Use the exact values you enter to make later calculations. Two capacitors C 1 =

Question

Use the exact values you enter to make later calculations.

Two capacitors C1 = 5 µF and C2 = 20 µF are independently charged to voltages V1 = 24 V and V2 = 72 V respectively. (When entering units, use micro for the metric system prefix µ.)

(a) The capacitors are then connected in parallel as shown in the diagram below.

(i) Once equilibrium is reached, what are the magnitudes of the charges on the two capacitors?

(ii) What is the potential difference across each capacitor?

(b) Instead of connecting the positive plate of C1 to the positive plate of C2 the two capacitors are connected as shown below. Note that the positive plate of one is connected to the negative plate of the other.

(i) Once equilibrium is reached, what are the magnitudes of the charges on the two capacitors now?

(ii) What is the potential difference across each capacitor now?

Explanation / Answer

C1 = 5 µF and C2 = 20 µF, V1 = 24 V, V2 = 72 V

(a) Q = CV

Q_total = C1V1 +C2V2

Q_total = (5*24) +(20*72) = 1560 uC

Now, the capacitors are connected in parallel, so the total charge is shared by the two; the equivalent capacitance is

C_total = 5+20 = 25 uF

V = Q/C = 1560/25= 62.4 V

in parallel combination potential is constant

Q1 = V*C1 = 62.4 *5 = 312 uC

Q2 = V*C2 = 62.4*20 = 1248 uC

(ii) V1 = V2 = 62.4 V

(b) (i) in series combination

V_total = V1+V2 = 24+72 = 96 V

C-total = C1C2/C1+C2 = (5*20)/(5+20)

C_total = 4 uF

Q_total = V*C = 96*4 = 384 uC

In series combination charge is constant

Q1 = Q2 = 384 uC

(ii) V1 = Q1/C1 = 384/5 = 76.8 V

V2 = 384/20 =19.2 V

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