Answer to Question #166543 in Electricity and Magnetism for JaJay

Question

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

Answer 1

Given:

$V=75\:\rm V$

$C_1=\rm 1.25\:\mu F,\; C_2=\rm 0.6\:\mu F$

(a) The total capacitance

$$C=\frac{C_1C_2}{C_1+C_2}=\frac{1.25*0.6}{1.25+0.6}=0.41\:\rm\mu F$$

The total charge

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

For the series connection of capacitors we have

$$q_1=q_2=q$$

(b)

$$V_1=\frac{q_1}{C_1}=\frac{30.4}{1.25}=24.3\:\rm V$$

$$V_2=\frac{q_2}{C_2}=\frac{30.4}{0.6}=50.7\:\rm V$$