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Physics · Alternating currents, peak and RMS value of alternating current/voltage

In an a.c. circuit and I are given by volt and \mathrm{mA} . \) The power dissip

In an a.c. circuit \( \mathrm{V} \) and I are given by \( V=50 \sin 50 t \) volt and \( I= \) \( 100 \sin (50 t+\pi / 3) \mathrm{mA} . \) The power dissipated in the circuit

  • A. 2.5 kw
  • B. 1.25 kw
  • C. \( 5.0 \mathrm{kw} \)
  • D. 500 watt

Step-by-step solution

The power dissipated in an AC circuit is given by P = V_rms I_rms cos φ, where φ is the phase difference between voltage and current. Here V = 50 sin(50t) V, I = 100 sin(50t + π/3) A (assuming current in amperes, as mA leads to a value not among options). V_rms = 50/√2 V, I_rms = 100/√2 A, cos φ = cos(π/3) = 0.5. Thus P = (50/√2)(100/√2)(0.5) = (5000/2)(0.5) = 2500 × 0.5 = 1250 W = 1.25 kW.
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