Two capacitors of equal capacitance (C_1 = C_ | vedclass

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

Question

(D) When the switch $S$ is closed,the two capacitors $C_1$ and $C_2$ are connected in parallel with the battery.Since they are in parallel,the potenti

Vedclass · Accepted Answer

$U_0 = U_1 + U_2$

Vedclass · Answer

(D) When the switch $S$ is closed,the two capacitors $C_1$ and $C_2$ are connected in parallel with the battery.Since they are in parallel,the potential difference across both capacitors must be equal,so $V_1 = V_2$. This makes option $C$ correct.Given $C_1 = C_2$ and $V_1 = V_2$,the charges on the capacitors must be equal,$Q_1 = C_1 V_1 = C_2 V_2 = Q_2$. This makes option $B$ correct.Since $Q_1 = Q_2$,we have $\frac{1}{2}(Q_1 + Q_2) = \frac{1}{2}(Q_1 + Q_1) = Q_1$. Since the total charge is conserved and distributed between the two capacitors,$Q_0 = Q_1 + Q_2$ is not necessarily true in the context of a battery being present. However,if we consider the final state,$Q_0$ (initial charge) is not equal to $\frac{1}{2}(Q_1 + Q_2)$.Regarding energy,the initial energy is $U_0 = \frac{1}{2} C_1 V^2$. After closing the switch,the final energy is $U_1 + U_2 = \frac{1}{2} C_1 V^2 + \frac{1}{2} C_2 V^2 = C_1 V^2$. Thus,$U_0 
eq U_1 + U_2$. The statement $U_0 = U_1 + U_2$ is $INCORRECT$.

Similar Questions

To establish an instantaneous current of $2\,A$ through a $1\,\mu F$ capacitor,the potential difference across the capacitor plates should be changed at the rate of:

Initially $n$ identical capacitors are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then

$A$ soap bubble of radius $R$ and surface tension $T$ is formed in a vacuum. It is slowly charged so that it slowly expands. It stops charging when the radius becomes $2R$. Find the amount of charge given to the bubble.

Vedclass Test Series

Exam Paper Generator

Online Exam Module