Question

As shown in the figure two identical capacitors are connected to a battery of V volts in parallel. When capacitors are fully charged, their stored energy is $U _ { 1 }$ . If the key K is opened and a material of dielectric constant $K = 3$ is inserted in each capacitor, their stored energy is now $U _ { 2 }$. $\frac { U _ { 1 } } { U _ { 2 } }$ will be

• A. $\frac { 3 } { 5 }$
• B. $\frac { 5 } { 3 }$
• C. 3
• D. $\frac { 1 } { 3 }$

Answer 1

Answer: (A)

$\frac { 3 } { 5 }$

Answer 2

Initially potential difference across both the capacitor is same hence energy of the system is

$U _ { 1 } = \frac { 1 } { 2 } C V ^ { 2 } + \frac { 1 } { 2 } C V ^ { 2 } = C V ^ { 2 }$ ……..(i)

In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is hence energy of the system now is

$= \frac { 10 } { 6 } C V ^ { 2 }$ …….(ii)

So, $\frac { U _ { 1 } } { U _ { 2 } } = \frac { 3 } { 5 }$