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Question: For an E-R-C series circuit \( E = 10volt \) ; \( C = 1mF \) ; \( R = 100\Omega \) then pick correct...

For an E-R-C series circuit E=10voltE = 10volt ; C=1mFC = 1mF ; R=100ΩR = 100\Omega then pick correct statements. Switch is closed at t=0t = 0 . (initially capacitor was uncharged)
(A) Rate of energy stored in capacitor is maximum at t=0.0693sec⁡t = 0.0693\sec
(B) Rate of energy stored in capacitor is maximum at t=0.053sec⁡t = 0.053\sec
(C) Potential across capacitor and resistor become same at t=0.0693sec⁡t = 0.0693\sec
(D) Potential across capacitor and resistor become same at t=0.053sec⁡t = 0.053\sec

Explanation

Solution

We need to use the equation for potential and charge during the charging of a capacitor. Then using the value of the potential, we can find the energy of the capacitor. By doing the double differentiation of energy and equation it to zero, we will get the time. By substituting the time we will get the potential across the capacitor and the resistance.

Formula used: In this solution we will be using the following formula,
V=E(1−e−tCR)V = E\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right)
Where VV is the potential, CC is the capacitance and RR is the resistance.
q=qo(1−e−tCR)q = {q_o}\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right)
where qq is the charge.
U=12CV2U = \dfrac{1}{2}C{V^2} where UU is the stored energy in the capacitor.

Complete step by step solution:
In a series E-R-C circuit, the potential and the charge of the capacitor during its charging is given by the equations
V=E(1−e−tCR)V = E\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right) and q=qo(1−e−tCR)q = {q_o}\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right)
Now using the equation of the potential we can find the energy that is stored in the capacitor while its charging as the energy is given by, U=12CV2U = \dfrac{1}{2}C{V^2}
So substituting we get,
U=12CE2(1−e−tCR)2U = \dfrac{1}{2}C{E^2}{\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right)^2}
The rate of the energy that is stored is given by differentiating this value of energy stored by time. So we have,
dUdt=ddt[12CE2(1−e−tCR)2]\dfrac{{dU}}{{dt}} = \dfrac{d}{{dt}}\left[ {\dfrac{1}{2}C{E^2}{{\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right)}^2}} \right]
On performing the differentiation we get,
dUdt=12CE2×2(1−e−tCR)×(−e−tCR)×(−1CR)\dfrac{{dU}}{{dt}} = \dfrac{1}{2}C{E^2} \times 2\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right) \times \left( { - {e^{ - \dfrac{t}{{CR}}}}} \right) \times \left( { - \dfrac{1}{{CR}}} \right)
On calculation and cancelling the common terms we have,
dUdt=E2R×(1−e−tCR)(e−tCR)\dfrac{{dU}}{{dt}} = \dfrac{{{E^2}}}{R} \times \left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right)\left( {{e^{ - \dfrac{t}{{CR}}}}} \right)
That is,
dUdt=E2R(e−tCR−e−2tCR)\dfrac{{dU}}{{dt}} = \dfrac{{{E^2}}}{R}\left( {{e^{ - \dfrac{t}{{CR}}}} - {e^{ - \dfrac{{2t}}{{CR}}}}} \right)
Now to find the maximum rate of stored energy, we take the double differentiation and then we can equate it to zero. So we have,
d2Udt2=0\dfrac{{{d^2}U}}{{d{t^2}}} = 0
So substituting we have,
ddt[E2R(e−tCR−e−2tCR)]=0\dfrac{d}{{dt}}\left[ {\dfrac{{{E^2}}}{R}\left( {{e^{ - \dfrac{t}{{CR}}}} - {e^{ - \dfrac{{2t}}{{CR}}}}} \right)} \right] = 0
So differentiating we get,
E2R(e−tCR×(−1CR)−e−2tCR×(−2CR))=0\dfrac{{{E^2}}}{R}\left( {{e^{ - \dfrac{t}{{CR}}}} \times \left( { - \dfrac{1}{{CR}}} \right) - {e^{ - \dfrac{{2t}}{{CR}}}} \times \left( { - \dfrac{2}{{CR}}} \right)} \right) = 0
On simplifying we get it as,
E2R(−e−tCRCR+2e−2tCRCR)=0\dfrac{{{E^2}}}{R}\left( { - \dfrac{{{e^{ - \dfrac{t}{{CR}}}}}}{{CR}} + \dfrac{{2{e^{ - \dfrac{{2t}}{{CR}}}}}}{{CR}}} \right) = 0
Since the value of E2R\dfrac{{{E^2}}}{R} is not equal to zero. So we can write,
−e−tCRCR+2e−2tCRCR=0- \dfrac{{{e^{ - \dfrac{t}{{CR}}}}}}{{CR}} + \dfrac{{2{e^{ - \dfrac{{2t}}{{CR}}}}}}{{CR}} = 0
So on simplifying we get,
−e−tCR+2e−2tCR=0- {e^{ - \dfrac{t}{{CR}}}} + 2{e^{ - \dfrac{{2t}}{{CR}}}} = 0
On taking one term to the RHS
2(e−tCR)2=e−tCR2{\left( {{e^{ - \dfrac{t}{{CR}}}}} \right)^2} = {e^{ - \dfrac{t}{{CR}}}}
Now on cancelling we get,
e−tCR=12{e^{ - \dfrac{t}{{CR}}}} = \dfrac{1}{2}
On taking inverse,
etCR=2{e^{\dfrac{t}{{CR}}}} = 2
We can take log on both the sides we get,
tCR=ln⁡2\dfrac{t}{{CR}} = \ln 2
So we get the time as,
t=CRln⁡2t = CR\ln 2
In the question we are given C=1mF=10−3FC = 1mF = {10^{ - 3}}F and R=100ΩR = 100\Omega
So substituting we get,
t=10−3×100×0.693t = {10^{ - 3}} \times 100 \times 0.693
On calculating we have,
t=0.0693st = 0.0693s
Substituting t=CRln⁡2t = CR\ln 2 in the potential across resistance VR=E(1−e−tCR){V_R} = E\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right) we have,
VR=E(1−e−CRln⁡2CR){V_R} = E\left( {1 - {e^{ - \dfrac{{CR\ln 2}}{{CR}}}}} \right) . So we get,
VR=E(1−e−ln2){V_R} = E\left( {1 - {e^{ - ln2}}} \right) Since, e−ln2{e^{ - ln2}} is 12\dfrac{1}{2} . So we have VR=E(1−12){V_R} = E\left( {1 - \dfrac{1}{2}} \right)
So, VR=E2{V_R} = \dfrac{E}{2}
Substituting t=CRln⁡2t = CR\ln 2 in the potential across capacitance VC=qoC(1−e−tCR){V_C} = \dfrac{{{q_o}}}{C}\left( {1 - {e^{ - \dfrac{t}{{CR}}}}} \right) we have,
VC=qoC(1−e−ln⁡2){V_C} = \dfrac{{{q_o}}}{C}\left( {1 - {e^{ - \ln 2}}} \right) . Since qoC=E\dfrac{{{q_o}}}{C} = E and e−ln2{e^{ - ln2}} is 12\dfrac{1}{2} we get,
VC=E2{V_C} = \dfrac{E}{2}
So VC=VR{V_C} = {V_R} at t=0.0693st = 0.0693s .
So the options A and C are correct.

Note:
A capacitor has the ability to store electrical charge. When a capacitor is fully charged, there is a potential difference present between the two plates. The larger is the area of the plates or the smaller is the separation of the plates, the more charge the capacitor can hold.