A 493 H inductor with negligible resistance is connected acr

A 4.93 H inductor with negligible resistance is connected across an ac source whose voltage amplitude is kept constant at 58.0 V but whose frequency can be varied. Find the current amplitude when the angular frequency is 1060 rad/s.

Solution

angular frequency, w = 1060 rad/s

reactance of inductance, XL = w L = 1060 x (4.93) = 5225.8 ohm

there in only inductor in circuit.

hence impedance , Z = XL = 5225.8 ohm

I = V / Z = 58 / 5225.8 = 0.0111 A    Or 11.1 mA ......Ans