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(a) Obtain the expression for the resultant capacitance when the three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected in (i) parallel and then (ii) in series.

(B) In the circuit shown in the figure, the charge on the capacitor of 4 $\mu F$ is 16 $\mu C$ . Calculate the energy stored in the capacitor of 12 $\mu F$ capacitance.

Answers (1)

(a) (i) Capacitors are connected in parallel

We know that,

Charge Q = CV

where C = Capacitance

V = Voltage

in parallel connection potential, the difference across all capacitors are the same. That means $V= V_{1}=V_{2}=V_{3}$

But the charge on each capacitance is not the same.

So, $Q_{1}=C_{1}V, Q_{2}= C_{2}V , Q_{3}=C_{3}V$

Therefore,

Total charge $Q= Q_{1}+Q_{2}+Q_{3}$

$CV = C_{1}V+C_{2}V+C_{3}V$

Total capacitance in parallel,

$C = C_{1}+C_{2}+C_{3}$

(ii) Capacitance is connected in series

In series connection charges on each capacitance are same.

That is $Q= Q_{1}+Q_{2}+Q_{3}$

But the potential drop on each capacitance is different. Total potential drop is the sum of each capacitance.

That is, $V= V_{1}+V_{2}+V_{3}$

$\frac{Q}{C}=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}}$

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$

Total capacitance in series,

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$

(b)

Given that,

Charge on the capacitor of

$4\mu F = 16 \mu C$

We know that,

Energy stored in the capacitor $=\frac{1}{2}CV^{2}=\frac{1}{2}QV$

Charge in the capacitor Q = CV

Therefore, voltage 4 $\mu F$ capacitor $V = \frac{Q}{C}$

$= \frac{16 \mu C}{4\mu F}= 4V$

In parallel connection potential difference across the capacitance is same.

So the voltage across $20\mu F$ capacitor = 4V

V = Potential difference across capacitors 12 $\mu F$ + potential difference across combination of capacitor

$12V= V +4V$

Voltage across 12 $\mu F$ capacitor = 8V

So energy stored in the capacitor of 12 $\mu F$ = $\frac{1}{2}\times 12\times 10^{-6}\times 8\times 8= 384\times 10^{-6}J$

Posted by

Safeer PP

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