Question

Determine the amplitude of the line current in a three voltage of 300 V that supplies 1200 W to a delta connected load at a lagging PF of 0.8; then find the phase impedance​

Answer 1

Explanation:

Let us again consider a single phase. It draws 400 W, 0.8 lagging PF, at a 300 V line voltage. Thus,

400=300\left(I_{p}\right)(0.8)400=300(I p )(0.8)

and

l_{p}=1.667 AIp

=1.667A

and the relationship between phase currents and line currents yields

I_{L}=\sqrt{3}(1.667)=2.89 AI

L = 3

(1.667)=2.89A

Next, the phase angle of the load is \cos ^{-1}(0.8)=36.9^{\circ}cos

−1 (0.8)=36.9 ∘

, and therefore the impedance in each phase must be

Z _{p}=\frac{300}{1.667} \angle 36.9^{\circ}=180 \angle36.9^{\circ} \Omega

Zp= 1.667/ 300 ∠36.9 ∘

=180∠36.9 ∘ Ω

PRACTICE

12.7 Each phase of a balanced three-phase Δ-connected load consists of a 200 mH inductor in series with the parallel combination of a 5 μF capacitor and a 200 Ω resistance. Assume zero line resistance and a phase voltage of 200 V at ω = 400 rad/s. Find (a) the phase current; (b) the line current; (c) the total power absorbed by the load.

Ans: 1.158 A; 2.01 A; 693 W.