A capacitor, is charged to a potential difference of 5 V using a battery. The battery is reoved and another capacitor, is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equil

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Q.
A capacitor, $C_{1} = 6 μ F$ is charged to a potential difference of $V_{0} =$ 5 V using a 5 V battery. The battery is removed and another capacitor, $C_{2} = 12 μ F$ is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges ( $q_{1}$ and $q_{2}$ ) on the capacitors $C_{1}$ and $C_{2}$ when equilibrium condition is reached [2025]

1 $q_{1} = 15 μ C, q_{2} = 30 μ C$

2 $q_{1} = 30 μ C, q_{2} = 15 μ C$

3 $q_{1} = 10 μ C, q_{2} = 20 μ C$

4 $q_{1} = 20 μ C, q_{2} = 10 μ C$

Ans.
(3)

Initially Finally

$q_{1}^{'} = 6 \times 5 = 30 μ C$
$6 V_{C} + 12 V_{C} = 30 + 0$

$18 V_{C} = 30$

$\Rightarrow V_{C} = \frac{30}{18} = \frac{5}{3} Volt$

$\Rightarrow q_{1} = \frac{6 \times 5}{3} = 10 μ C$

and $q_{2} = \frac{12 \times 5}{3} = 20 μ C$

Solution

1 $q_{1} = 15 μ C, q_{2} = 30 μ C$

2 $q_{1} = 30 μ C, q_{2} = 15 μ C$

3 $q_{1} = 10 μ C, q_{2} = 20 μ C$

4 $q_{1} = 20 μ C, q_{2} = 10 μ C$

Ans.
(3)

Initially Finally

$q_{1}^{'} = 6 \times 5 = 30 μ C$
$6 V_{C} + 12 V_{C} = 30 + 0$

$18 V_{C} = 30$

$\Rightarrow V_{C} = \frac{30}{18} = \frac{5}{3} Volt$

$\Rightarrow q_{1} = \frac{6 \times 5}{3} = 10 μ C$

and $q_{2} = \frac{12 \times 5}{3} = 20 μ C$

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Solution

1 q1=15 μC, q2=30 μC

2 q1=30 μC, q2=15 μC

3 q1=10 μC, q2=20 μC

4 q1=20 μC, q2=10 μC

Ans. (3) Initially Finally q1'=6×5=30 μC 6VC+12VC=30+0 18VC=30 ⇒ VC=3018=53 Volt ⇒ q1=6×53=10 μC and q2=12×53=20 μC

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1 $q_{1} = 15 μ C, q_{2} = 30 μ C$

2 $q_{1} = 30 μ C, q_{2} = 15 μ C$

3 $q_{1} = 10 μ C, q_{2} = 20 μ C$

4 $q_{1} = 20 μ C, q_{2} = 10 μ C$

Ans.
(3)

Initially Finally

$q_{1}^{'} = 6 \times 5 = 30 μ C$
$6 V_{C} + 12 V_{C} = 30 + 0$

$18 V_{C} = 30$

$\Rightarrow V_{C} = \frac{30}{18} = \frac{5}{3} Volt$

$\Rightarrow q_{1} = \frac{6 \times 5}{3} = 10 μ C$

and $q_{2} = \frac{12 \times 5}{3} = 20 μ C$