Question

A capacitor of 1 mF withstands a maximum voltage 6KV while another capacitor 2 mF withstands a maximum voltage of 4KV. If the capacitors are connected in series, the system will withstand a maximum voltage of
A. 2 KV
B. 4 KV
C. 6 KV
D. 9 KV

Answer 1

Hint:On connecting the two capacitors in series, the charge on each capacitor becomes equal. The maximum charge on the capacitor combination is equal to charge on the first capacitor. Determine the equivalent capacitance of the series combination of two capacitors. Use the formula for the potential difference across the capacitor.

Formula used:
Charge, \[Q = CV\]
where, C is the capacitance and V is the voltage.

Complete step by step answer:
We know that the charge remains conserved in the series combination. Therefore, when we connect two capacitors of different capacitors in series, the charge on both the capacitors becomes equal.
We have, the maximum charge on the first capacitor is,
\[{Q_1} = {C_1}{V_1}\]
Here, \[{C_1}\] is the capacitance of the first capacitor and \[{V_1}\] is the maximum voltage that the first capacitor can withstand.

When we connect \[{C_1}\] and \[{C_2}\] in series, the maximum charge on the combination that it can withstand will be the charge on \[{C_1}\], since, numerically \[{C_1}{V_1} < {C_2}{V_2}\].
Therefore, the maximum charge on the combination of two capacitors will be,
\[{Q_{\max }} = {C_1}{V_1}\]
Substituting \[{C_1} = 1\,{\text{mF}}\] and \[{V_1} = 6\,{\text{KV}}\] in the above equation, we get,
\[{Q_{\max }} = \left( {1 \times {{10}^{ - 3}}\,{\text{F}}} \right)\left( {6 \times {{10}^3}\,{\text{V}}} \right)\]
\[ \Rightarrow {Q_{\max }} = 6\,{\text{C}}\]

We have the formula for the equivalent capacitance of the series combination of capacitors,
\[\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\]
\[ \Rightarrow {C_{eq}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}\]
Substituting \[{C_1} = 1\,{\text{mF}}\] and \[{C_2} = 2\,{\text{mF}}\] in the above equation, we get,
\[{C_{eq}} = \dfrac{{\left( 1 \right)\left( 2 \right)}}{{1 + 2}}\]
\[ \Rightarrow {C_{eq}} = \dfrac{2}{3}\,{\text{mF}}\]

Now, the maximum voltage that the combination of capacitors can withstand is,
\[{V_{\max }} = \dfrac{{{q_{\max }}}}{{{C_{eq}}}}\]
Substituting \[{Q_{\max }} = 6\,{\text{C}}\] and \[{C_{eq}} = \dfrac{2}{3}\,{\text{mF}}\] in the above equation, we get,
\[{V_{\max }} = \dfrac{6}{{\dfrac{2}{3}}}\]
\[ \Rightarrow {V_{\max }} = \dfrac{{18}}{2}\]
\[ \therefore {V_{\max }} = 9\,{\text{V}}\]
Therefore, the maximum voltage that the two series capacitors can withstand is 9 V.

So, the correct answer is option D.

Note:When the capacitors are connected in series, the charge on each capacitor becomes equal regardless of the capacitance since the current has to pass through each capacitor. In this case, the charge on each capacitor will be equal to the charge on the first capacitor in the path of current. Do not get confused between the formulae for equivalent capacitance and equivalent resistance since the formulae are quite similar but are actually opposite for series combination and parallel combination.