Question

1) Three capacitors each of capacitance $9pE$ are connected in series.
A) What is the total capacitance of the combination?

2) Three capacitors each of capacitance $9pF$ are connected in series.
B) What is the potential difference across each capacitor if the combination is connected to a $120V$ supply?

Answer 1

1) Given, $C=9pF$

For series-combination:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

As $C_1=C_2=C_3=C$


Putting values in equation (1)

$\frac{1}{C_{eq}}=\frac{3}{C}$

$C_{eq}=\frac{C}{3}$ ...(2)

Putting values in $C$ in equation (2)

$C_{eq}=\frac{9pF}{3}$

$=3pF$

2) Charge flown by the battery:

$q=C_{eq}V$

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$q=\left[\frac{C}{3}\right]\left[V\right]=\frac{CV}{3}$

As in series combination charge flown through all the capacitor is same.

Potential-difference across capacitor:

$V^{\prime }=\frac{q}{C_1}$

As $q\&C$ are same, hence potential drop across each capacitor is same i.e., $[V_1=V_2=V_3=V^{\prime }]$

$V^{\prime }=\frac{\left[\frac{CV}{3}\right]}{C}$

$=\frac{V}{3}=\frac{120V}{3}=40V$