Question

2) a) How many 6.25-4F capacitors connected in parallel would it take to store a total charge of 1.05 mC if the potential difference across each capacitor is 10.5 V? (Round your answer to the nearest whole number.) b) If the capacitors in Part a) are discharged, connected in series, and then energized until the potential difference across each is equal to 10.5 V, find the charge on each capacitor and the potential difference across the connection.

Answer 1

Q=105 mc V = 10.5 volt. using Q=Gq LOS X163- 1005 = 100x10 Now, parallel combination formula, G=9+4+G + - +G. Cnc a n = n 100X10 – 16 Capacitors n=16 capacitors @ Equivalent capacitance in roins, t2 = I-16 a les 6-256156 Cop+ Gods = 0.391 MF Q=GV = Q = 4.102 4c charge on each capacitor. Now voltage on each Capacitor, 9:v 4.109 x 15 6-250166 V= 0.656 volt