Question

A capacitor C₁ of capacitance 5 μF is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C₂ of capacitance 10 μF as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C₂ is ____ μC.

Fig.

Answer 1

Correct Answer: 100

To solve the problem, follow these steps: 

Initially, capacitor C₁ is charged to a potential of 30 V. The initial charge on C₁ is given by:

Q₁ = C₁ × V = 5 μF × 30 V = 150 μC

When the switch is closed, charge is conserved. The total charge in the system is shared between C₁ and C₂:

Total charge, Q_total = 150 μC

The potential difference across both capacitors at equilibrium is the same. Let V be this potential. Using charge conservation, we have:

(C₁ + C₂) × V = Q_total

V = 150 μC / (5 μF + 10 μF) = 150 μC / 15 μF = 10 V

Now, calculate the charge on C₂:

Q₂ = C₂ × V = 10 μF × 10 V = 100 μC

The charge on capacitor C₂ at equilibrium is 100 μC, which falls within the expected range of 100,100.

Thus, the charge on C₂ is 100 μC.

Answer 2

The correct answer is 100

Fig.

Let the charge q is flown in the circuit.
So using Kirchoff’s law
\(\frac{q}{10}=\frac{150−q}{5}\)
q = 100 μC