Aseries 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?

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=

Given Applied voltage,

Hence

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

                                                                           =41.196≈41.2V

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.

nlgqer

step-by-step