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In a network, four capacitors $C_{1}$ , $C_{2}$ , $C_{3}$ and $C_{4}$ are connected as shown in the figure.

(a) Calculate the net capacitance in the circuit.

(b) If the charge on the capacitor $C_{1}$ is $6\mu C,$

(i) calculate the charge on the capacitors $C_{3}$ and $C_{4}$ , and

(ii) net energy stored in the capacitors $C_{3}$ and $C_{4}$ connected in series.

Answers (1)

The given capacitors are :

$C_{1}=3\; \mu F$ ____Connetced in parallel

$C_{2}=6\; \mu F$

$C_{3}=4\; \mu F$

$C_{4}=12\; \mu F$ _____connected in series

(a) Let the capacitance across $C_{2}$ and $C_{1}$ is ;

$C_{12}=6+3=9\mu F$

the capacitance across $C_{3}$ and $C_{4}$ is ;

$C_{34}=\frac{12\times 4}{16}=3\mu F$

Hence, the net capacitance would be ;

$C_{net}=\frac{9\times 3}{12}=\frac{9}{4}\mu F$

(b) The total charge flowing through the circuit

$C_{net}V=\frac{9}{4}\times8=18\mu C$

(ii) Since, $Q=18\mu C$

and $C_{34}=3\mu F$

Then energy $E_{34}=\frac{1}{2}\frac{Q^{2}}{C_{34}}$

$E_{34}=\frac{1}{2}\times \frac{(18\times 10^{-6})^{2}}{3\times 10^{-6}}$

$E_{34}=54\times 10^{-6}Joule$

Posted by

Safeer PP

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