Answer:
Maximum current in the inductor will be 3.824 A
Explanation:
In first case inductive reactance = 59.2 ohm
Frequency = 60 Hz
We know that inductive reactance is given by [tex]X_L=\omega L[/tex]
[tex]59.2=2\pi f\times L[/tex]
[tex]59.2=2\times 3.14\times 60\times L[/tex]
[tex]L=0.157H[/tex]
In second case frequency f = 45 Hz
Now inductive reactance [tex]X_L=\omega L =2\times 3.14\times 45\times .157=44.368ohm[/tex]
Now current [tex]i=\frac{V}{X_L}=\frac{120}{44.368}=2.70A[/tex]
Maximum current [tex]i_{max}=\sqrt{2}i=1.414\times 2.70=3.824A[/tex]
