A source of potential difference V is connected to the combinatio… - Vidyadip

MCQ

A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :

A
$\frac{1}{10}$
B
$\frac{2}{5}$
✓
$\frac{5}{13}$
D
$\frac{5}{26}$

✓

Answer

Correct option: C.

$\frac{5}{13}$

c
$(1)$ Switch is closed

$C _{ eq }=2 C$

Energy $E_{1}=\frac{1}{2} C_{\text {oq }} V^{2}$

$=\frac{1}{2} 2 C ^{2} \times V ^{2}$

$E _{1}= CV ^{2}$

$(ii)$ When switch is opened charge on right capacitor remain CV while potential on left capacitor remain same

Dielectric $K =5$

$C ^{\prime}= KC$

$C ^{\prime}=5\,C$

$E _{2}=\frac{1}{2}(5 C ) V ^{2}+\frac{( CV )^{2}}{2(5 C )}$

$E _{2}=\frac{5 CV ^{2}}{2}+\frac{ CV ^{2}}{10}$

$E _{2}=\frac{13 CV ^{2}}{5}$

$\frac{ E _{1}}{ E _{2}}=\frac{ CV ^{2}}{\frac{13 CV ^{2}}{5}}=\frac{5}{13}$

$\frac{ E _{1}}{ E _{2}}=\frac{5}{13}$

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