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Alternating Current question

2020 · 7 Jan · Shift 1 · Q53
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Alternating Current question

2020 · 7 Jan · Shift 1 · Q53

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:
  1. A
    L ↔\leftrightarrow↔ k, C ↔\leftrightarrow↔ b, R ↔\leftrightarrow↔ m
  2. B
    L ↔\leftrightarrow↔ m, C ↔\leftrightarrow↔ k, R ↔\leftrightarrow↔ b
  3. C
    L ↔\leftrightarrow↔ m, C ↔\leftrightarrow↔ 1k{1 \over k}k1​, R ↔\leftrightarrow↔ b
  4. D
    L ↔\leftrightarrow↔ 1b{1 \over b}b1​, C ↔\leftrightarrow↔ 1m{1 \over m}m1​, R ↔\leftrightarrow↔ 1k{1 \over k}k1​
View written solutionFree

Correct answer: C

  1. Write the differential equation for a damped mechanical oscillator

For a mass-spring system with damping constant bbb:

mx¨+bx˙+kx=0m\ddot{x} + b\dot{x} + kx = 0mx¨+bx˙+kx=0

where:

  • mmm = mass
  • bbb = damping constant
  • kkk = spring constant
  1. Write the differential equation for an LCR circuit

For charge qqq in a series LCR circuit:

Lq¨+Rq˙+qC=0L\ddot{q} + R\dot{q} + \frac{q}{C} = 0Lq¨​+Rq˙​+Cq​=0

where:

  • LLL = inductance
  • RRR = resistance
  • CCC = capacitance
  1. Compare the two equations term by term

Mechanical oscillator:

mx¨+bx˙+kx=0m\ddot{x} + b\dot{x} + kx = 0mx¨+bx˙+kx=0

Electrical oscillator:

Lq¨+Rq˙+1Cq=0L\ddot{q} + R\dot{q} + \frac{1}{C}q = 0Lq¨​+Rq˙​+C1​q=0

Matching corresponding coefficients:

  • m↔Lm \leftrightarrow Lm↔L
  • b↔Rb \leftrightarrow Rb↔R
  • k↔1Ck \leftrightarrow \dfrac{1}{C}k↔C1​

Therefore,

L↔m,R↔b,C↔1kL \leftrightarrow m, \quad R \leftrightarrow b, \quad C \leftrightarrow \frac{1}{k}L↔m,R↔b,C↔k1​

  1. Check the options
  • A: L↔k, C↔b, R↔mL \leftrightarrow k,\ C \leftrightarrow b,\ R \leftrightarrow mL↔k, C↔b, R↔m ❌
  • B: L↔m, C↔k, R↔bL \leftrightarrow m,\ C \leftrightarrow k,\ R \leftrightarrow bL↔m, C↔k, R↔b ❌ because actually 1C↔k\dfrac{1}{C} \leftrightarrow kC1​↔k
  • C: L↔m, C↔1k, R↔bL \leftrightarrow m,\ C \leftrightarrow \dfrac{1}{k},\ R \leftrightarrow bL↔m, C↔k1​, R↔b ✅
  • D: Incorrect correspondence ❌
  1. Final answer

The correct equivalence is:

L↔m,C↔1k,R↔bL \leftrightarrow m, \quad C \leftrightarrow \frac{1}{k}, \quad R \leftrightarrow bL↔m,C↔k1​,R↔b

So the correct option is C.

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