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Capacitor question

2012 · Shift 0 · Q54
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Capacitor question

2012 · Shift 0 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant τ\tauτ of this circuit lies between AIEEE 2012 Physics - Capacitor Question 108 English
  1. A
    100 sec and 150 sec
  2. B
    0 and 50 sec
  3. C
    50 sec and 100 sec
  4. D
    150 sec and 200 sec
View written solutionFree

Correct answer: A

  1. For a discharging capacitor in an RCRCRC circuit, V(t)=V0e−t/RC=V0e−t/τV(t)=V_0 e^{-t/RC}=V_0 e^{-t/\tau}V(t)=V0​e−t/RC=V0​e−t/τ where τ=RC\tau=RCτ=RC is the time constant.

  2. The time constant has a standard graphical meaning:

    • At t=τt=\taut=τ, the voltage falls to V=V0e≈0.37V0V=\frac{V_0}{e}\approx 0.37V_0V=eV0​​≈0.37V0​
    • Equivalently, the tangent to the curve at t=0t=0t=0 meets the time axis at t=τt=\taut=τ.

  3. From the given discharge plot, the decay is such that the capacitor voltage reaches about 37%37\%37% of its initial value between 100 s100\,\text{s}100s and 150 s150\,\text{s}150s.

  4. Hence, 100 s<τ<150 s100\,\text{s} < \tau < 150\,\text{s}100s<τ<150s

  5. Therefore the correct option is: A\boxed{\text{A}}A​

  6. Comparison with stored answer:

    • Derived answer: A
    • Stored correct answer: A
    • They agree.
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