Question

A coil has an inductance of 0.1 H and resistance 12Omega It is connected to 220 V, 50 Hz line. Find the impedance, power factor and power.

Answer 1

Answer:

hey mate here is ur answer

Explanation:

Impedance of the circuit is given by:

Z=

R

2

+(2πfL)

2

Z=

220

2

+(2π×50×0.7)

2

Z≈220

2

Ω

Power factor is given by:

cosϕ=

Z

R

=

4

π

Total current in the circuit flowing is:

I=

Z

V

=

220

2

220

=

2

1

A

Wattless component of current in the circuit is:

I

a

=Isin(ϕ)

I

a

=

2

1

×

2

1

=0.5A

hope it helps plz mark as brainliest

Answer 2

Answer:

Z = 33.64Ω

cosΦ = 0.3567

P = 513.25 W

Explanation:

L= 0.1 H, R = 12Ω, f = 50 Hz, V = 220 V

Z = $\sqrt{ R^2 + X_{L} ^{2}$    where   $X_{L} = (2\pi f)L$

Z = $\sqrt{12^{2}+ (2* \pi *50*0.1)^{2} }$

Z = $\sqrt{144 + 987.7}$

Z = $\sqrt{1331.755}$ = 33.64 Ω

cosФ =  R÷Z = 12÷33.64 = 0.3567

I = V÷Z = 220÷33.64 = 6.54 A

Power = V×I×cosФ = 220×6.54×0.3567

          = 513.25 W