A capacitor C is fully charged with voltage V | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Initial charge on capacitor $C$ is $Q = CV_{0}$.Initial energy stored in the capacitor is $U_{i} = \frac{1}{2}CV_{0}^{2}$.When connected in parall

Vedclass · Accepted Answer

$0.166$

Vedclass · Answer

(A) Initial charge on capacitor $C$ is $Q = CV_{0}$.Initial energy stored in the capacitor is $U_{i} = \frac{1}{2}CV_{0}^{2}$.When connected in parallel,the common potential $V$ is given by $V = \frac{Q_{total}}{C_{total}} = \frac{CV_{0}}{C + C/2} = \frac{CV_{0}}{3C/2} = \frac{2}{3}V_{0}$.The final energy stored in the system is $U_{f} = \frac{1}{2}(C + C/2)V^{2} = \frac{1}{2}(\frac{3C}{2})(\frac{2}{3}V_{0})^{2} = \frac{1}{2}(\frac{3C}{2})(\frac{4}{9}V_{0}^{2}) = \frac{1}{3}CV_{0}^{2}$.The energy loss is $\Delta U = U_{i} - U_{f} = \frac{1}{2}CV_{0}^{2} - \frac{1}{3}CV_{0}^{2} = \frac{1}{6}CV_{0}^{2}$.Since $\frac{1}{6} \approx 0.166$,the correct option is $A$.

Similar Questions

$A$ capacitor of $10\,\mu F$ charged up to $250\,V$ is connected in parallel with another capacitor of $5\,\mu F$ charged up to $100\,V$. The common potential is.....$V$.

Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

$A$ capacitor having capacitance $C$ is charged to a voltage $V$. It is then removed and connected in parallel with another identical capacitor which is uncharged. The new charge on each capacitor is now

$A$ 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 the $4\,\mu F$ capacitor is.......$mJ$.

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module