Question

As shown in figure two identical capacitors are connected to a battery of $V\text{volts}$ in parallel. When capacitors are fully charged, their stored energy is $U_1$. If the key $K$ is opened and material of dielectric constant ${\epsilon }_r=3$ is inserted in each capacitors their stored energy is now $U_2$. Then $\frac{U_1}{U_2}$ will be:

• A. $\frac{3}{5}$
• B. $\frac{5}{3}$
• C. $\frac{4}{5}$
• D. $\frac{5}{4}$

Answer 1

The correct option is A $\frac{3}{5}$


According to the question,
Initial total energy $U_1=U_A+U_B$
$=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2$
$U_1=C{V}^2...\left(i\right)$

Now switch is open, then energy of capacitor $A$,
$E_A=\frac{1}{2}KC{V}^2$ {$K$ = dielectric constant}
$E_A=\frac{1}{2}\times 3C{V}^2$
$E_A=\frac{3}{2}C{V}^2.....\left(ii\right)$

The energy of capacitor $B$,
$E_B=\frac{1}{2}KC(V^{\prime }{)}^2$
$(V^{\prime }=\frac{V}{K})$
$E_B=\frac{1}{2}KC{\left(\frac{V}{K}\right)}^2$
$E_B=\left(\frac{1}{2}C{V}^2\right).\frac{1}{3}....\left(iii\right)$

From equation (ii) and (iii):
$U_2=E_A+E_B$
$=(3+\frac{1}{3}).\frac{1}{2}C{V}^2$
$U_2=\frac{10}{3}.\frac{1}{2}C{V}^2...\left(iv\right)$

Compare equation (i) with (iv):
$\frac{U_1}{U_2}=\frac{3}{5}$