Question

A capacitor of capacitance \[{{C}_{1}}=1\mu F\] withstand the maximum voltage \[{{V}_{1}}=6.0kV\] while the capacitor of capacitance \[{{C}_{2}}=2\mu F\], the maximum voltage \[{{V}_{2}}=4.0kV\]. What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?
$\begin{align}
  & \left( A \right)7 \\
 & \left( B \right)9 \\
 & \left( C \right)8 \\
 & \left( D \right)6 \\
\end{align}$

Answer 1

Hint: To solve this problem, first find the maximum charge capacitance of both the capacitors. Then use the formula for equivalent capacitance, substitute the values and find the equivalent capacitance of the system. Then, find the maximum charge that both the capacitors can withstand. Now, take the ratio of this maximum charge and the equivalent capacitance. This will give the voltage which the system of two capacitors can withstand if they are connected in series.

Formula used: $Q=CV$
${{V}_{\max }}=\dfrac{{{Q}_{\max }}}{{{C}_{R}}}$
Where:
$Q$- charge stored by capacitor
$C$- capacitance of capacitor
$V$- voltage
${{C}_{R}}$- resultant capacitance of capacitor

Complete step by step answer:
Given: ${C}_{1}= 1\mu F$
            ${V}_{1}= 6.0kV$
            ${C}_{2}= 2\mu F$
            ${V}_{2}= 4.0kV$
We know, capacitance of a conductor is given by,
$ C = \dfrac {Q}{V}$
Rearranging the above equation we get,
$Q = CV$
Maximum charge capacitance capacitor ${C}_{1}$ can withstand is given by,
$Q = {C}_{1}{V}_{1}$
Substituting the values in above equation we get,
$Q= 1 \times {10}^{-6} \times 6 \times {10}^{3}$
$\Rightarrow Q = 6000 \mu F$
Maximum charge capacitance capacitor ${C}_{2}$ can withstand is given by,
$Q = {C}_{2}{V}_{2}$
Substituting the values in above equation we get,
$Q= 2 \times {10}^{-6} \times 4 \times {10}^{3}$
$\Rightarrow Q = 8000 \mu F$
When the capacitors are connected in series, the resultant capacitance is given by,
$\dfrac {1}{{C}_{eq}} = \dfrac {1}{{C}_{1}} + \dfrac {1}{{C}_{2}}$
Substituting the values in above equation we get,
$\dfrac {1}{{C}_{eq}}= \dfrac {1}{1 \times {10}^{-6}} + \dfrac {1}{2 \times {10}^{-6}}$
$\Rightarrow \dfrac {1}{{C}_{eq}}= {10}^{-6} + 0.5 \times {10}^{-6}$
$\Rightarrow \dfrac {1}{{C}_{eq}}= 1.5 \times {10}^{-6}$
$\Rightarrow {C}_{eq} = 0.67 \times {10}^{6}$
Now, as we know, in series combination, charge on both the capacitors should be the same. Maximum charge that can be the same for both the capacitors is $6000 \mu F$ i.e. $ {C}_{1}{V}_{1}$.
So, the maximum charge that both the capacitors can withstand is,
${Q}_{max} = {C}_{1}{V}_{1}$
As ${C}_{1}{V}_{1} < {C}_{2}{V}_{2}$
$\Rightarrow {Q}_{max} = 6 \times {10}^{-3} C$
Maximum voltage that two capacitors cam withstand is given by,
${V}_{max}= \dfrac {{Q}_{max}}{{C}_{eq}}$
Substituting the values in above equation we get,
${V}_{max}= \dfrac {6 \times {10}^{-3}} {0.67 \times {10}^{6}}$
$\Rightarrow {V}_{max}= 9 \times {10}^{3} V$
$\Rightarrow {V}_{max}= 9kV$
Hence, the system of these two capacitors will withstand 9kV voltage.

So, the correct answer is “Option B”.

Note: Students must remember that when the capacitors are connected in series, the total capacitance is less than at least any one of the series capacitors individual capacitance. When capacitors are connected in parallel, the total capacitance is the sum of all the capacitors’ capacitances. Students should remember that the formula for total capacitance is not the same as that for total resistance. So students should not get confused between the formula for capacitance and resistance in series and capacitance and resistance in parallel.