{2} \times 4 \times {10^{ – 6}}{(50)^2} = 0.005\,J$
Energy lost by this body = $U_i -U_f = 7.8 \,mJ$
Standard 12
Physics
