Question

An inductor 20mH, a capacitor 50μF and a resistor 40Ω are connected in series across a source of emf V=10sin340t. The power loss in A.C. circuit is :

Answer 1

Answer:

840 Watts

Explanation:

V=10Sin340t

V=V°sinωt so ω=340 V°=10

Inductive reactance=ωI=340*20*$10^{-3}$=6.8Ω

Capacitive reactance=1/ωC=1/340*50*$10^{-6}$=58.8Ω

Z=$\sqrt{R^{2}+(Xc-Xl)}^{2}$

so Z=52Ω

POWER=Irms * Vrms *cos∅

cos∅=R/Z=40/52=0.76

Vrms = V°/$\sqrt{2}$ =340/$\sqrt{2}$

Irms=Vrms/Z =6.5/$\sqrt{2}$

power=(340/$\sqrt{2}$)*(6.5/$\sqrt{2}$)*0.76

=839.8

=840W (approx.)