Question:
Two parallel plate capacitors, Capacitor A and Capacitor B, are connected in series to a voltage source of 50 V. Capacitor A has a capacitance of 2 μF, while Capacitor B has a capacitance of 4 μF. The capacitors have no initial charge. Calculate the potential difference across each capacitor and the total energy stored in the capacitors.
Solution:
Given:
First, we can find the equivalent capacitance (Ceq) of the two capacitors connected in series using the formula:
1/Ceq = 1/C1 + 1/C2
Substituting the values:
1/Ceq = 1/(2 × 10^-6 F) + 1/(4 × 10^-6 F)
1/Ceq = 3/(4 × 10^-6 F)
Ceq = (4 × 10^-6 F)/3
Ceq ≈ 1.333 μF
Using the formula: Q = Ceq × V, we can find the charge (Q) stored in the capacitors.
Q = (1.333 × 10^-6 F) × 50 V
Q ≈ 6.67 × 10^-5 C
Now, we can calculate the potential difference (V1) across Capacitor A:
V1 = Q/C1
V1 = (6.67 × 10^-5 C)/(2 × 10^-6 F)
V1 ≈ 33.33 V
Similarly, we can calculate the potential difference (V2) across Capacitor B:
V2 = Q/C2
V2 = (6.67 × 10^-5 C)/(4 × 10^-6 F)
V2 ≈ 16.67 V
Finally, we can find the total energy (U) stored in the capacitors using the formula:
U = (1/2) × C1 × V1^2 + (1/2) × C2 × V2^2
U = (1/2) × (2 × 10^-6 F) × (33.33 V)^2 + (1/2) × (4 × 10^-6 F) × (16.67 V)^2
U ≈ 1.11 × 10^-3 J
Therefore, the potential difference across Capacitor A is approximately 33.33 V, the potential difference across Capacitor B is approximately 16.67 V, and the total energy stored in the capacitors is approximately 1.11 × 10^-3 J.