Question

The potential difference measured across a coil is 20 V
when a direct current of 2 A is passed through it. With
an alternating current of 2 A at 40 Hz, the p.d. across
the coil is 140 V. If the coil is connected to a 230 V,
50 Hz supply, calculate: (a) the current; (b) the active
power; (c) the power factor

Answer 1

R = E/I = 20/2 = 10 Ohms.

Xl = E/I = 140/2 = 70 Ohms @ 40 Hz.= The
inductive reactance.

Xl == 2pi*F*L = 70 Ohms.
6.28*40*L = 70.
L = 0.279 Henrys. = The inductance.

a. Xl = 2pi*F*L = 6.28*50*0.279 = 87.5
Ohms.
I = E/Xl = 230/87.5 = 2.63 Amps.

b. Tan A = Xl/R = 87.5/10 = 8.75.
A = 83.5o = Phase shift angle.

P = E*I*Cos A =

Answer 2

A coil takes a current of 10.0 A and dissipates 1410 W
when connected to a 230 V, 50 Hz sinusoidal supply.
When another coil is connected in parallel with it,
the total current taken from the supply is 20.0 A at a
power factor of 0.866. Determine the current and the
overall power factor when the coils are connected in
series across the same supply