Question

calculate the potential difference and the energy stored in the capacitor C2 in the circuit.given potential at A is 90v,C1=20microF,C2=30 micro F,C3=15microF.

the circuit is in series with a at one end and the capacitors connecetd in series
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Answer 1

Answer: $V_{2} = 20 V$ and $E = 6\times10^{-3}\ J$

Explanation:

Given that,

Potential at A $V_{1} = 90\ V$

Capacitor $C_{1} = 20\mu F$

Capacitor $C_{2} = 30\ \mu F$

Capacitor $C_{3} = 15\ \mu F$

Now, all capacitors joined in series

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

$\dfrac{1}{C} = \dfrac{1}{20\mu F} +\dfrac{1}{30\mu F} +\dfrac{1}{15\mu F}$

$C = \dfrac{60}{9}\mu F$

Now, the charge is

$q = CV$

$q = \dfrac{60}{9}\times10^{-6}\ F\times 90\ V$

$q = 600\times10^{-6} C$

Now, the charge will be same in$C_{2}$

So, the potential

$V_{2} = \dfrac{q}{C_{2}}$

$V_{2} = \dfrac{600\times10^{-6}\ C}{30\times10^{-6}\ F}$

$V_{2} = 20 V$

Now, the energy stored is

$W = \dfrac{q^{2}}{2C_{2}}$

$W = \dfrac{(600\times10^{-6}\ C)^{2}}{2\times 30\times10^{-6}\ F}$

$W = 6\times10^{-3}\ J$

Hence, the energy stored is $6\times10^{-3}\ J$

Answer 2

1/Cs = 1/C1 + 1/C2 + 1/C3 + 1/Cn

C1 = 20 uF

C2 = 30 uF

C3 = 15 uF

Cs = 20/3 uF

Q = CV

V = 90 V

Q2 = 600 C

V = Q/C

V2 = 20 V

U = (1/2)CV^2

U2 = 6000 J.