A LCR circuit behaves like a damped harmonic oscillator. Comparin… - Vidyadip

MCQ

A $LCR$ circuit behaves like a damped harmonic oscillator. Comparing it with a physical springmass damped oscillator having damping constant $\mathrm{b}$, the correct equivalence would be:

✓
$\mathrm{L} \leftrightarrow \mathrm{m}, \mathrm{C} \leftrightarrow \frac{1}{\mathrm{k}}, \mathrm{R} \leftrightarrow \mathrm{b}$
B
$\mathrm{L} \leftrightarrow \frac{1}{\mathrm{b}}, \mathrm{C} \leftrightarrow \frac{1}{\mathrm{m}}, \mathrm{R} \leftrightarrow \frac{1}{\mathrm{k}}$
C
$\mathrm{L} \leftrightarrow \mathrm{m}, \mathrm{C} \leftrightarrow \mathrm{k}, \mathrm{R} \leftrightarrow \mathrm{b}$
D
$\mathrm{L} \leftrightarrow \mathrm{k}, \mathrm{C} \leftrightarrow \mathrm{b}, \mathrm{R} \leftrightarrow \mathrm{m}$

✓

Answer

Correct option: A.

$\mathrm{L} \leftrightarrow \mathrm{m}, \mathrm{C} \leftrightarrow \frac{1}{\mathrm{k}}, \mathrm{R} \leftrightarrow \mathrm{b}$

a
By $kVL$

$-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}-\frac{\mathrm{q}}{\mathrm{C}}-\mathrm{iR}=0$

$\mathrm{L} \frac{\mathrm{d}^{2} \mathrm{q}}{\mathrm{dt}^{2}}+\frac{1}{\mathrm{C}} \mathrm{q}+\mathrm{R} \frac{\mathrm{dq}}{\mathrm{dt}}=0$

for damped oscillator

net force $=-\mathrm{kx}-\mathrm{bv}=\mathrm{ma}$

$\frac{m d^{2} x}{d t^{2}}+k x+\frac{b d x}{d t}=0$

by comparing : Equivalence is

$\mathrm{L} \rightarrow \mathrm{m}: \mathrm{C} \rightarrow \frac{1}{\mathrm{K}} ; \mathrm{R} \rightarrow \mathrm{b}$

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