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 μ F. The total impedance, and phase difference between the voltage across the source and the current will respectively be: (a) 10 Ω and tan ⁻¹ (5/3) (b) 7 Ω and 45° (c) 10 Ω and tan ⁻¹ (8/3) (d) 7 Ω and tan ⁻¹ (5/3)

Answer: (b). XL = 320 × 25×10⁻³ = 8 Ω XC = 1320 × 10⁻³ ≈ 3.1 Ω, so XL - XC ≈ 4.9 ≈ 5 Ω. Z = √(5² + 5²) ≈ 7 Ω tan φ = 5/5 = 1 ⇒ φ = 45° So the answer is (b) 7 Ω and 45°.

A sinusoidal voltage of peak value 283 V and angular frequency 320/s… | JEE Main 2017 (9 Apr, Shift 1)

Q57JEE Main 2017 · 9 APR · SHIFT 1Difficult

Series LCR Circuit

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

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Answer: (b)

$X_L = 320 \times 25\times10^{-3} = 8\ \Omega$
$X_C = \dfrac{1}{320 \times 10^{-3}} \approx 3.1\ \Omega$ , so $X_L - X_C \approx 4.9 \approx 5\ \Omega$ .
$Z = \sqrt{5^2 + 5^2} \approx 7\ \Omega$
$\tan\phi = \dfrac{5}{5} = 1 \Rightarrow \phi = 45^\circ$
So the answer is (b) $7\ \Omega$ and $45^\circ$ .

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