Question

A 0.50-μF and a 1.4-μF capacitor (C1 and C2, respectively) are connected in series to a 22-V battery.

Part A: Calculate the potential difference across each capacitor.

Part B: Calculate the charge on each capacitor.

Part C: Calculate the potential difference across each capacitor assuming the two capacitors are in parallel.

Part D: Calculate the charge on each capacitor assuming the two capacitors are in parallel.

Answer 1

Because they are in series, each capacitor has the same charge Q. We know that generally Q = CV so Q/C = V

Therefore:

Q/C1 + Q/C2 = 22 volts

=> QC1+ QC2 = 22C1C2 = 1.54e-11

=> Q(C1+C2) = 1.54e-11

=> Q = 1.54e-11/(C1+C2) = 8.11e-6 C

A) Calculate the potential difference across each capacitor.

Vc1 = Q/C1 = 8.11e-6/0.5e-6 = 16.21V <-----

Vc2 = Q/C2 = 8.11e-6/1.4e-6 = 5.79V <-----

B) Calculate the charge on each capacitor. Q = 1.54e-11/(C1+C2) = 8.11e-6 C <-----

C) Calculate the potential difference across each capacitor assuming the two capacitors are in parallel.

Since they are in parallel, they have the same voltage V and the charge Q is distributed accordingly

C1V+C2V = Q1+Q2 = Q

=> V = Q/(C1+C2) = 8.11E-6/[(1.4+0.5)*1e-6] = 4.27V <-----

D) Calculate the charge on each capacitor assuming the two capacitors are in parallel.

Q1 = C1*V = 0.5e-6*4.27V = 2.13e-6 C <-----

Q2 = C2*V = 1.4e-6*4.27V = 5.98e-5 C <-----

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