Answer:
voltage drop across 40Ω inductor is 41.2V
Step-by-step explanation:
We are given that a series RL circuit contains two resistors 27Ω, 47Ω and two inductors which have reactances 50Ω,40Ω.
Hence total resistance in the circuit =27+47=74Ω
Total reactance=50+40=90Ω
Since all of these elements are in series, same current goes through all the elements.
And that current I=[tex]\frac{E_{T}} {Z}[/tex]
Given Applied voltage,[tex]E_{T}=120V[/tex]
Hence [tex]I=\frac{120}{\sqrt{74^{2}+90^{2}}} =\frac{120}{\sqrt{13576}} =1.03A[/tex]
Therefore voltage drop on the inductor that has 40Ω of reactance=40X1.03
=41.196≈41.2V
