Question

A coil possessing both inductance L and resistance R is connected to a 24V dc supply having negligible internal resistance. The dc current in this circuit is found to be 3A. When the coil is connected to a 24V, 50Hz ac supply with negligible internal....?

....impedance, the circuit current is found to be 0.8 A. Find:

(a) the resistance of the coil.
(b) the inductance of the coil.

Answer 1

As we are considering inductance `L` and resistance `R` of the same coil, it becomes a series `LR` circuit.

(a) `24\ V` dc supply connected.
Inductance does not play any role. The current in the coil is given by

`I_(DC)=V_(DC)/R`

Inserting given values we get

`3=24/R`
`=>R=24/3=8\ Omega`

(b) `24\ V, 50\ Hz` ac supply connected.
Circuit impedance `Z` is given by the expression

`Z^2=R^2+X_L^2`
where `X_L=omegaL`, is inductive reactance of the coil. Now the current in the coil becomes
`I_(AC)=V_(AC)/Z`

Inserting various values we get

`0.8=24/sqrt(8^2+(2pixx50xxL)^2)`

Rearranging and squaring both sides

`8^2+(2pixx50xxL)^2=(24/0.8)^2`
`=>(2pixx50xxL)^2=(24/0.8)^2-8^2`
`=>L=sqrt(836)/(100pi)`
`=>L=0.9\ H`