JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 14)
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 :
_26th_July_Evening_Shift_en_14_1.png)
$$\frac{1}{10}$$
$$\frac{2}{5}$$
$$\frac{5}{13}$$
$$\frac{5}{26}$$
Explanation
$${E_1} = {1 \over 2}(2C){V^2}$$
$$ \Rightarrow {E_1} = C{V^2}$$ ...... (i)
$${E_2} = {1 \over 2}(5C){V^2} + {1 \over 2}{{{{(CV)}^2}} \over {5C}}$$
$$ = {{13} \over 5}C{V^2}$$
$$ \Rightarrow {{{E_1}} \over {{E_2}}} = {5 \over {13}}$$
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