Question

A 4.00 muF capacitor and a 6.00muF capacitor are connected in parallel across a 660 V supply line (a) Find the charge on each capacitor and the voltage across each. (b) The charged capacitors are disconnected from the line and from each other, and reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each.

Answer 1

The final charge is q = = 13.2×10^−3 C  and voltage across each capacitor is 132 V.

Explanation:

(a) Voltage "V" = 660 V across capacitors .

Now, q = CV for both

q1 = C1 V = 4 x 10^-6 x 660 V

q1 = 264 x 10^-5 C

q1 = 2.64 x 10^-3 C

q2 = C2V = 6 x 10^-6 x 660

q2 = 396 x 19^-5 C = 3.96 x 10^-3 C

(b) q net = q1+q2

= (3.96 − 2.64) × 10 k^3 C

= 13.2×10^−3 C

Now find the common potential

V = Total charge / Total capacity = 13.2 × 10^−3 / 10 × 10^−6

V = 132 V

Now apply q = CV for both capacitors.

q1 = C1V = 4 x 10^-6

q2 = C2V = 6 x 10^-6

q = q1 + q2 = (6 + 4) x 10^-6

q = 13.2 × 10^−3

q1 = 4 x 10^-6 x 132 = 528 x 10^-6 C

q2 = 6 x 10^-6 x 132 = 792 x 10^-6 C

Thus the final charge is q = = 13.2×10^−3 C  and voltage across each capacitor is 132 V.