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:
Solution
For spring mass damped oscillator
<br><br>ma = - kx - bv
<br><br>$\Rightarrow$ ma + kx + bv = 0
<br><br>$\Rightarrow$ $m{{{d^2}x} \over {d{t^2}}}$ + b${{dx} \over {dt}}$ + kx = 0 ....(1)
<br><br>For LCR circuit
<br><br>L${{di} \over {dt}}$ + iR + ${q \over C}$ = 0
<br><br>$\Rightarrow$ L${{{d^2}q} \over {d{t^2}}}$ + R${{dq} \over {dt}}$ + ${q \over C}$ = 0 .....(2)
<br><br>Comparing (1) and (2), we get
<br><br>L $\leftrightarrow$ m, C $\leftrightarrow$ ${1 \over k}$, R $\leftrightarrow$ b
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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