College circuit theory test questions

Solution: Italic letters indicate phasors. Let UA = 100 ∠ 0 V, then UB = 100∞- 120, UC =100 ∠120 V.

za = 4+J3 = 5∠36.87(ω)，Zb = 8+j6 = 10∠36.87(ω)，Zc = 5∠36.87ω。

Ia = ua/za =100 ∠ 0/5 ∠ 36.87 = 20 ∞-36.87 =16-j12 (a), that is, Ia=20A.

IB = UB/ZB = 100∞- 120/ 10∠36.87 = 10∞- 156.87 =-9.65438。

IC = UC/ZC = 100∠ 120/5∠36.87 = 20∠83. 13 = 2.392+j 19.856(A)。

Neutral current: I = ia+IB+IC =16-j12-9.196-j3.928+2.392+j19.856 = 9.1.

A-phase active power: pa = ua× ia× cos φ a =100× 20× cos (0+36.87) =1600 (w);

Phase B active power: Pb = UB× IB× COS φ b =100×10× COS (-120+156.87) = 800 (w);

Active power of phase c: PC = UC× IC× cos φ c =100× 20× cos (120-83.13) =1600 (w).

Total active power of the circuit: p = pa+Pb+PC =1600+800+1600 = 4000 (w).

Or: P=Pa+Pb+Pc=Ia? ×Re(Za)+Ib？ ×Re(Zb)+Ic？ ×Re(Zc)=20？ ×4+ 10? ×8+20? × 4 =1600+800+1600 = 4000 (width).

If the neutral line is disconnected, the neutral point potential is no longer zero.

ya = 1/Za = 0.2∞-36.87 = 0. 16-j 0. 12(S)，Yb = 0. 1∞-36.87 = 0.08-j 0.06(S)，Yc = 0.2∞-36.87 = 0. 16-j 0. 12

Neutral point potential: u 'o = (ua× ya+ub× Yb+UC× YC)/(ya+Yb+YC) = (100 ∠ 0× 0.2 ∞-36.87+100 ∞. (0. 16-j 0. 12+0.08-j 0.06+0. 16-j 0. 12)=(20∞-36.87+ 10∞- 156.87+20∠83. 13)/(0.4-j 0.3)=。

Phase B load voltage: u 'bo = UB-u 'o =100 ∞-120 ∠ 60 =-50-j86.6-10-j17.32 =

The phase B load voltage rises (from 100V to 120V), so the phase B current increases.