A parallel plate capacitor of capacitance 5 μ F is charged to 120 V and then connected to another uncharged capacitor. If the potential falls to 40 V, and capacitance of the second capacitor is

Question

A parallel plate capacitor of capacitance 5 μ F is charged to 120 V and then connected to another uncharged capacitor. If the potential falls to 40 V, and capacitance of the second capacitor is (A) 5 μ F (B) 10 μ F (C) 15 μ F (D) 20 μ F

Tardigrade Admin · Accepted Answer

Change, Q=C V Given, C=5 μ F=5 × 10-6 F , V=120 V Q=5 × 10-6 × 120 Also, 5 × 10-6 × 120=40 × C ⇒ C=(5 × 10-6 × 120/40) =15 × 10-6 F ⇒ C=15 μ F

Q. A parallel plate capacitor of capacitance $5 μ F$ is charged to $120 V$ and then connected to another uncharged capacitor. If the potential falls to $40 V,$ and capacitance of the second capacitor is

$5 μ F$

$10 μ F$

$15 μ F$

$20 μ F$

Change, $Q = C V$
Given, $C = 5 μ F = 5 \times 1 0^{- 6} F, V = 120 V$
$Q = 5 \times 1 0^{- 6} \times 120$
Also, $5 \times 1 0^{- 6} \times 120 = 40 \times C$
$\Rightarrow C = \frac{5 \times 1 0 ^{- 6} \times 120}{40}$
$= 15 \times 1 0^{- 6} F$
$\Rightarrow C = 15 μ F$

Q. A parallel plate capacitor of capacitance 5μF is charged to 120V and then connected to another uncharged capacitor. If the potential falls to 40V, and capacitance of the second capacitor is

5μF

10μF

15μF

20μF