Question

A capacitor ${C_1} = 1.0\,{\mu F}$ can withstand a maximum voltage ${V_1} = 6.0\,{\text{kV}}$ . Another capacitor of capacitance ${C_2} = 2.0\,{\mu F}$ can withstand a maximum voltage of ${V_2} = 4.0\,{\text{kV}}$ . If the capacitors are connected in series, the combination can withstand a maximum voltage of:
(A) $10\,{\text{kV}}$
(B) $9\,{\text{kV}}$
(C) $8\,{\text{kV}}$
(D) $6\,{\text{kV}}$

Answer 1

First of all, we will find out the charge stored by the capacitors individually. Then we will choose the magnitude of charge which is in common. After that we will find the maximum voltage just by adding the individual voltage which is followed by manipulation.

Complete step by step solution:
In the given question, we are supplied with the following data:
There is a capacitor which can withstand a voltage of $6.0\,{\text{kV}}$ . The capacity of the capacitor is $1.0\,{\mu F}$ .There is another capacitor which can withstand a voltage of $4.0\,{\text{kV}}$ . The capacity of the capacitor is $2.0\,{\mu F}$ .The two capacitors are connected in series with one another.We are asked to find out the maximum voltage that the two capacitors can withstand when both of them are connected in series.
To begin with, we will discuss a few things: each capacitor has its own capacity to hold maximum charge for the given conditions of voltage and its capacity. Again, if a number of capacitors are connected in series, the total capacitance is somewhat less than either one of the series capacitor’s capacity.Now, we will find the charge present in the capacitors for the given capacity and the applied voltage.
For the first capacitor:
The formula which gives the charge of a capacitor is given by:
${Q_1} = {C_1}{V_1}$ …… (1)
Where,
${Q_1}$ indicates the amount of charge present in the capacitor.
${C_1}$ indicates the capacity of the capacitor.
${V_1}$ indicates the voltage applied across the capacitor.
Now, we substitute the required values in the equation (1), we get:
${Q_1} = {C_1}{V_1} \\\ \Rightarrow {Q_1} = 1 \times {10^{ - 6}} \times 6 \times {10^3} \\\ \Rightarrow {Q_1} = 6 \times {10^{ - 3}}\,{\text{C}}$
Therefore, the charge on the first capacitor is $6 \times {10^{ - 3}}\,{\text{C}}$ .
Again, we will calculate the charge on the second capacitor.
${Q_2} = {C_2}{V_2} \\\ \Rightarrow {Q_2} = 2 \times {10^{ - 6}} \times 4 \times {10^3} \\\ \Rightarrow {Q_2} = 8 \times {10^{ - 3}}\,{\text{C}}$
The charge on the second capacitor is found to be $8 \times {10^{ - 3}}\,{\text{C}}$ .
We know, when we connect a number of capacitors in series, the amount of current flowing through the capacitors are equal, and hence the charge remains the same.
Charge present on both the capacitors is $6 \times {10^{ - 3}}\,{\text{C}}$ .
So, the combined voltage $\left( V \right)$ can be written as the sum of the individual voltage of the two capacitors with charge $6 \times {10^{ - 3}}\,{\text{C}}$ in common.
$V = \dfrac{{6 \times {{10}^{ - 3}}}}{{1 \times {{10}^{ - 6}}}} + \dfrac{{6 \times {{10}^{ - 3}}}}{{2 \times {{10}^{ - 6}}}} \\\ \Rightarrow V = 6 \times {10^3} + 3 \times {10^3} \\\ \Rightarrow V = 9 \times {10^3}\,{\text{V}}$
Hence, the maximum voltage that the two capacitors can withstand is $9 \times {10^3}\,{\text{V}}$ i.e. $9\,{\text{kV}}$ .

The correct option is (B).

Note: In the given problem, always remember that if a number of capacitors are connected in a circuit in parallel, then the net capacity of the circuit increases while in series connection the net capacity decreases. Charge depends on the voltage across the capacitor. Higher the values of the voltage are, higher is the charge stored in it.