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Capacitor question

2021 · 27 Jul · Shift 1 · Q53
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  5. /2021 · 27 Jul · Shift 1 · Q53

Capacitor question

2021 · 27 Jul · Shift 1 · Q53

JEE MainPhysicsCapacitorMCQ+4 / −1
JEE Main 2021 (Online) 27th July Morning Shift Physics - Capacitor Question 85 English A capacitor of capacitance C = 1 μ\muμ F is suddenly connected to a battery of 100 volt through a resistance R = 100 Ω\OmegaΩ. The time taken for the capacitor to be charged to get 50 V is : [Take ln 2 = 0.69]
  1. A
    1.44 ×\times× 10 −-− 4 s
  2. B
    3.33 ×\times× 10 −-− 4 s
  3. C
    0.69 ×\times× 10 −-− 4 s
  4. D
    0.30 ×\times× 10 −-− 4 s
View written solutionFree

Correct answer: C

  1. Charging equation of a capacitor

For a capacitor charging through a resistor from a battery of emf V0V_0V0​, the potential across the capacitor at time ttt is

VC=V0(1−e−t/RC).V_C = V_0\left(1-e^{-t/RC}\right).VC​=V0​(1−e−t/RC).

Here,

  • V0=100 VV_0 = 100\,\text{V}V0​=100V
  • R=100 ΩR = 100\,\OmegaR=100Ω
  • C=1 μF=1×10−6 FC = 1\,\mu\text{F} = 1 \times 10^{-6}\,\text{F}C=1μF=1×10−6F
  • Required VC=50 VV_C = 50\,\text{V}VC​=50V
  1. Substitute the given values

50=100(1−e−t/RC).50 = 100\left(1-e^{-t/RC}\right).50=100(1−e−t/RC).

Divide by 100100100:

12=1−e−t/RC.\frac{1}{2} = 1 - e^{-t/RC}.21​=1−e−t/RC.

So,

e−t/RC=1−12=12.e^{-t/RC} = 1 - \frac{1}{2} = \frac{1}{2}.e−t/RC=1−21​=21​.

Taking natural log,

−tRC=ln⁡(12)=−ln⁡2.-\frac{t}{RC} = \ln\left(\frac{1}{2}\right) = -\ln 2.−RCt​=ln(21​)=−ln2.

Hence,

t=RCln⁡2.t = RC\ln 2.t=RCln2.

  1. Calculate RCRCRC

RC=100×1×10−6=10−4 s.RC = 100 \times 1 \times 10^{-6} = 10^{-4}\,\text{s}.RC=100×1×10−6=10−4s.

Therefore,

t=10−4×0.69=0.69×10−4 s.t = 10^{-4} \times 0.69 = 0.69 \times 10^{-4}\,\text{s}.t=10−4×0.69=0.69×10−4s.

  1. Match with the options

t=0.69×10−4 s.t = 0.69 \times 10^{-4}\,\text{s}.t=0.69×10−4s.

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.

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