In the figure below, a potential difference V = 200 V is applied across a capaci

Question

In the figure below, a potential difference V = 200 V is applied across a capacitor arrangement with capacitances C1 = 11.0 µF, C2 = 7.00 µF, and C3 = 16.0 µF. Find the following values.

(a) the charge on capacitor 3
C

(b) the potential difference across capacitor 3
V

(c) the stored energy for capacitor 3
J

(d) the charge on capacitor 1
C

(e) the potential difference across capacitor 1
V

(f) the stored energy for capacitor 1
J

(g) the charge on capacitor 2
C

(h) the potential difference across capacitor 2
V

(i) the stored energy for capacitor 2
J

Explanation / Answer

C1 and C2 are in parallel, totaling 18 µF
in series with C3, that is 16*18/34 = 8.47 µF
Total charge is Q = CV = 1694 µC
This is on both C3, and the parallel combo, as they are in series.
Voltage across the parallel pair V = Q/C = 1694/18 = 94.1 volts, leaving 105.9 V on C3
Charge on C1 is Q = CV = 11 x 94.1 = 1035.1 µC
Charge on C2 is 7 x 94.1 = 658.7 µC

Now a) The charge on C3 = 1694 µC

(b) the potential difference across C3 = V3 = 105.9 V

(c) the stored energy for C3 = U3 = 0.5*C*V^2 = 0.5*16*10^-6*105.9^2 = 0.0897 J

d) The charge on C1 = 1035.1 µC

(e) the potential difference across C1 = V1 = 94.1 V

(f) the stored energy for C1 = U1 = 0.5*C*V^2 = 0.5*11*10^-6*94.1^2 = 0.0487 J

g) The charge on C2 = 658.7 µC

(h) the potential difference across C2 = V2 = 94.1 V

(i) the stored energy for C2 = U2 = 0.5*C*V^2 = 0.5*7*10^-6*94.1^2 = 0.03099 J

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