In an A.C. circuit, the current flowing is I=5 sin (100 t - π/2) ampere and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to

Question

In an A.C. circuit, the current flowing is I=5 sin (100 t - π/2) ampere and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to (A) 20 W (B) 0 W (C) 1000W (D) 40 W.

Tardigrade Admin · Accepted Answer

Current (I) = 5 sin (100 t - π/2) and voltage (V) = 200 sin (100 t). Comparing the given equation, with the standard equation, we find that between current and voltage is φ=(π/2)=90°. Power consumption P=IrmsVrmscos φ =IrmsVrmscos90°=0

Q. In an A.C. circuit, the current flowing is I=5sin(100t−π/2) ampere and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to

20 W

0 W

1000W

40 W.

Current (I)=5sin(100t−π/2) and voltage (V) =200sin(100t). Comparing the given equation, with the standard equation, we find that between current and voltage is ϕ=2π​=90∘. Power consumption P=Irms​Vrms​cosϕ =Irms​Vrms​cos90∘=0

Q. In an A.C. circuit, the current flowing is $I = 5 sin (100 t - π /2)$ ampere and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to