Answer:
Current amplitude, I = 5.57 A
Explanation:
It is given that,
Effective resistance, R = 22 ohms
Inductive reactance, L = 72 ohms
Voltage of the alternating source, V = 420 volt
The total impedance of the RL circuit is given by :
[tex]Z=\sqrt{R^2+L^2}[/tex]
[tex]Z=\sqrt{(22)^2+(72)^2}[/tex]
Z = 75.28 ohms
From Ohm's law,
[tex]V=I\times Z[/tex]
[tex]I=\dfrac{V}{Z}[/tex]
[tex]I=\dfrac{420}{75.28}[/tex]
I = 5.57 A
So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.