A coil takes a current of 10.0 A and dissipates 1410 W

Question

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.

Answer 1

230*10*CosA = 1410.
CosA = 0.6130 = Power factor.
A = 52.2o. = Phase angle.

I^2*R = 1410W.
10^2 * R = 1410.
R1 = 14.1 Ohms. = Resistance of coil.

Tan52.2 = Xl/R = Xl/14.1
Xl = 14.1*Tan52.2 = 18.2 Ohms=Reactance
of coil.

Z1 = R1/CosA = 14.1/Cos52.2 = 23 Ohms.

Parallel:
Zp = 230/20 = 11.5 Ohms.

CosA = 0.866.
A = 30o

Zp = 11.5[30o] Ohms.
Rp = 11.5*Cos30 = 9.96 Ohms.
Xp = 11.5*sin30 = 5.75 Ohms.

1/14.1 + 1/R2 = 1/9.96.
1/R2 = 1/9.96-1/14.1 = 0.0295.
R2 = 33.9 Ohms.

1/X2 + 1/18.2 = 1/5.75.
1/X2 = 1/5.75 - 1/18.2 = 0.1190.
X2 = 8.41 Ohms.

Series:
Z = (R1+R2) + j(X1+X2).
Z = (14.1+33.9) + j(18.2+8.41).
Z = 48 + j26.6 = 54.9 Ohms[29o].

I = E/Z = 230/54.9 = 4.2 Amps.

Pf = Cos29o = 0.875.

Answer 2

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