Question

A coil of self inductance 0.5h and resistance 100 ohm is connected to a 240v,50hz ac supply a)what is the maximum current in circuit

Answer 1

Given, L= 0.5H, R= 100Ω,

Supplied voltage, E= 240V, and frequency f= 50hz

We know, E= E₀/√2,

                 E₀= E×√2= 240×1.414= 339.36V

Total Impedance of the circuit

X= √(R²+XL²)

=>X=√{(100²+(ωL)²}

     = √{(100²+(100π×0.5)²}

     = 186Ω

Now, maximum Current, I₀=E₀/X

                                            =339.36/ 186

                                            = 1.82 A