Question

A capacitor of capacity $2\mu F$ is charged to $100V$. What is the heat generated when this capacitor is connected in parallel to an another capacitor of same capacity?

• A. $2.5mJ$
• B. $5.0mJ$
• C. $10mJ$
• D. $4mJ$

Answer 1

Answer: (B)

$5.0mJ$

Initial charge is $Q_i=CV$
Let $V_c$ be the common potential when another capacitance C is connected in parallel.
now the final charge is $Q_f=(C+C)V_c=2C{V}_c$
as total charge is conserved, $Q_i=Q_f\Rightarrow CV=2C{V}_c$
or $V_c=V/2$
generated heat $=$ potential energy loss $=U_i-U_f$
$=\frac{1}{2}C{V}^2-\frac{1}{2}(C+C)V_c^2=\frac{1}{2}C{V}^2-\frac{1}{2}\left(2C\right)(V/2{)}^2$
$=\frac{1}{4}C{V}^2=\frac{1}{4}\times 2\times {10}^{-6}\times (100{)}^2=5\times {10}^{-3}J=5mJ$