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 E₁. 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 E₂. The ratio E₁/E₂ will be: (a) 1/10 (b) 2/5 (c) 5/13 (d) 5/26

Answer: (c). K closed: both C in parallel across V: E₁ = 1/2(2C)V² = CV². K opened: left capacitor stays on the battery; with dielectric 5C at V: U = 1/2(5C)V² = 2.5CV². Right capacitor is isolated with q = CV; with dielectric: U = (q²)/(2(5C)) = 0.1CV². E₂ = 2.6CV², so (E₁)/(E₂) = 1/2.6 = 5/13. So the answer is (c) 5/13.

A source of potential difference V is connected to the combination of… | JEE Main 2022 (26 Jul, Shift 2)

Q55JEE Main 2022 · 26 JUL · SHIFT 2Medium

Combination of Capacitors

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:

Show answer and solution

Answer: (c)

$K$ closed: both $C$ in parallel across $V$ : $E_1 = \frac{1}{2}(2C)V^2 = CV^2$ .
$K$ opened: left capacitor stays on the battery; with dielectric $5C$ at $V$ : $U = \frac{1}{2}(5C)V^2 = 2.5CV^2$ .
Right capacitor is isolated with $q = CV$ ; with dielectric: $U = \frac{q^2}{2(5C)} = 0.1CV^2$ .
$E_2 = 2.6CV^2$ , so $\frac{E_1}{E_2} = \frac{1}{2.6} = \frac{5}{13}$ .
So the answer is (c) $\frac{5}{13}$ .

Report a mistake

Combination of Capacitors question from JEE Main 2022 (26 Jul, Shift 2)

More questions on Combination of Capacitors