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Alternating Current question

2020 · 3 Sep · Shift 1 · Q60
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Alternating Current question

2020 · 3 Sep · Shift 1 · Q60

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω\OmegaΩ, an inductance of 0.1803 H and a capacitance of 10 μ\muμ F all in series. The time in which the resistance (heat capacity 2 J/oC) will get heated by 10oC. (assume no loss of heat to the surroudnings) is close to :
  1. A
    348 s
  2. B
    418 s
  3. C
    245 s
  4. D
    365 s
View written solutionFree

Correct answer: A

  1. Given data
  • Frequency: f=750 Hzf = 750\,\text{Hz}f=750Hz
  • RMS voltage: V=20 VV = 20\,\text{V}V=20V
  • Resistance: R=100 ΩR = 100\,\OmegaR=100Ω
  • Inductance: L=0.1803 HL = 0.1803\,\text{H}L=0.1803H
  • Capacitance: C=10 μF=10×10−6 FC = 10\,\mu\text{F} = 10\times 10^{-6}\,\text{F}C=10μF=10×10−6F
  • Heat capacity of resistance: 2 J/∘C2\,\text{J}/^\circ\text{C}2J/∘C
  • Required temperature rise: 10∘C10^\circ\text{C}10∘C

So, heat required is Q=(2)(10)=20 J.Q = (2)(10) = 20\,\text{J}.Q=(2)(10)=20J.

  1. Find angular frequency

ω=2πf=2π(750)=1500π rad/s.\omega = 2\pi f = 2\pi(750) = 1500\pi\,\text{rad/s}.ω=2πf=2π(750)=1500πrad/s.

Numerically, ω≈4712.39 rad/s.\omega \approx 4712.39\,\text{rad/s}.ω≈4712.39rad/s.

  1. Calculate reactances

Inductive reactance: XL=ωL=(4712.39)(0.1803)≈849.34 Ω.X_L = \omega L = (4712.39)(0.1803) \approx 849.34\,\Omega.XL​=ωL=(4712.39)(0.1803)≈849.34Ω.

Capacitive reactance: XC=1ωC=1(4712.39)(10×10−6)≈21.22 Ω.X_C = \frac{1}{\omega C} = \frac{1}{(4712.39)(10\times10^{-6})} \approx 21.22\,\Omega.XC​=ωC1​=(4712.39)(10×10−6)1​≈21.22Ω.

Net reactance: X=XL−XC=849.34−21.22=828.12 Ω.X = X_L - X_C = 849.34 - 21.22 = 828.12\,\Omega.X=XL​−XC​=849.34−21.22=828.12Ω.

  1. Find impedance of series RLC circuit

Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L-X_C)^2}Z=R2+(XL​−XC​)2​ Z=1002+828.122Z = \sqrt{100^2 + 828.12^2}Z=1002+828.122​ Z≈10000+685786.73Z \approx \sqrt{10000 + 685786.73}Z≈10000+685786.73​ Z≈695786.73≈834.14 Ω.Z \approx \sqrt{695786.73} \approx 834.14\,\Omega.Z≈695786.73​≈834.14Ω.

  1. Find current in the circuit

I=VZ=20834.14≈0.02398 A.I = \frac{V}{Z} = \frac{20}{834.14} \approx 0.02398\,\text{A}.I=ZV​=834.1420​≈0.02398A.

  1. Power dissipated in the resistor

Only the resistor dissipates real power: P=I2RP = I^2RP=I2R P=(0.02398)2(100)≈0.0575 W.P = (0.02398)^2(100) \approx 0.0575\,\text{W}.P=(0.02398)2(100)≈0.0575W.

  1. Find time required to produce 20 J heat

Since Q=Pt,Q = Pt,Q=Pt, we get t=QP=200.0575≈347.8 s.t = \frac{Q}{P} = \frac{20}{0.0575} \approx 347.8\,\text{s}.t=PQ​=0.057520​≈347.8s.

Thus the required time is close to 348 s.\boxed{348\,\text{s}}.348s​.

  1. Option check
  • A: 348 s348\,\text{s}348s ✅
  • B: 418 s418\,\text{s}418s ❌
  • C: 245 s245\,\text{s}245s ❌
  • D: 365 s365\,\text{s}365s ❌

Therefore, the correct option is A.

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