Question

Hi everyone,
Three capacitors C1, C2, and C3 have capacitances of 3 F, 2F, and 5 F, respectively. They are connected in a circuit as shown.

Calculate:

a. The equivalent capacitance

b. The potential difference across each

capacitor

c. The charge on each capacitor

Answer 1

Given that,

Capacitor $C_{1}=3\ F$

Capacitor $C_{2}=2\ F$

Capacitor $C_{3}=5\ F$

We know that,

In series, the charge are same on all capacitors.

(a). We need to calculate the equivalent capacitance

Using formula of series

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

Put the value into the formula

$\dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{5}$

$C=0.98\ F$

(c). We need to calculate the charge on the capacitors

Using formula of charge

$Q = CV$

Put the value into the formula

$Q=0.98\times12$

$Q=11.76\ C$

(b). We need to calculate the potential difference across each capacitor

Using formula of potential

For first capacitor,

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{11.76}{3}$

$V_{1}=3.92\ V$

For second capacitor,

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

Put the value into the formula

$V_{2}=\dfrac{11.76}{2}$

$V_{2}=5.88\ V$

For third capacitor,

$V_{3}=\dfrac{Q}{C_{3}}$

Put the value into the formula

$V_{3}=\dfrac{11.76}{5}$

$V_{3}=2.352\ V$

Hence, (a). The equivalent capacitance is 0.98 F.

(b). The potential difference across each  capacitor is 3.92 V, 5.88 V and 2.352 V.

(c). The charge on each capacitor is 11.76 C.