In the given circuit switch K is open. The charge on the capacitor is ' C ' in steady state is q 1. Now key is closed and steady state charge on C is q 2 . The ratio of charges q 1 / q 2 is-

Question

In the given circuit switch K is open. The charge on the capacitor is ' C ' in steady state is q 1. Now key is closed and steady state charge on C is q 2 . The ratio of charges q 1 / q 2 is- <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/84d553b4c3ebbd993f4e4bcf581c62a1-.png" /> (A) 3 / 2 (B) 2 / 3 (C) 1 (D) 1 / 2

Tardigrade Admin · Accepted Answer

Initially when switch is open

Finally when switch is closed

I =( E /3 R ); ∴ V A - V B = I 2 R =( E / 3 R ) × 2 R =2 E / 3 ∴ Charge on capacitor = CV ; q 2=( C 2 E /3) ∴ ( q 1/ q 2)=( CE /2 / 3 CE ) =3: 2

Q. In the given circuit switch $K$ is open. The charge on the capacitor is ' $C$ ' in steady state is $q_{1}$ . Now key is closed and steady state charge on $C$ is $q_{2} .$ The ratio of charges $q_{1} / q_{2}$ is-

$3/2$

$2/3$

$1$

$1/2$

Initially when switch is open

Finally when switch is closed

$I = \frac{E}{3 R}$ ;
$∴ V_{A} - V_{B} = I 2 R = (E /3 R) \times 2 R = 2 E /3$
$∴$ Charge on capacitor $= C V; q_{2} = \frac{C 2 E}{3}$
$∴ \frac{q _{1}}{q _{2}} = \frac{CE}{2/3 CE}$ $= 3 : 2$

Q. In the given circuit switch K is open. The charge on the capacitor is ' C ' in steady state is q1​. Now key is closed and steady state charge on C is q2​. The ratio of charges q1​/q2​ is-

3/2

2/3

1

1/2

Initially when switch is open Finally when switch is closed I=3RE​; ∴VA​−VB​=I2R=(E/3R)×2R=2E/3 ∴ Charge on capacitor =CV;q2​=3C2E​ ∴q2​q1​​=2/3CECE​ =3:2

Q. In the given circuit switch $K$ is open. The charge on the capacitor is ' $C$ ' in steady state is $q_{1}$ . Now key is closed and steady state charge on $C$ is $q_{2} .$ The ratio of charges $q_{1} / q_{2}$ is-

$3/2$

$2/3$

$1$

$1/2$

Initially when switch is open

Finally when switch is closed

$I = \frac{E}{3 R}$ ;
$∴ V_{A} - V_{B} = I 2 R = (E /3 R) \times 2 R = 2 E /3$
$∴$ Charge on capacitor $= C V; q_{2} = \frac{C 2 E}{3}$
$∴ \frac{q _{1}}{q _{2}} = \frac{CE}{2/3 CE}$ $= 3 : 2$