Question

(a) Obtain a relation for equivalent capacitance of the series combination of capacitors. Draw a circuit diagram.
(b) $10$ capacitors each of capacity $10\mu F$ are joined first in series and then in parallel. Write the value of product of equivalent capacitances.
(c) What will be the value of capacitance of a $4\mu F$ capacitor if a dielectric of dielectric constant $2$ is inserted fully between the plates of parallel plate capacitor.

Answer 1

(a)

Capacitors in series connected to a voltage source is shown in the attached figure.

Charge across capacitors is equal. Hence, $Q=Q_1=Q_2$

By KVL, $V=V_1+V_2$

From definition of capacitance, $Q=CV,Q_1=Q=C_1{V}_1,Q_2=Q=C_2{V}_2$

where $C$ is equivalent capacitance

From above equations:

$\frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2}$

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

(b)

For capacitors is series, equivalent capacitance is:

$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+.....+\frac{1}{C_{10}}$

$\frac{1}{C_s}=\frac{1}{10}+\frac{1}{10}+....10times$

$C_s=1\mu F$

For capacitors is parallel, equivalent capacitance is:

$C_p=C_1+C_2+.....+C_{10}$

$=10+10+......10times$

$=100\mu F$

Product of equivalent capacitance is $C_s{C}_p=100\times {10}^{-6}\times 1\times {10}^{-6}={10}^{-10}{F}^2$

(c)

Capacitor of capacitance with a dielectric is:

$C=\frac{K{\epsilon }_{o}A}{d}$

$=2\times 4=8\mu F$