Question #166543

1.      4.2 A potential difference of 75V is applied to a combination of 1.25microFarad and 0.6microfarad capacitor connected in series (a) what is the charge on each capacitor? (b) what is the potential difference across the 1.25microFaradcapacitor


1
Expert's answer
2021-02-28T07:24:14-0500

Given:

V=75VV=75\:\rm V

C1=1.25μF,  C2=0.6μFC_1=\rm 1.25\:\mu F,\; C_2=\rm 0.6\:\mu F

(a) The total capacitance

C=C1C2C1+C2=1.250.61.25+0.6=0.41μFC=\frac{C_1C_2}{C_1+C_2}=\frac{1.25*0.6}{1.25+0.6}=0.41\:\rm\mu F

The total charge

q=CV=0.4175=30.4μCq=CV=0.41*75=30.4\:\rm \mu C

For the series connection of capacitors we have

q1=q2=qq_1=q_2=q

(b)

V1=q1C1=30.41.25=24.3VV_1=\frac{q_1}{C_1}=\frac{30.4}{1.25}=24.3\:\rm V

V2=q2C2=30.40.6=50.7VV_2=\frac{q_2}{C_2}=\frac{30.4}{0.6}=50.7\:\rm V

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