The circuit shows two capacitors A and B of c | vedclass

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

Question

(D) $1$. Initially,the capacitors are in series. The equivalent capacitance is $C_{eq} = \frac{C \cdot 2C}{C + 2C} = \frac{2C}{3}$.$2$. The charge on

Vedclass · Accepted Answer

$c$ alone is correct

Vedclass · Answer

(D) $1$. Initially,the capacitors are in series. The equivalent capacitance is $C_{eq} = \frac{C \cdot 2C}{C + 2C} = \frac{2C}{3}$.$2$. The charge on each capacitor when fully charged is $Q = C_{eq}E = \frac{2CE}{3}$.$3$. When the cell is removed and plates of opposite polarities are connected,the net charge becomes $Q_{net} = Q_B - Q_A = \frac{2CE}{3} - \frac{2CE}{3} = 0$. Since the net charge is zero,the final charge on each capacitor will be zero.$4$. Statement $(a)$ and $(b)$ are incorrect as the final charge is zero.$5$. Initial energy $U_i = \frac{1}{2} C_{eq} E^2 = \frac{1}{2} (\frac{2C}{3}) E^2 = \frac{CE^2}{3}$.$6$. Final energy $U_f = 0$ (since $Q=0$).$7$. Loss of energy $\Delta U = U_i - U_f = \frac{CE^2}{3}$. Thus,statement $(c)$ is correct.

Similar Questions

Two capacitors $C_1$ and $C_2$ are connected in parallel. If a charge $Q$ is given to the combination,the charge gets shared. Then the ratio of charge on $C_1$ to charge on $C_2$ is . . . . . . .

$A$ capacitor of capacity $C_1$ is charged up to $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. On bringing them in contact,there is:

$A$ $10\,\mu F$ capacitor is charged to a potential difference of $50\;V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20\;V$. The capacitance of the second capacitor is....$\mu F$

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module