Question

Consider the following. (Let C1 = 42.40 µF and C2 = 36.40 µF.)

A rectangular circuit contains a battery and four capacitors. The bottom side has a 9.00 V battery with the positive terminal on the left. The left and right sides of the circuit each contain a capacitor labeled C1. The top side splits into two parallel horizontal branches, which recombine before reaching the top right corner. There is a 6.00 µFcapacitor on the upper branch and a capacitor labeled C2 on the lower branch.

(a) Find the equivalent capacitance of the capacitors in the figure.
µF

(b) Find the charge on each capacitor.
on the right 42.40 µF capacitor
on the left 42.40 µF capacitor
on the 36.40 µF capacitor
on the 6.00 µF capacitor


(c) Find the potential difference across each capacitor.
on the right 42.40 µF capacitor
on the left 42.40 µF capacitor
on the 36.40 µF capacitor
on the 6.00 µF capacitor

Answer 1

Step-by-step explanation:

Resultant capacitance of parallel capacitor

= 6µF. + 36.4 µF.

= 42.4 µF.

Thus three capacitance each of value 42.40 is attached with voltage source of 9 v . So 9V will be equally distributed among them .

voltage over each capacitor = 3 V .

charge on left capacitor = C V ,C is capacitance , V is volt

= 42.4 x 3 = 127.2 µC

charge on right capacitor = C V

= 42.4 x 3 = 127.2 µC

potential over each of parallel capacitor = 3 V

potential over 6 µF = 3 V

potential over 36.4 µF = 3 V

charge on 6 µF

= 6 x 3 = 18 µC

charge on 36.4 µF

= 36.4 x 3 = 109.2 µC.