A 220 V source of AC supply is applied across a pure inductor of 35 mH. If the frequency of the source is 50 Hz, then find the rms current in the circuit.

CONCEPT:

Inductive reactance:

• The inductive reactance is the opposition offered by the inductor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The inductive reactance is given as,

⇒ XL = 2πfL

Where f = frequency of ac current and L = self-inductance of the coil

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure inductor, the inductive reactance is equal to the impedance.

AC voltage applied to an inductor:

• When an AC voltage is applied to an inductor, the current in the circuit is given as,

• In a pure inductor circuit, the current reaches its maximum value later than the voltage by one-fourth of a period.

CALCULATION:

Given V_rms = 220 V, L = 35 mH = 35×10-3 H, and f = 50 Hz

• We know that the inductive reactance is given as,

⇒ XL = 2πfL

⇒ XL = 11 Ω      -----(1)

When an AC voltage is applied to an inductor, the current in the circuit is given as,

     -----(2)

By equation 1 and equation 2, the rms current in the circuit is given as,

⇒ I_rms = 20 A

• Hence option 3 is correct.