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Circuit theory test questions

Italics indicate phasors.

Induced current: IL = IS-I = 2 ∠ 0-1∠ 30 = 2-∠ 3/2-j0.5 = (2-∠ 3/2)-j0.5 (a).

UL = il×j40 =[(2-√3/2)-j 0.5]×j40 = 20+j(80-20√3)(V).

Capacitance voltage: UC = i× (-j40) =1∠ 30× 40 ∞-90 = 40 ∞-60 = 20-j20 √ 3 (v).

The network terminal voltage n: u = UC-ul = 20-j20 √ 3-20-j (80-20 √ 3) =-j80 (v).

The current of n is phasor IL=(2-√3/2)-j0.5(A), and the complex number of * * * yoke is IL'=(2-√3/2)+j0.5(A). The voltage phasor is U=-j80V, so:

Complex power of n: s' = u× il' =-j80× [(2-√ 3/2)+j0.5] = 40-j (160-40 √ 3) = p+jq.

So: P=40W, choose D. You can calculate without a calculator.