Question

A series combination of two capacitances of value \[0.1\,\mu F\] and \[1\,\mu F\] is connected with a source of voltage 500 volts. The potential difference in volts across the capacitor of value \[0.1\,\mu F\] will be:
A. 50
B. 500
C. 45.5
D. 454.5

Answer 1

Hint: In series circuit, the charge through the circuit does not change. Use the formula for voltage across the capacitor in terms of capacitance and charge to express the charge across both capacitors. In a series circuit, the total potential difference is the potential difference across each component.

Formula used:
\[Q = CV\]
Here, Q is the charge, C is capacitance and V is the voltage across the capacitor.

Complete step by step answer:
We have given that two capacitors are connected in series. We know that, according to the law of conservation of charge, in series connection the charges remain conserved. Therefore, the charge across both capacitors is the same.

We know the charge flowing through the capacitor is,
\[Q = CV\]

Here, Q is the charge and V is the voltage across the capacitor.

We assume the first capacitor is \[{C_1} = 0.1\,\mu F\] and the second capacitor is \[{C_2} = 1\,\mu F\].

Therefore, we can write,
\[Q = {C_1}{V_1} = {C_2}{V_2}\]
\[ \Rightarrow {V_2} = \dfrac{{{C_1}{V_1}}}{{{C_2}}}\] …… (1)

This is the potential difference across the\[1\,\mu F\] capacitor.

Now, we know that the total potential difference in the circuit is the sum of potential differences across \[{C_1}\] and \[{C_2}\].
Therefore, we can write,
\[V = {V_1} + {V_2}\]

Here, \[{V_1}\] is the potential difference across \[{C_1}\].

We can use equation (1) to rewrite the above equation.
\[V = {V_1} + \dfrac{{{C_1}{V_1}}}{{{C_2}}}\]
\[ \Rightarrow V = {V_1}\left( {1 + \dfrac{{{C_1}}}{{{C_2}}}} \right)\]
\[ \Rightarrow V = {V_1}\left( {\dfrac{{{C_2} + {C_1}}}{{{C_2}}}} \right)\]
\[ \Rightarrow {V_1} = \dfrac{{{C_2}V}}{{{C_1} + {C_2}}}\]

Substitute \[1\,\mu F\] for \[{C_2}\], 500 V for V and \[0.1\,\mu F\] for \[{C_1}\] in the above equation.
\[ \Rightarrow {V_1} = \dfrac{{\left( 1 \right)\left( {500} \right)}}{{0.1 + 1}}\]
\[ \Rightarrow {V_1} = 454.5\,V\]

So, the correct answer is “Option D”.

Note:
To answer such types of questions, the key is to remember voltage, charge and current in series and parallel circuits. In a parallel circuit, the voltage across the capacitor remains constant but charge does not remain conserved. In a series circuit, the voltage drops at each component but the charge remains conserved.