Find the following. (In the figure, use C 1 = 21.80 ?F and C 2 = 15.80 ?F.) (a)

Question

Find the following. (In the figure, use

C1 = 21.80 ?F

and

C2 = 15.80 ?F.)

(a) the equivalent capacitance of the capacitors in the figure above
?F

(b) the charge on each capacitor

(c) the potential difference across each capacitor

on the right 21.80-?F capacitor     ?C on the left 21.80-?F capacitor     ?C on the 15.80-?F capacitor     ?C on the 6.00-?F capacitor     ?C Find the following. (In the figure, use C1 = 21.80 ?F and C2 = 15.80 ?F.) (a) the equivalent capacitance of the capacitors in the figure above ?F (b) the charge on each capacitor on the right 21.80-?F capacitor ?C on the left 21.80-?F capacitor ?C on the 15.80-?F capacitor ?C on the 6.00-?F capacitor ?C (c) the potential difference across each capacitor on the right 21.80-?F capacitor V on the left 21.80-?F capacitor V on the 15.80-?F capacitor V on the 6.00-?F capacitor

Explanation / Answer

Answer a). capacitor c1 is inseries with itself(c1) and capacitor c2 is in parallel with 6.0micro farads capacitor so their sum is       21.80+21.80=43.60,(say c3)    6.00+15.80=21.80(sayc4). Now capacitor c3 and c4 are in series so their capacitance is given as 1/c3+1/c4 which gives 14.53microF.

b)     Charge on each capacitor is given as

Q=C(capacitance in farad)*v(volts)

c)    Q=C*V

means on left and right of 21.80microF capacitor potential difference is 19.62*10-5/9=2.18*10-5volt

Across 15.80 potential difference is 14.22*10-5/9=1.58*10-5volt

Across 6.00 potential difference is 54*10-6/9=6.00*10-5volts

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