The diagram below illustrates an a.c.source of 50V(r.m.s.), \(\frac{100}{\pi}\)Hz connected in series with an inductor of inductor of inductance L and a resistor of resistance R. The current in the circuit is 2A and the p.d across L and R are 30V and 40V respectively. Calculate the power factor of the circuit

Ekene1245

3 years ago

First of all they said find power factor...
Co¢=R/z
But phase angle across RLC circle is given as Tan¢=Z/R...so don't get it twisted...
So let's proceed shall we
Look the diagram well they said Voltage follow across the inductance
VL=30V and that of the resistance is V=40V
So the total current in the circuit is I=2A so to get impendance across RL circuit is given as
Z=√R²+(xl)²
Let solve
Recall that voltage across the inductance is given as
VL=IXL
30=2xl
Xl=30/2
Xl=15ohms
Voltage across the resistance
V=iR
40=2R
R=20ohms
Z=√R²+(xl)²
Z=√20²+15²
Z=25ohms
So therefore power factor
Cos¢=R/z
Cos¢=20/25
Cos¢=0.8... QED..!

gideon_simon

5 years ago

Vr = I × R
R = Vr/2 = 40/2 = 20 ohms

Vl = I × Xl
Xl = Vl/ I = 30/2 = 15 ohms

Z = √(R² + Xl²)
= √20² + 15² = √625
= 25ohms

Power factor = cos ¢ = R/z
= 20/25
= 0.8

oyinadeadesina

6 years ago

1,2,3 I choose d right answer nd I was marked wrong