Question

A 60-Hz 240-V-rms source supplies power to a load consisting of a resistance in series with an inductance. The real poweris 1500W, and the apparent power is 2500 VA. Determine the value of the resistance and the value of the inductance.

Answer 1

Resistance will be 13.824 ohm

And inductance will be 0.048 Henry

Step-by-step explanation:

We have given real power P = 1500 watt

And apparent power S = 2500 VA

Voltage V = 240 volt

Frequency f = 60 Hz

Power factor
`cos\Phi` is given by

`cos\Phi =(P)/(S)=(1500)/(2500)=0.6`

And real power is given by
`P=VIcos\Phi`

So
`1500=240* I* 0.6`

I = 10.4166 A

Now impedance
`Z=(V)/(I)=(240)/(10.4166)=23.04ohm`

And
`cos\Phi =(R)/(Z)`

So
`0.6=(R)/(23.04)`

R = 13.824 ohm

Impedance is given by
`Z=√(R^2+X_L^2)`

`23.04=√(13.824^2+X_L^2)`

Squaring both side

`530.84=191.10+X_L^2`

`X_L=18.43ohm`

`18.43=2* 3.14* 60* L`

L = 0.048 Henry