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Alternating Current question

2020 · 7 Jan · Shift 2 · Q55
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Alternating Current question

2020 · 7 Jan · Shift 2 · Q55

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 Ω\OmegaΩ resistor. The ratio of the currents at time t =∞\infty∞ and at t = 40 s is close to : (Take e2 = 7.389)
  1. A
    1.06
  2. B
    0.84
  3. C
    1.15
  4. D
    1.46
View written solutionFree

Correct answer: C: 1.15

  1. Identify the circuit and formula

A DC emf V=20 VV = 20\,\text{V}V=20V is suddenly applied to a series RLRLRL circuit with:

  • Inductance L=10 mH=0.01 HL = 10\,\text{mH} = 0.01\,\text{H}L=10mH=0.01H
  • Resistance R=5 ΩR = 5\,\OmegaR=5Ω

For growth of current in an RLRLRL circuit,

i(t)=I∞(1−e−t/τ)i(t) = I_\infty \left(1 - e^{-t/\tau}\right)i(t)=I∞​(1−e−t/τ)

where

I∞=VR,τ=LRI_\infty = \frac{V}{R}, \qquad \tau = \frac{L}{R}I∞​=RV​,τ=RL​
  1. Find the steady current
I∞=205=4 AI_\infty = \frac{20}{5} = 4\,\text{A}I∞​=520​=4A

So at t=∞t = \inftyt=∞,

i(∞)=4 Ai(\infty) = 4\,\text{A}i(∞)=4A
  1. Find the time constant
τ=LR=0.015=0.002 s\tau = \frac{L}{R} = \frac{0.01}{5} = 0.002\,\text{s}τ=RL​=50.01​=0.002s
  1. Interpret the given time

The question asks for current at t=40 st = 40\,\text{s}t=40s, but with τ=0.002 s\tau = 0.002\,\text{s}τ=0.002s, we get

tτ=400.002=20000\frac{t}{\tau} = \frac{40}{0.002} = 20000τt​=0.00240​=20000

which makes e−20000≈0e^{-20000} \approx 0e−20000≈0, so the ratio would be essentially 111, and none of the options match.

Since the question provides the hint "Take e2=7.389e^2 = 7.389e2=7.389", it strongly indicates they intend

tτ=2\frac{t}{\tau} = 2τt​=2

which happens if t=4 mst = 4\,\text{ms}t=4ms, not 40 s40\,\text{s}40s.

So this is almost certainly a misprint, and the intended time is t=4 mst = 4\,\text{ms}t=4ms.

  1. Calculate current at the intended time t=4 mst = 4\,\text{ms}t=4ms
t=4 ms=0.004 st = 4\,\text{ms} = 0.004\,\text{s}t=4ms=0.004s

Then

tτ=0.0040.002=2\frac{t}{\tau} = \frac{0.004}{0.002} = 2τt​=0.0020.004​=2

Thus,

i(0.004)=4(1−e−2)i(0.004) = 4\left(1 - e^{-2}\right)i(0.004)=4(1−e−2)

Using e2=7.389e^2 = 7.389e2=7.389,

e−2=17.389≈0.1353e^{-2} = \frac{1}{7.389} \approx 0.1353e−2=7.3891​≈0.1353

so

i(0.004)=4(1−0.1353)=4(0.8647)=3.4588 Ai(0.004) = 4(1 - 0.1353) = 4(0.8647) = 3.4588\,\text{A}i(0.004)=4(1−0.1353)=4(0.8647)=3.4588A
  1. Find the required ratio

The ratio of currents at t=∞t = \inftyt=∞ and at t=4 mst = 4\,\text{ms}t=4ms is

i(∞)i(0.004)=43.4588≈1.156\frac{i(\infty)}{i(0.004)} = \frac{4}{3.4588} \approx 1.156i(0.004)i(∞)​=3.45884​≈1.156

This is close to

1.151.151.15
  1. Check options
  • A: 1.061.061.06
  • B: 0.840.840.84
  • C: 1.151.151.15 ✅
  • D: 1.461.461.46

So the correct option should be C.

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