Calculate the self-inductance of the coil by direct method by using the following data :

`{(1,1.0,1.5,0.4,1.5),(2, 1.3, 2.0, 0.6, 2.0):}`
Frequency of AC `=50` Hz.

Explanation

Formula : `L = (sqrt(Z^(2)-R^(2)))/(2 pi f)`
Using `R=V/I`, Calculating `R_(1) = 1.50 Omega`
`R_(2)=1.54 Omega`
& Mean `R = 1.52 Omega`
Using `Z= V/I`, Calculating `Z_(1) = 3.75 Omega`
`Z_(2)=3.33 Omega`
Mean `Z=3.54 Omega`
Calculation of `L=0.01019` H
Deltailed Answer :
For trial number (1) D.C. part,
`R=V/I`
`rArr R_(1)=(15)/(1) = 1.5 Omega`
and for trial number (2) D.C. part
`R_(2)=(2.0)/(1.3) = 1.54 Omega`
Mean `R=(1.5+1.54)/(2) = 1.52 Omega`
for A.C. part
`Z_(1) = (1.5)/(0.4) = 3.75 Omega`
`Z_(2)=(2.0)/(0.6) = 3.33 Omega`
Mean `Z=3.54 Omega`
Now `L= (sqrt(Z^(2)-R^(2)))/(2 pi f)`
`rArr L = (sqrt((3.54)^(2)-(1.54)^(2)))/(2 xx 3.14 xx 50)`
`L=0.01019 H`.