Question

A capacitor is connected to a 12V battery through a resistance of \[10\,\Omega \]. It is found that the potential difference across the capacitor rises to \[4.0\,{\text{V}}\] in \[1\,{\mu s}\]. Find the capacitance of the capacitor.
(A) \[\dfrac{{100}}{{\log _{\text{e}}^{1.5}}}\,{\mu F}\]
(B) \[\dfrac{{10}}{{\log _{\text{e}}^{1.5}}}\,{\mu F}\]
(C) \[\dfrac{1}{{\log _{\text{e}}^{1.5}}}\,{\mu F}\]
(D) \[\dfrac{{0.1}}{{\log _{\text{e}}^{1.5}}}\,{\mu F}\]

Answer 1

Hint: First of all, we will use the equation of charge at any instant. Then we will use the general formula of charge in a capacitor. We will equate both the equations. After modification we will substitute the required values followed by manipulation to obtain the desired result.

Complete step by step solution:
In the given question, we are supplied with the following data:
There is a capacitor, which is connected to a \[12\,{\text{V}}\] battery.
The capacitor is connected through a resistor of \[10\,\Omega \] .
The potential difference across the capacitor rises to \[4.0\,{\text{V}}\] in \[1\,{\mu s}\] .
We are asked to find the capacitance of the capacitor.
To begin with, we need to use the expression for the charge in a circuit containing capacitance. There is a magnitude of resistance given along with the time duration. We can use those quantities to find the capacitance.
We know,
\[1\,{\mu s} = {\text{1}} \times {\text{1}}{{\text{0}}^{ - {\text{6}}}}\,{\text{s}}\]
Let us proceed to solve the problem.
In a circuit containing capacitance, the equation of the charge contained in it is given by:
\[Q = {Q_0}\left( {1 - {{\text{e}}^{\dfrac{{ - t}}{{RC}}}}} \right)\] …… (1)
Where,
\[Q\] indicates the charge after an interval of time.
\[{Q_0}\] indicates the initial charge.
\[t\] indicates the time interval.
\[R\] indicates resistance of the resistor.
\[C\] indicates the capacitance of the capacitor.
Again, we know, the formula of charge in a capacitor is given by:
\[Q = CV\] …… (2)
Where,
\[Q\] indicates the charge in the capacitor.
\[C\] indicates the capacitance of the capacitor.
\[V\] indicates the potential difference across the capacitor.
Now, we use the equation (2) in equation (1), we get:
$Q = {Q_0}\left( {1 - {{\text{e}}^{\dfrac{{ - t}}{{RC}}}}} \right) \\
\Rightarrow CV = C{V_0}\left( {1 - {{\text{e}}^{\dfrac{{ - t}}{{RC}}}}} \right) \\
\Rightarrow V = {V_0}\left( {1 - {{\text{e}}^{\dfrac{{ - t}}{{RC}}}}} \right) \\
\Rightarrow 4 = 12\left( {1 - {{\text{e}}^{\dfrac{{ - 1 \times {{10}^{ - 6}}}}{{10C}}}}} \right) \\$
Again, we manipulate:
$\dfrac{1}{3} = 1 - {{\text{e}}^{\dfrac{{ - 1 \times {{10}^{ - 6}}}}{{10C}}}} \\
\Rightarrow {{\text{e}}^{\dfrac{{ - 1 \times {{10}^{ - 6}}}}{{10C}}}} = 1 - \dfrac{1}{3} \\
\Rightarrow {{\text{e}}^{\dfrac{{ - 1 \times {{10}^{ - 6}}}}{{10C}}}} = \dfrac{2}{3} \\
\Rightarrow {{\text{e}}^{\dfrac{{{{10}^{ - 6}}}}{{10C}}}} = \dfrac{3}{2} \\$
We simplify further:
${{\text{e}}^{\dfrac{{{{10}^{ - 6}}}}{{10C}}}} = 1.5 \\
\Rightarrow \ln \left( {{{\text{e}}^{\dfrac{{{{10}^{ - 6}}}}{{10C}}}}} \right) = \ln 1.5 \\
\Rightarrow \dfrac{{{{10}^{ - 6}}}}{{10C}} = \log _{\text{e}}^{1.5} \\
\Rightarrow C = \dfrac{{{{10}^{ - 6}}}}{{10\log _{\text{e}}^{1.5}}}\,{\text{F}} \\
\Rightarrow C = \dfrac{{0.1 \times {{10}^{ - 6}}}}{{\log _{\text{e}}^{1.5}}}\,{\text{F}} \\
\therefore C = \dfrac{{0.1}}{{\log _{\text{e}}^{1.5}}}\,{\mu F} \\$

Hence, the capacitance of the capacitor is \[\dfrac{{0.1}}{{\log _{\text{e}}^{1.5}}}\,{\mu F}\] .The correct option is (D).

Note:While solving this problem, \[12\,{\text{V}}\] is the emf of the battery which remains constant while \[4.0\,{\text{V}}\] is the potential difference after a definite interval of time. Don’t get confused with that. After the result is obtained in farads, don’t forget to convert it into microfarads, as the answer in the options is in microfarads.