Question

As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 11.5 V battery and measures a current of 0.620 A. The student then connects the coil to a 24.0 V (rms), 60.0 Hz generator and measures an rms current of 0.570 A. What is the inductance?L = mH

Answer 1

L = 100.24 mH

Step-by-step explanation:

Given that

V= 11.5 V

I= 0.62 A

V(rms) = 24 V

f= 60 Hz

I(rms) = 0.57 A

resistance R

V= I R

11.5 = 0.62 R

R= 18.54 ohm

impedance Z given

V(rms) = I(rms) Z

24 = 0.57 Z ,Z= 42.10 ohm

inductor impedance
`X_L`

`X_L=√(Z^2-R^2)`

`X_L=√(42.10^2-18.54^2)`

`X_L=37.79`

`X_L=2\pi f L`

`L=(X_L)/(2\pi f)`

`L=(37.79)/(2\pi * 60)`

L= 0.10024 H

L = 100.24 mH

This is the inductance .