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Alternating Current question

2020 · 4 Sep · Shift 2 · Q56
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Alternating Current question

2020 · 4 Sep · Shift 2 · Q56

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches (1n)\left( {{1 \over n}} \right)(n1​) times of its maximum value, is :
  1. A
    LRln⁡(nn+1){L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)RL​ln(n​+1n​​)
  2. B
    LRln⁡(nn−1){L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)RL​ln(n​−1n​​)
  3. C
    LRln⁡(n+1n−1){L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)RL​ln(n​−1n​+1​)
  4. D
    LRln⁡(n−1n){L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)RL​ln(n​n​−1​)
View written solutionFree

Correct answer: B

  1. Current growth in an LLL-RRR series circuit

When a battery of emf VVV is connected to a series LLL-RRR circuit at t=0t=0t=0, the current grows as

I(t)=VR(1−e−Rt/L).I(t)=\frac{V}{R}\left(1-e^{-Rt/L}\right).I(t)=RV​(1−e−Rt/L).

The maximum steady current is

I0=VR.I_0=\frac{V}{R}.I0​=RV​.

So,

I(t)=I0(1−e−Rt/L).I(t)=I_0\left(1-e^{-Rt/L}\right).I(t)=I0​(1−e−Rt/L).
  1. Energy stored in the inductor

The magnetic energy stored in the inductor at time ttt is

U(t)=12LI2(t).U(t)=\frac{1}{2}LI^2(t).U(t)=21​LI2(t).

Its maximum value is when current becomes I0I_0I0​:

Umax⁡=12LI02.U_{\max}=\frac{1}{2}LI_0^2.Umax​=21​LI02​.

Given that at time ttt,

U(t)=1nUmax⁡.U(t)=\frac{1}{n}U_{\max}.U(t)=n1​Umax​.

Substitute:

12LI2(t)=1n⋅12LI02.\frac{1}{2}LI^2(t)=\frac{1}{n}\cdot \frac{1}{2}LI_0^2.21​LI2(t)=n1​⋅21​LI02​.

Thus,

I2(t)=I02nI^2(t)=\frac{I_0^2}{n}I2(t)=nI02​​

or

I(t)=I0n.I(t)=\frac{I_0}{\sqrt n}.I(t)=n​I0​​.
  1. Use the current expression

We have

I0(1−e−Rt/L)=I0n.I_0\left(1-e^{-Rt/L}\right)=\frac{I_0}{\sqrt n}.I0​(1−e−Rt/L)=n​I0​​.

Cancel I0I_0I0​:

1−e−Rt/L=1n.1-e^{-Rt/L}=\frac{1}{\sqrt n}.1−e−Rt/L=n​1​.

So,

e−Rt/L=1−1n=n−1n.e^{-Rt/L}=1-\frac{1}{\sqrt n}=\frac{\sqrt n-1}{\sqrt n}.e−Rt/L=1−n​1​=n​n​−1​.

Taking natural log,

−RtL=ln⁡(n−1n).-\frac{Rt}{L}=\ln\left(\frac{\sqrt n-1}{\sqrt n}\right).−LRt​=ln(n​n​−1​).

Hence,

t=−LRln⁡(n−1n)=LRln⁡(nn−1).t=-\frac{L}{R}\ln\left(\frac{\sqrt n-1}{\sqrt n}\right) =\frac{L}{R}\ln\left(\frac{\sqrt n}{\sqrt n-1}\right).t=−RL​ln(n​n​−1​)=RL​ln(n​−1n​​).
  1. Match with options

This is exactly

LRln⁡(nn−1)\boxed{\frac{L}{R}\ln\left(\frac{\sqrt n}{\sqrt n-1}\right)}RL​ln(n​−1n​​)​

which corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored answer.

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