Question

A 11.9 μ F capacitor and a 18 μ F capacitor are connected in parallel across the terminals of a 6.0 V battery.

1)What is the equivalent capacitance of this combination?

2)What is the potential difference across the 11.9μF capacitor?

3)What is the potential difference across the 18μF capacitor?

4)What is the charge on the 11.9 μF capacitor?

5)What is the charge on the 18 μF capacitor?

6)Find the energy stored in the 11.9 μF capacitor.

7)Find the energy stored in the 18 μF capacitor.

Answer 1

C1=11.9 μ F

C2= 18 μ F

V=6 V

when Capacitors are in parallel the net C = C1+C2

1) C equivalent = 29.9 μ F

2) the voltage remain same in parallel combination,

hence the potential difference across the 11.9μF capacitor = 6 V

3) the potential difference across the 18μF capacitor = 6V

4) charge = CV = 11.9 μF*6V = 71.4 μC

5)the charge on the 18 μF capacitor = 18 μF*6V =108 μC

6) energy stored = U= 1/2*C*V²

energy stored in the 11.9 μF capacitor = 1/2*11.9*6²=214.2 μJoules = 0.214*10^-3J

7) the energy stored in the 18 μF capacitor = 1/2*18*6²=324 μJoules = 0.324*10^-3J