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NEET PHYSICSALTERNATING CURRENTMedium

Question

The instantaneous values of alternating current and voltages in a circuit are given as i=12sin⁡(100πt)i = \frac{1}{\sqrt{2}} \sin(100\pi t) ampere and e=12sin⁡(100πt+π/3)e = \frac{1}{\sqrt{2}} \sin(100\pi t + \pi/3) volt. The average power in Watts consumed in the circuit is:

A

1/4

B

√3/4

C

1/2

D

1/8

Step-by-Step Solution

The average power (PP) consumed in an AC circuit is determined by the RMS values of voltage and current and the cosine of the phase difference (power factor). The formula is P=VrmsIrmscos⁡ϕP = V_{rms} I_{rms} \cos \phi .

  1. Identify Peak Values: From the given equations i=Imsin⁡(ωt)i = I_m \sin(\omega t) and e=Vmsin⁡(ωt+ϕ)e = V_m \sin(\omega t + \phi), we have peak current Im=1/2I_m = 1/\sqrt{2} A and peak voltage Vm=1/2V_m = 1/\sqrt{2} V.
  2. Calculate RMS Values: Irms=Im2=1/22=12I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} A. Vrms=Vm2=1/22=12V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} V.
  3. Identify Phase Difference: The phase difference ϕ\phi is given as π/3\pi/3 radians (or 60∘60^\circ).
  4. Calculate Power: P=(12)(12)cos⁡(π3)P = (\frac{1}{2}) (\frac{1}{2}) \cos(\frac{\pi}{3}) P=14×12=18P = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} W.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSALTERNATING CURRENTinstantaneousvaluesalternatingcurrentvoltages

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