Two capacitors C 1 = 6 µF and C 2 = 24 µF are independently charged to voltages

Question

Two capacitors C1 = 6 µF and C2 = 24 µF are independently charged to voltages V1 = 18 V and V2 = 54 V respectively. The capacitors are connected in parallel.

Once equilibrium is reached, what are the magnitudes of the charges on the two capacitors now?
Q1 =
Q2 =

What is the potential difference across each capacitor now?

Two capacitors C1 = 6 mu F and C2 = 24 mu F are independently charged to voltages V1 = 18 V and V2 = 54 V respectively. The capacitors are connected in parallel. Once equilibrium is reached, what are the magnitudes of the charges on the two capacitors now? Q1 = Q2 = What is the potential difference across each capacitor now?

Explanation / Answer

First, lets find the charge of each capacitor before they are connected.

C = q/v for each individually,

6e-6 F = q1 / 18 V therefore q1 = 1.08e-4 C

24e-6 F = q2 / 54 V therefore q2 = .0013 C

Notice when we connect them, the negative of one capacitor is in contact with the positive of another, so when they are connected, the total charge will redistribute into equilibrium, but the total charge becomes:

.0013 C MINUS 1.08e-4 C = .001188 C

Now, with the new total charge, we find the new equilivalent total capacitance. In parallel, capacitors simply add, so 24 + 6 = 30 F

C = q/v

30e-6 F = .001188 C / v

v = 39.6 V. FOR BOTH since they are in parallel. This is the answer to the second part. Now, the first part:

c = q/v for each individual capacitor:

6e-6 F = q1 / 39.6 V therefore q1 = 2.376e-4 C

24e-6 F = q2 / 39.6 V therefore q2 = 9.504e-4 C

Notice the addition of these two charges equals the total, which was .001188 C

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