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Alternating Current question

2019 · 10 Apr · Shift 2 · Q62
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Alternating Current question

2019 · 10 Apr · Shift 2 · Q62

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A coil of self inductance 10 mH and resistance 0.1 Ω\OmegaΩ is connected through a switch to a battery of internal resistance 0.9 Ω\OmegaΩ. After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [take ln 5 = 1.6]
  1. A
    0.324 s
  2. B
    0.002 s
  3. C
    0.103 s
  4. D
    0.016 s
View written solutionFree

Correct answer: D

  1. Identify the RL circuit parameters

A coil has:

  • Self inductance: L=10 mH=10×10−3 H=0.01 HL = 10\ \text{mH} = 10 \times 10^{-3}\ \text{H} = 0.01\ \text{H}L=10 mH=10×10−3 H=0.01 H
  • Coil resistance: 0.1 Ω0.1\ \Omega0.1 Ω
  • Battery internal resistance: 0.9 Ω0.9\ \Omega0.9 Ω

So total resistance is

R=0.1+0.9=1.0 ΩR = 0.1 + 0.9 = 1.0\ \OmegaR=0.1+0.9=1.0 Ω
  1. Use the current growth equation in an RL circuit

When the switch is closed, current grows as

I=I0(1−e−t/τ)I = I_0\left(1-e^{-t/\tau}\right)I=I0​(1−e−t/τ)

where the time constant is

τ=LR\tau = \frac{L}{R}τ=RL​

Thus,

τ=0.011=0.01 s\tau = \frac{0.01}{1} = 0.01\ \text{s}τ=10.01​=0.01 s
  1. Given current reaches 80% of saturation value

So,

I=0.8I0I = 0.8 I_0I=0.8I0​

Substitute into the equation:

0.8I0=I0(1−e−t/τ)0.8I_0 = I_0(1-e^{-t/\tau})0.8I0​=I0​(1−e−t/τ)

Cancelling I0I_0I0​,

0.8=1−e−t/τ0.8 = 1 - e^{-t/\tau}0.8=1−e−t/τ e−t/τ=0.2=15e^{-t/\tau} = 0.2 = \frac{1}{5}e−t/τ=0.2=51​

Taking natural log:

−tτ=ln⁡(15)=−ln⁡5-\frac{t}{\tau} = \ln\left(\frac{1}{5}\right) = -\ln 5−τt​=ln(51​)=−ln5 tτ=ln⁡5\frac{t}{\tau} = \ln 5τt​=ln5

Given ln⁡5=1.6\ln 5 = 1.6ln5=1.6,

t=τln⁡5=0.01×1.6=0.016 st = \tau \ln 5 = 0.01 \times 1.6 = 0.016\ \text{s}t=τln5=0.01×1.6=0.016 s
  1. Match with the options
t=0.016 s\boxed{t = 0.016\ \text{s}}t=0.016 s​

So the correct option is D.

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