Question

A series RL circuit contains two resistors and two inductors. The resistors are 27Ω and 47Ω. The inductors have inductive reactances of 50 Ω and 40 Ω. The applied voltage is 120 V. What is the voltage drop on the inductor that has 40 Ω of reactance?

Answer 1

voltage drop across 40Ω inductor is 41.2V

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=
`\frac{E_(T)} {Z}`

Given Applied voltage,
`E_(T)=120V`

Hence
`I=\frac{120}{\sqrt{74^(2)+90^(2)}} =(120)/(√(13576)) =1.03A`

Therefore voltage drop on the inductor that has 40Ω of reactance=40X1.03

=41.196≈41.2V