Question

A mega ohm resistor and an uncharged \[1\,\mu {\text{F}}\] capacitor are connected in a single loop circuit with a constant source of 4 volt. At one second after the connection is made what are the rates at which energy is being stored in the capacitor:
A.\[\dfrac{{16}}{3}\left( {1 - {e^{ - 1/3}}} \right){e^{ - 1/3}}\,\mu \cdot {\text{J/s}}\]
B. \[\dfrac{{16}}{3}\left( {1 - {e^{ - 2/3}}} \right)\,\mu \cdot {\text{J/s}}\]
C. \[\dfrac{{16}}{3}{e^{ - 2/3}}\,\mu \cdot {\text{J/s}}\]
D. None of these

Answer 1

Hint: Use the formulae for the charge in the capacitor, the increasing charge in the capacitor and energy stored in the capacitor. First determine the initial charge on the capacitor then deduce the equation for the charge on capacitor at time t and finally determine the equation for rate of energy being stored in the capacitor.

Formula used:
The charge \[q\] stored in capacitor is given by
\[q = CV\] …… (1)
Here, \[C\] is the capacitance and \[V\] is the potential difference.
The increasing charge on a capacitor is given by
\[q = {q_0}\left( {1 - {e^{ - t/RC}}} \right)\] …… (2)
Here, \[{q_0}\] is the initial charge, \[t\] is the time, \[R\] is the resistance and \[C\] is the capacitance.
The energy \[E\] stored in the capacitor is given by
\[E = \dfrac{{{q^2}}}{{2C}}\] …… (3)
Here, \[q\] is the charge and \[C\] is the capacitance.

Complete step by step answer:
The capacitance of the capacitor is \[1\,\mu {\text{F}}\] and potential difference across its plates is \[4\,{\text{V}}\].
\[C = 1\,\mu {\text{F}}\]
\[V = 4\,{\text{V}}\]

Determine the initial charge \[{q_0}\] stored on the capacitor.

Substitute \[1\,\mu {\text{F}}\] for \[C\] and \[4\,{\text{V}}\] for \[V\] in equation (1).
\[{q_0} = \left( {1\,\mu {\text{F}}} \right)\left( {4\,{\text{V}}} \right)\]
\[ \Rightarrow {q_0} = 4\,\mu {\text{C}}\]
\[ \Rightarrow {q_0} = 4 \times {10^{ - 6}}\,{\text{C}}\]

Hence, the initial charge on the capacitor is \[4 \times {10^{ - 6}}\,{\text{C}}\].

Determine the equation for the charge on the capacitor for time \[t\].

Substitute \[4 \times {10^{ - 6}}\,{\text{C}}\] for \[{q_0}\] in equation (2).
\[q = \left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right)\]
\[ \Rightarrow q = \left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right)\]

This is the expression for the charge on the capacitor at time \[t\].

Determine the energy stored by the capacitor.

Substitute \[\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right)\] for \[q\] in equation (3).
\[E = \dfrac{{{{\left[ {\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right)} \right]}^2}}}{{2C}}\]
\[ \Rightarrow E = \dfrac{{{{\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{2C}}{\left( {1 - {e^{ - t/RC}}} \right)^2}\]

Determine the rate at which energy is stored in the capacitor.

Differentiate the above equation with respect to time \[t\].
\[\dfrac{{dE}}{{dt}} = \dfrac{d}{{dt}}\left[ {\dfrac{{{{\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{2C}}{{\left( {1 - {e^{ - t/RC}}} \right)}^2}} \right]\]
\[ \Rightarrow \dfrac{{dE}}{{dt}} = \dfrac{{{{\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{2C}}2\left( {1 - {e^{ - t/RC}}} \right)\dfrac{d}{{dt}}\left( {1 - {e^{ - t/RC}}} \right)\]
\[ \Rightarrow \dfrac{{dE}}{{dt}} = \dfrac{{{{\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{2C}}2\left( {1 - {e^{ - t/RC}}} \right)\left( {0 - {e^{ - t/RC}}\left( { - \dfrac{1}{{RC}}} \right)} \right)\]
\[ \Rightarrow \dfrac{{dE}}{{dt}} = \dfrac{{{{\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{R{C^2}}}\left( {1 - {e^{ - t/RC}}} \right){e^{ - t/RC}}\]

Substitute \[1\,{\text{s}}\] for \[t\], \[1 \times {10^{ - 6}}\,{\text{F}}\] for \[C\] and \[1 \times {10^6}\,\Omega \] for \[R\] in the above equation.
\[ \Rightarrow \dfrac{{dE}}{{dt}} = \dfrac{{{{\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{\left( {1 \times {{10}^6}\,\Omega } \right){{\left( {1 \times {{10}^{ - 6}}\,{\text{F}}} \right)}^2}}}\left( {1 - {e^{ - \left( {1\,{\text{s}}} \right)/\left( {1 \times {{10}^6}\,\Omega } \right)\left( {1 \times {{10}^{ - 6}}\,{\text{F}}} \right)}}} \right){e^{ - 1\,{\text{s}}/\left( {1 \times {{10}^6}\,\Omega } \right)\left( {1 \times {{10}^{ - 6}}\,{\text{F}}} \right)}}\]
\[ \Rightarrow \dfrac{{dE}}{{dt}} = 16 \times {10^{ - 6}}\left( {1 - {e^{ - 1}}} \right){e^{ - 1}}\]
\[ \Rightarrow \dfrac{{dE}}{{dt}} = 16\left( {1 - {e^{ - 1}}} \right){e^{ - 1}}\,\mu \cdot {\text{J/s}}\]

Therefore, the rate at which energy is being stored in the capacitor is \[16\left( {1 - {e^{ - 1}}} \right){e^{ - 1}}\,\mu \cdot {\text{J/s}}\].

So, the correct answer is “Option D”.

Note:
The initial charge is less than the charge stored by the capacitor initially. Convert the units of all physical quantities in the SI system of units while determining the expression for energy storing rate of capacitor.