Firstly, we will draw
RL series circuit
ET = 120 V, R = 40 Ω, and XL = 30 Ω
Firstly, we will find current
[tex]I=\frac{E_T}{R+jX_L}[/tex]
now, we can plug values
and we get
[tex]I=\frac{120}{40+j30}[/tex]
now, we can find voltage across L
[tex]E_L=IjX_L[/tex]
so, we get
[tex]E_L=\frac{120}{40+j30}(j30)[/tex]
now, we can find absolute value
[tex]|E_L|=120\frac{30}{\sqrt{40^2+30^2} }[/tex]
[tex]|E_L|=120\frac{30}{ 50 }[/tex]
[tex]|E_L|=120\frac{3}{ 5 }[/tex]
[tex]|E_L|=72V[/tex]...............Answer
