JEE Main 2017 · 9 Apr · Shift 1 · Physics Q20

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A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R = 5 Ω, L = 25 mH and C = 1000 μF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :

10 Ω and

\tan^{-1}\left(\dfrac53\right)

7 Ω and 45°

10 Ω and

\tan^{-1}\left(\dfrac83\right)

7 Ω and

\tan^{-1}\left(\dfrac53\right)

Answer and worked solution

Official answer: B

$X_L=320\times0.025=8$ Ω and $X_C=\dfrac1{320\times10^{-3}}=3.125$ Ω, so $X\approx4.9$ Ω and
$Z=\sqrt{5^2+4.9^2}\approx7\ \Omega,\qquad\phi=\tan^{-1}\frac{4.9}5\approx45^\circ.$

Topics:Reactance and impedance LCR circuits and resonance

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 20