Question

A pure inductive coil allows a current of 10A to flow from 230V AC and 50Hz AC supply .Find inductive reactance and inductance of coil respectively.

Answer 1

Explanation:

Here, L=50mH=50×10−3HL=50mH=50×10-3H

Ev=220V,V=50HzEv=220V,V=50Hz

XL=ωL=2πvLXL=ωL=2πvL

=2×3.14×50×50×10−3ohm=2×3.14×50×50×10-3ohm

=15700×10−3ohm=15.7ohm=15700×10-3ohm=15.7ohm

Z=XL=15.7ohmZ=XL=15.7ohm

Iv=EVZ=22015.7amp=14.0AIv=EVZ=22015.7amp=14.0A

I0=2–√Iv=1.414×14.0=19.8A

Answer 2

Given,

Current=10A,
Voltage=230V,

Frequency=50Hz

To Find,

Inductive reactance,

Inductance of the coil.

Solution,

As per the given information,

here, Ev=230V,Iv=10A,f=50Hz

Inductive is written as X L.

So, XL=Ev/Iv

          =230V/10A

          =23 ohm

Now, we have the formula for inductive reactance as 2πfL

Inductance of the coil, L= XL/2πf

                                         =23/2π×50

                                         =0.073H

Hence, inductive reactance of the coil is 23ohm and inductance of the coil is 0.073H.