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Electronic Devices question

2020 · 9 Jan · Shift 2 · Q51
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Electronic Devices question

2020 · 9 Jan · Shift 2 · Q51

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm) JEE Main 2020 (Online) 9th January Evening Slot Physics - Semiconductor Question 150 English
  1. A
    QA = CVe{{CV} \over e}eCV​, QB = VC2{{VC} \over 2}2VC​
  2. B
    QA = CV2{{CV} \over 2}2CV​, QB = VCe{{VC} \over e}eVC​
  3. C
    QA = VC, QB = VCe{{VC} \over e}eVC​
  4. D
    QA = VC, QB = CV
View written solutionFree

Correct answer: C

  1. Initial charge on each capacitor

Each capacitor has capacitance CCC and is initially charged to potential V=5 VV=5\text{ V}V=5 V. So initial charge on each is Q0=CV.Q_0=CV.Q0​=CV.

We must find the charge at time t=CRt=CRt=CR in the two given circuits.

  1. Circuit A

In circuit A, the capacitor is connected in such a way that there is no discharge path through the resistor for the capacitor plates. Hence the capacitor remains isolated with its initial charge unchanged.

Therefore, QA=CV.Q_A=CV.QA​=CV.

  1. Circuit B

In circuit B, the capacitor discharges through resistance RRR. For a discharging capacitor, Q(t)=Q0e−t/RC.Q(t)=Q_0 e^{-t/RC}.Q(t)=Q0​e−t/RC.

Here,

  • Q0=CVQ_0=CVQ0​=CV
  • t=CRt=CRt=CR
  • time constant RCRCRC

So, QB=CV e−(CR)/(RC)=CVe−1=CVe.Q_B = CV\, e^{-(CR)/(RC)} = CV e^{-1} = \frac{CV}{e}.QB​=CVe−(CR)/(RC)=CVe−1=eCV​.

Thus, QB=CVe.Q_B=\frac{CV}{e}.QB​=eCV​.

  1. Compare with options

We found: QA=CV,QB=CVe.Q_A=CV, \qquad Q_B=\frac{CV}{e}.QA​=CV,QB​=eCV​.

This matches Option C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored correct answer.

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