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A 0.50-μF and a 1.4-μF capacitor (C1 and C2, respectively) are connected in series to a 6.0-V battery.

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

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

Answer :

adioabiola

The potential to difference across each capacitor assuming the two capacitors are in parallel are;

  • p.d(0.50-μF) = 4.42V
  • p.d(1.4-μF) = 1.58V

The charge on each capacitor assuming the two capacitors are in parallel are:

  • Q(0.5-μF) = 0.58C

  • Q(1.4-μF) = 1.64C

Capacitors in series and parallel connections

C) The potential difference, p.d for the capacitors connected in series are inversely proportional to the capacitance and are as follows;

  • p.d(0.50-μF) = (1.4)/(1.9) × 6 = 4.42V

  • p.d(1.4-μF) = (0.5)/(1.9) × 6 = 1.58V

D) When the connection is parallel, the charge on each capacitor is shared as follows;

First, Total charge, Q = C(total) × V.

where, C(total) = (0.5×1.4)/(1.9) = 0.37-μF.

  • Q = 0.37 × 6

Q = 2.22C

Hence, the charge is shared as follows;

  • Q(0.5-μF) = (0.5/1.9) × 2.22 = 0.58C

  • Q(1.4-μF) = (1.4/1.9) × 2.22 = 1.64C

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