Question

A capacitor of capacitance $1\mu F$is discharging through a resistance. Initial potential difference across the capacitor is 12V. At t=0.1 sec, potential difference across the capacitor becomes 6V. Find the value of resistance R.
(A) $1.443 \times {10^4}\Omega $
(B) $2.886 \times {10^4}\Omega $
(C) $1.443 \times {10^5}\Omega $
(D) $1.443 \times {10^6}\Omega $

Answer 1

Hint:Capacitor is a device used to store electrical energy i.e. it stores charges. A charged capacitor with a potential difference V when connected in series with a resistance produces a complete circuit like that with a cell/battery. It loses its charged particles in the form of current and slowly the potential difference V comes to zero.

Formula used:
\[
\Rightarrow\dfrac{{dQ}}{{dt}}R = - \dfrac{Q}{C} \\
\Rightarrow Q = {Q_o}\exp ( - t/RC) \\
\Rightarrow Q = CV \\
\Rightarrow V = {V_o}\exp ( - t/RC) \\
\]

Complete step by step answer:
 When a capacitor is connected to a cell/battery, charge starts to build up on both the metallic faces. When the potential difference between them becomes equal to the potential of the cell/battery, the flow of current stops.
Now this capacitor charged with potential difference V when connected to a resistance in series and the circuit is closed, a current starts to flow. This current flows till the capacitor is fully discharged. At any time t, the potential difference across the resistor R is equal to the potential difference across the capacitor of capacitance C. Capacitance is defined as the charge stored Q per unit potential difference V. So applying Kirchhoff Law,
\[
\Rightarrow C = \dfrac{Q}{V} \Rightarrow V = \dfrac{Q}{C} \\
\Rightarrow IR + \dfrac{Q}{C} = 0 \\
\]
And current is nothing but the charge flowing per unit time,
\[
\Rightarrow \dfrac{{dQ}}{{dt}}R = - \dfrac{Q}{C} \\
\Rightarrow\dfrac{{dQ}}{Q} = - \dfrac{{dt}}{{RC}} \\
\]
On integrating the above equation, we have,
\[
\Rightarrow\int {\dfrac{{dQ}}{Q}} = - \dfrac{1}{{RC}}\int {dt} \\
\Rightarrow\ln Q = - \dfrac{t}{{RC}} + \ln c \\
\]
Where ln c in the constant of integration. At time t=0, the capacitor is fully charged so,
\[
\Rightarrow t = 0 \\
\Rightarrow Q = {Q_o} \\
\Rightarrow\ln {Q_o} = \ln c \\
\Rightarrow \ln Q = - \dfrac{t}{{RC}} + \ln {Q_o} \\
\Rightarrow Q = {Q_o}\exp \left( { - \dfrac{t}{{RC}}} \right) \\
\]
Since capacitance is constant so, by the relation Q=CV we have,
\[\Rightarrow V = {V_o}\exp \left( { - \dfrac{t}{{RC}}} \right)\]
Now substituting the data from the question we have,
\[
\Rightarrow 6 = 12\exp \left( { - \dfrac{{0.1}}{{{{10}^{ - 6}}R}}} \right) \\
\Rightarrow \ln 2 - \dfrac{1}{{{{10}^{ - 5}}R}} = 0 \\
\Rightarrow R = \dfrac{{{{10}^5}}}{{\ln 2}} = 1.443 \times {10^5}\Omega \\
\]
The correct answer is option C.

Note:The formula has been derived specifically to not get the students confused. Most of the time, in a hurry, students use the formula $Q = {Q_o}(1 - \exp ( - t/RC))$ which is wrong in this context. This formula is when the capacitor is charged while in the question the capacitor is being discharged. You can always check the formula for the correct situation by substituting boundary conditions i.e. t= zero and t=infinity.