Two identical condensers are joined as shown | vedclass

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

Question

(A) When the switch $S$ is closed,both capacitors are in parallel with the battery of voltage $V$. The equivalent capacitance is $C_{eq} = C + C = 2C$

Vedclass · Accepted Answer

$3: 1$

Vedclass · Answer

(A) When the switch $S$ is closed,both capacitors are in parallel with the battery of voltage $V$. The equivalent capacitance is $C_{eq} = C + C = 2C$. The total energy stored is $U_1 = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (2C) V^2 = CV^2$.When the switch $S$ is opened,the battery is disconnected. The charge on each capacitor remains $Q = CV$. Now,both capacitors are filled with a dielectric of constant $K = 3$. The new capacitance of each capacitor becomes $C' = KC = 3C$.The total energy of the system is $U_2 = \frac{Q^2}{2C'} + \frac{Q^2}{2C'} = \frac{Q^2}{C'} = \frac{(CV)^2}{3C} = \frac{C^2 V^2}{3C} = \frac{CV^2}{3}$.Therefore,the ratio $\frac{U_1}{U_2} = \frac{CV^2}{CV^2 / 3} = 3: 1$.

Similar Questions

$A$ parallel plate capacitor has a capacitance of $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $K$ of the oil is:

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?

The function of a dielectric in a capacitor is

Vedclass Test Series

Exam Paper Generator

Online Exam Module