Question

A coil is connected to a 250V, 50Hz sinusoidal supply takes a current of 10A at a phase angle of 30degrees. Calculate Resistance and;Inductive reactance of the coil​

Answer 1

cos(30°)=R/Z

square on b/s,

3/4=R^2 /Z^2

R^2 =3/4 Z^2

Z=V/I=250/10=25 ohm

R=21.65 ohm

Z^2=R^2+XL^2

Z^2=3/4 Z^2 +XL^2

Z^2 /4=XL^2

XL=Z/2=25/2=12.5 ohm

Answer 2

Answer:

Resistance of the Coil = 43.30 ohm

Reactance of the Coil = 25 ohm

Explanation:

Given that: Voltage of Sinusoidal AC source = 250V

r.m.s voltage(V r.m.s) = 250/√2 volt =

Frequency of AC source (f) = 50 Hz

Angular Frequency (ω) = 2πf = 100π

Current (I0) = 10 A

r.m.s current (I r.m.s) = 10/√2 A

Now, let the Reactance of Inductor = XL

$xl \: = \frac{V \: rms}{I \: rms} \\ = \frac{176.78}{7.071} \\ = 25.000 \: ohm$

So, the Inductive Reactance of the Coil

(XL) = 25 ohm

Let the resistance of coil = r

We know Phase angle in Inductive Circuit;

$Φ = { tan }^{ - 1} ( \frac{xl}{r} )$

Given that: Phase angle(Φ) = 30°

We get;

$\tan(Φ) = \frac{xl}{r} \\ or \: \tan(30) = \frac{xl}{r} \\ or \: 0.57735 = \frac{25}{r} \\ or \: r = \frac{25 }{0.57735} \: ohm \\ r = 43.30 \: ohm$