Question

In the given network, calculate (i) equivalent capacitance between A and B, (ii) if a voltage of 15
is applied between A and B calculate the PD between the plates of 6 uF capacitance.​

Answer 1

To Find :

• Equivalent Capacitance between A and B
• If voltage of 15V is applied between A and B calculate the Potential Difference between the plates of 6μF

Formula Used :

We know that in capacitor -

• Parallel Combination -

$\bullet\underline{\boxed{\sf C_{eq}=C_1+C_2.....so\: on}}$

• Series Combination -

$\bullet\underline{\boxed{\sf \dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}.....so \; on }}$

Solution :

(i) Equivalent Capacitance between A and B

• 2μF and 3μF are connected in series -

$\implies{\sf \dfrac{1}{C'}=\dfrac{1}{2}+\dfrac{1}{3} }$

$\implies{\sf \dfrac{3+2}{2\times 3} }$

$\implies{\sf \dfrac{5}{6}\mu F }$

$\implies{\bf C' = \dfrac{6}{5}\mu F }$

• Now , C' and 1.8μF are in parallel combination -

$\implies{\sf C'' = C' + 1.8 }$

$\implies{\sf C'' = \dfrac{6}{5}+1.8 }$

$\implies{\bf C'' = 3\mu F }$

• Now , C" and 6μF are in series combination

$\implies{\sf \dfrac{1}{C'''}=\dfrac{1}{C''}+\dfrac{1}{6} }$

$\implies{\sf \dfrac{1}{3}+\dfrac{1}{6} }$

$\implies{\sf \dfrac{6+3}{18}}$

$\implies{\sf \dfrac{9}{18}}$

$\implies{\sf C'''=\dfrac{18}{9} }$

$\implies{\bf C'''=2\mu F }$

(ii) Potential Difference between plates of 6μF Capacity -

We know that ,

$\implies{\sf Q \propto V }$

$\implies{\bf Q = CV }$

Q = Charge , C = Capacitance , V = Voltage

Now ,

$\implies{\sf Q = C''' \times 15}$

$\implies{\sf Q = 2 \times 15 }$

$\implies{\bf Q = 30\mu C }$

Potential Difference between plate 6μF

$\implies{\sf 30 = 6 \times V }$

$\implies{\sf V = \dfrac{30}{6} }$

$\implies{\bf V = 5\: V}$

Answer :

(i) Equivalent Capacitance between A and B is 2μF

(ii) Potential Difference between 6μF is 5V