Question

A current of 1A flows through a coil when it is connected across a DC battery of 100V. If the DC battery is replaced by an AC source of 100 V and angular frequency of 100 rad s^-1, the current reduces to 0.5 A. Find

1. the impedance of the circuit.
2. self-inductance of coil.
3. Phase difference between the voltage and the current.

Answer 1

Given: I_DC = 1A, V = 100V, ω = 100 rad/sec.

I_DC = `V/R ⇒ 100/R = 1`

R = 100 Ω

The a.c. flowing,

I_AC = 0.5 A

1. Impedance: Z = `sqrt(R^2 + (omegaL)^2`
   when AC is applied.
   Z = `V/I = 100/0.5 = 200 Omega`
2. Self-inductance of the coil:
   We know `Z^2 = R^2 + X_L^2`
   `X_L^2 = Z^2 - R^2`
   = `(200)^2 - (100)^2`
   = 40000 - 10000
   X_L = 173.21 Ω
   X_L = ωL
   ⇒ L = `X_L/omega = 173.21/100`
   L = 1.73 H
3. The phase difference between V and I, (cosΦ):
   
   `cosphi = R/L = 100/1.73`
   `cosphi = 57.73`
   `phi = cos^-1(57.73)`