Question

consider the circuit shown below. The capacitors have the following capacitances: C1=6

3. Consider the circuit shown below. The capacitors have the following capacitances: C1-6 μF, C2 = 3μF and Δ V-20V. Ci is charged by closing the switch Si. Switch Sı is then opened and the charged capacitor is connected to the uncharged capacitor by closing S2. (ぺ s, (a) Calculate the initial charge on Cr (b) After opening Si and closing S2, how does the potential difference across C compare with C2? Why is this the case? (c) Calculate the final charge on each capacitor.

Answer 1

a)

charge stored by capacitor is given as

Q₁ = C₁ $\Delta$ V

Q₁ = 6 x 20

Q₁ = 120 uC

b)

the Potential on C₁ and C₂ will becomes equal by redistribution of charge since C₁ and C₂ are connected in Parallel and in parallel electric potential remains same

c)

Total Capacitance = C₁ + C₂ = 6 + 3 = 9 uF

Common Potential

V_c = Total charge / Total capacitance = Q / (C₁ + C₂) = 120 / 9 = 13.33 volts

charge on C₁ = q₁ = C₁V_c = 6 x 13.33 = 79.98

charge on C₂ = q₂ = C₂V_c = 3 x 13.33 = 39.99