Question

How would you connectged 8, 12 and 24muF capacitors to obtain (i) minimum capacitance (ii) maximum capacitance ? If a potential difference of 100 volt is applied across the system, what would be the charges on the capacitors in each case?

Answer 1

Answer:

Explanation:

(i) for minimum, connect the capacitors in series.

Thus, 1/Ceq = 1/8 + 1/12 + 1/24

thus, Ceq = 4 μF

then formula for finding charge = Ceq*V

therefore, charge = 4*100 = 400μC

(ii)  for maximum capacitance,

connect the capacitors in parallel

thus, Ceq = 8 + 12 + 24 = 44μF

thus, charge = 100*44 = 4400μC

HOPE THIS HELPS!!