When 100 V d.c. is applied across a coil, a current of 1 A flows through it. When 100 V a.c. of 50 Hz is applied to the same coil only 0.5 A flos. The inductance of the coil is

Options

A. 5.5 mH
B. 0.55 mH
C. 55 mH
D. 0.55 H

Answer

D

Explanation

Resistance of the coil,
`R=(V)/(I)=(100V(D.C))/(1A)=100Omega`
Now, with 100 V a.c., ile. Addition to the resistance of the coil, the inductance of the coil will also offer resistance to the flow of current.
Let L be the inductance of the coil.
`therefore`Inductive resistance, `X_(L)=omegaL=2pifL`
`therefore`Impendence`=Z=sqrt(R^(2)xxX_(L)^(2))=sqrt(R^(2)+(2pifL)^(2))`
`therefore`Current, `I=(V)/(Z)=sqrt(V)/(sqrt(R^(2)+(2pifL)^(2)))`
[Frequency, f=50Hz]
`therefore0.5=(100)/({(100)^(2)+(2pixx50xxL)^(2)}^(1//2))`
or, `2500+24649L^(2)=10000`
`thereforeL=sqrt((10000-2500)/(24649))=0.55H`.