Back to Directory
NEET PHYSICSMedium

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=VrmsIrmscosϕP = 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 6060^\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.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut