Question

In: Physics

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

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.

Expert Solution

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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genius_generous answered 1 year ago

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