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Question: A mega ohm resistor and an uncharged \[1\,\mu {\text{F}}\] capacitor are connected in a single loop ...

A mega ohm resistor and an uncharged 1 μF1\,\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.163(1−e−1/3)e−1/3 μ⋅J/s\dfrac{{16}}{3}\left( {1 - {e^{ - 1/3}}} \right){e^{ - 1/3}}\,\mu \cdot {\text{J/s}}
B. 163(1−e−2/3) μ⋅J/s\dfrac{{16}}{3}\left( {1 - {e^{ - 2/3}}} \right)\,\mu \cdot {\text{J/s}}
C. 163e−2/3 μ⋅J/s\dfrac{{16}}{3}{e^{ - 2/3}}\,\mu \cdot {\text{J/s}}
D. None of these

Explanation

Solution

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 qq stored in capacitor is given by
q=CVq = CV …… (1)
Here, CC is the capacitance and VV is the potential difference.
The increasing charge on a capacitor is given by
q=q0(1−e−t/RC)q = {q_0}\left( {1 - {e^{ - t/RC}}} \right) …… (2)
Here, q0{q_0} is the initial charge, tt is the time, RR is the resistance and CC is the capacitance.
The energy EE stored in the capacitor is given by
E=q22CE = \dfrac{{{q^2}}}{{2C}} …… (3)
Here, qq is the charge and CC is the capacitance.

Complete step by step answer:
The capacitance of the capacitor is 1 μF1\,\mu {\text{F}} and potential difference across its plates is 4 V4\,{\text{V}}.
C=1 μFC = 1\,\mu {\text{F}}
V=4 VV = 4\,{\text{V}}

Determine the initial charge q0{q_0} stored on the capacitor.

Substitute 1 μF1\,\mu {\text{F}} for CC and 4 V4\,{\text{V}} for VV in equation (1).
q0=(1 μF)(4 V){q_0} = \left( {1\,\mu {\text{F}}} \right)\left( {4\,{\text{V}}} \right)
⇒q0=4 μC\Rightarrow {q_0} = 4\,\mu {\text{C}}
⇒q0=4×10−6 C\Rightarrow {q_0} = 4 \times {10^{ - 6}}\,{\text{C}}

Hence, the initial charge on the capacitor is 4×10−6 C4 \times {10^{ - 6}}\,{\text{C}}.

Determine the equation for the charge on the capacitor for time tt.

Substitute 4×10−6 C4 \times {10^{ - 6}}\,{\text{C}} for q0{q_0} in equation (2).
q=(4×10−6 C)(1−e−t/RC)q = \left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right)
⇒q=(4×10−6 C)(1−e−t/RC)\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 tt.

Determine the energy stored by the capacitor.

Substitute (4×10−6 C)(1−e−t/RC)\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right) for qq in equation (3).
E=[(4×10−6 C)(1−e−t/RC)]22CE = \dfrac{{{{\left[ {\left( {4 \times {{10}^{ - 6}}\,{\text{C}}} \right)\left( {1 - {e^{ - t/RC}}} \right)} \right]}^2}}}{{2C}}
⇒E=(4×10−6 C)22C(1−e−t/RC)2\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 tt.
dEdt=ddt[(4×10−6 C)22C(1−e−t/RC)2]\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]
⇒dEdt=(4×10−6 C)22C2(1−e−t/RC)ddt(1−e−t/RC)\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)
⇒dEdt=(4×10−6 C)22C2(1−e−t/RC)(0−e−t/RC(−1RC))\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)
⇒dEdt=(4×10−6 C)2RC2(1−e−t/RC)e−t/RC\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 s1\,{\text{s}} for tt, 1×10−6 F1 \times {10^{ - 6}}\,{\text{F}} for CC and 1×106 Ω1 \times {10^6}\,\Omega for RR in the above equation.
⇒dEdt=(4×10−6 C)2(1×106 Ω)(1×10−6 F)2(1−e−(1 s)/(1×106 Ω)(1×10−6 F))e−1 s/(1×106 Ω)(1×10−6 F)\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)}}
⇒dEdt=16×10−6(1−e−1)e−1\Rightarrow \dfrac{{dE}}{{dt}} = 16 \times {10^{ - 6}}\left( {1 - {e^{ - 1}}} \right){e^{ - 1}}
⇒dEdt=16(1−e−1)e−1 μ⋅J/s\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(1−e−1)e−1 μ⋅J/s16\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.