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Alternating Current question

2021 · 20 Jul · Shift 2 · Q52
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  5. /2021 · 20 Jul · Shift 2 · Q52

Alternating Current question

2021 · 20 Jul · Shift 2 · Q52

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
For a series LCR circuit with R = 100 Ω\OmegaΩ, L = 0.5 mH and C = 0.1 pF connected across 220V −-− 50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :
  1. A
    0 ∘^\circ∘, resistive circuit
  2. B
    ≈\approx≈ 90 ∘^\circ∘, predominantly inductive circuit
  3. C
    0 ∘^\circ∘, resonance circuit
  4. D
    ≈\approx≈ 90 ∘^\circ∘, predominantly capacitive circuit
View written solutionFree

Correct answer: D

  1. Given data
  • Resistance: R=100 ΩR = 100\,\OmegaR=100Ω
  • Inductance: L=0.5 mH=0.5×10−3 HL = 0.5\,\text{mH} = 0.5 \times 10^{-3}\,\text{H}L=0.5mH=0.5×10−3H
  • Capacitance: C=0.1 pF=0.1×10−12 F=10−13 FC = 0.1\,\text{pF} = 0.1 \times 10^{-12}\,\text{F} = 10^{-13}\,\text{F}C=0.1pF=0.1×10−12F=10−13F
  • Supply: 220 V,  50 Hz220\,\text{V},\; 50\,\text{Hz}220V,50Hz

We need the phase angle and whether the circuit is inductive or capacitive.

  1. Angular frequency
ω=2πf=2π(50)=100π≈314 rad/s\omega = 2\pi f = 2\pi(50) = 100\pi \approx 314\,\text{rad/s}ω=2πf=2π(50)=100π≈314rad/s
  1. Inductive reactance
XL=ωL=314×0.5×10−3X_L = \omega L = 314 \times 0.5 \times 10^{-3}XL​=ωL=314×0.5×10−3 XL=0.157 ΩX_L = 0.157\,\OmegaXL​=0.157Ω
  1. Capacitive reactance
XC=1ωC=1314×10−13X_C = \frac{1}{\omega C} = \frac{1}{314 \times 10^{-13}}XC​=ωC1​=314×10−131​ XC≈3.18×1010 ΩX_C \approx 3.18 \times 10^{10}\,\OmegaXC​≈3.18×1010Ω

So, clearly,

XC≫XLX_C \gg X_LXC​≫XL​

Hence the net reactance is

X=XL−XC≈−XCX = X_L - X_C \approx -X_CX=XL​−XC​≈−XC​

which is negative, so the circuit is predominantly capacitive.

  1. Phase angle

For a series LCR circuit,

tan⁡ϕ=XL−XCR\tan \phi = \frac{X_L - X_C}{R}tanϕ=RXL​−XC​​

Substituting,

tan⁡ϕ=0.157−3.18×1010100\tan \phi = \frac{0.157 - 3.18\times 10^{10}}{100}tanϕ=1000.157−3.18×1010​

This is a very large negative number, so

ϕ≈−90∘\phi \approx -90^\circϕ≈−90∘

Negative phase angle means current leads voltage, i.e. the circuit is capacitive.

Thus, the magnitude of phase angle is approximately 90∘90^\circ90∘.

  1. Checking options
  • A: 0∘0^\circ0∘, resistive circuit →\rightarrow→ Incorrect
  • B: ≈90∘\approx 90^\circ≈90∘, predominantly inductive circuit →\rightarrow→ Incorrect
  • C: 0∘0^\circ0∘, resonance circuit →\rightarrow→ Incorrect
  • D: ≈90∘\approx 90^\circ≈90∘, predominantly capacitive circuit →\rightarrow→ Correct
  1. Final answer

The phase angle is approximately 90∘90^\circ90∘ in magnitude, with current leading voltage, so the circuit is predominantly capacitive.

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