A pure inductor A is connected to an AC source of 220 V. When the inductor A is replaced with another pure inductor B, then the current in the circuit becomes double. Find the ratio of the inductive reactance of the inductor A to inductor B.

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,

\(⇒ I=\frac{V}{X_L}\)

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

CALCULATION:

Given VA = V_B = V = 220 V, f_A = f_B = f, and IB = 2IA

• We know that the inductive reactance is given as,

⇒ XL = 2πfL      -----(1)

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

\(⇒ I=\frac{V}{X_L}\)     -----(2)

By equation 1 and equation 2,

\(⇒ I=\frac{V}{2πfL}\)     -----(3)

By equation 3, the current in inductor A is given as,

\(⇒ I_A=\frac{V}{2πfL_A}\)     -----(4)

By equation 3, the current in inductor B is given as,

\(⇒ I_B=\frac{V}{2πfL_B}\)     -----(5)

∵ IB = 2IA     -----(6)

By equation 4, equation 5, and equation 6,

\(⇒ \frac{V}{2πfL_B}=2\times\frac{V}{2πfL_A}\)

\(⇒ \frac{L_A}{L_B}=2\)

• Hence option 1 is correct.