Circuit phasor method topic 4, circuit phasor method topic

Solution: i1=2cos(-t-100°)=2sin[90°-(t-100°)]=2sin( t+190°);

i2=-4sin[90°-（t+190°）]=-4sin（-ωt-100°）=4sin（ωt+100°）；

i3=5sin（ωt+10°）。

So: i1 (phasor) = 2 2190° = 2 190° = 2 (cos190° + jsin190°) = 2 (;

i2 (phasor) = 4 2 100° = 2 2 (cos100° + jsin100°) = 2 (;

i3 (phasor) = 5 2 10° =.

By KCl, I (phasor) = I1 (phasor) + I2 (phasor) + I3 (phasor) = 2 (.

So: i=.

The phase angle is the subtraction of the numerator and denominator phase angles, and the modulus length is the division of the numerator and denominator phase angles, and the subtraction of the phase angles in this question is -240°, which is equivalent to 120°

<> solution: The result of the answer is correct, but the solution process is flawed, and the phasor diagram of the three currents should be the above figure.

i1 = i2 = i3, and i1 (phasor) + i2 (phasor) = i3 (phasor), so i1 (phasor), i2 (phasor), and i3 (phasor) form an equilateral triangle.

Regarding the position of i1 (phasor) and i2 (phasor), there is an error in the answer.

RL and RC branches are connected in parallel, the voltage at the social end is U (phasor), and naturally the phase of I1 (phasor) is ahead of U (phasor), and the phase of I2 (phasor) lags behind U (phasor), so the phase of I1 (phasor) is ahead of I2 (phasor), so the position of I1 (phasor) and I2 (phasor) should be as shown in the figure above. Namely:

Let i3 (phasor) = 10 0°A, then: i1 (phasor) = 10 60 °A, i2 (phasor) = 10 -60 °A.

So: U (phasor) = i1 (phasor) (r-jxc) = i2 (phasor) (r+jxl).

10∠60°×（r-jxc）=10∠-60°×（r+jxl）。

cos60°+jsin60°）（20-jxc）=（cos60°-jsin60°）（20+jxl）。

cos60°=,sin60°=,so:

xl=xc=20√3（ω）

U23 (phasor) = i2 (phasor) r + i3 (phasor) r = (10 -60° + 10 0°) 20 = 300 - j100 3 = 200 3 -30 ° (v).

So: u23=200 3(v).

U12 (phasor) = i1 (phasor) R-i2 (phasor) r = (10 60°-10 -60°) 20 = J200 3 = 200 3 90° (V), so: U12 = 200 3V.

U13 (phasor) = i1 (phasor) r + i3 (phasor) r = (10 60° + 10 0°) 20 = 300 + J100 3 = 200 3 30 ° (V).

Therefore: U31 (phasor) = 200 3 -150 ° (V). u31=200√3（v）。

From the phasor results, it can be seen that these three voltages constitute a triad symmetrical line voltage.

US (phasor) = i1 (phasor) (r-jxc) + i3 (phasor) r=10 60° (20-j20 3)+10 0° 20=10 60° 40 -60°+200=400+200=600 0°(v).

That is: us=600v.

Wherein, the parallel road voltage: u (phasor) = 400 0°.

Apparent power: s=

Active power: p=scos

Reactive power: q=ssin

The square of sin + the square of cos = 1, it can be seen from this formula that the reactive power is large and the active power is small, and the opposite is true.

The initial phase angle of the three-phase electricity is not the same, so it is possible to reverse the phase sequence, and it is OK.

It's rare to come across this kind of topic. XL=J1, XC=-J1, before T1, the circuit is in a parallel resonant state, and the voltage and current on R are all 0, so IL=US J1, that is, IL=, when T=T1, we can find T1+ =. When T1 is opened, the circuit is a first-order circuit, and IL has two components, one is the forced component (steady-state solution) and the other is the free component (initial value - the steady-state value at this time), similar to the three-element method.

Steady-state solution, IL'=us (1+J1), i.e. il'(t)= *cos(ωt+φ-45°)，il'(t1)=。The initial value question has been given il(t1)=, and the time constant t=l r=1 1=1 s. Finally, when t>t1.

il(t)=cos(ωt+φ-45°)+

Solution: BAI resistors consume power.

dup=2000=i1 r=i1 20, so: i1=10a, then: i=i1=i2=10a.

Let u2 (phase.

zhi quantity dao) version = u2 0°, then: i2 (phasor) = 10 -90° = -j10 (a).

According to the weight kcl: i (phasor) = i1 (phasor) + i2 (phasor), and i = i1 = i2, therefore, the three phasors form an equilateral triangle, and the phasor diagram is as follows:

Explanation: If the phase of U2 (phasor) is determined, then the phase of I2 (phasor) is also determined; But the relationship between i1 (phasor) and i (phasor) from an equilateral triangle may also be in the third quadrant; However, the phase of the current i1 (phasor) of the RC series branch is at most 90° ahead of the phase of its voltage u2 (phasor) = u2 0°, and cannot be larger, so i1 (phasor) cannot be in the third quadrant, and the triangle formed cannot be in the third quadrant.

Thus, we get: i1 (phasor) = 10 30° (a), i (phasor) = 10 -30° (a).

Therefore, we get: U1 (phasor) = I (phasor) (jxc1) = 10 -30° (J20) = 10 -30° 20 -90° = 200 -120° (V).

Due to the lack of conditions in the question, u2 (phasor) cannot be calculated, and therefore u (phasor) cannot be calculated.

I took a look and there is no better way than you, the total impedance calculated from the result is capacitive, and the condition given in the question for the capacitive nature of the circuit is redundant. If it is given sensibility, it is wrong.

<> solution: Let the node voltage be u (phasor). —The problem uses cosine phasor.

Because the capacitance and current are connected in series, so: UC (phasor) = is (phasor) (JXC) = 10 30° 20 -90° = 200 -60° (V).

Controlled voltage source voltage: phasor) = 100 -60°(v).

List the nodal voltage equations:

U (phasor) + phasor)] 10 + U (phasor) 10 = is (phasor).

2U (phasor) +100 -60°=10 10 30°.

U (phasor) = 50 30°-50 -60° = 50 ( 3 2 + J1 2-1 2 + J 3 2) = 50 [(3-1) 2 + J( 3 + 1) 2] = 50 2 75° (V).

So: i1 (phasor) = u (phasor) + phasor)] 10 = (50 2 75° + 100 -60°) 10 = 5 2 75° + 10 -60° =.

So: i1(t)=5 2 2cos(5000t-15°)=10cos(5000t-15°a).

i2 (phasor) = u (phasor) 10 = 50 2 75° 10 = 5 2 75° (a).

So: i2(t)=5 2 2cos(5000t+75°)=10cos(5000t+75°a).

You're a conceptual error.

1. The phasor method is a mathematical method that is artificially "created" in order to liberate you from the sum and difference of complex trigonometric functions and the product and difference when solving the problem of AC circuits (remember, this is just mathematics, there is no such physical quantity as "phasor" in real life), and phasor and complex numbers correspond one-to-one, so in the end, the phasor method becomes a complex number operation. In your first question, the phasor of the current i is a complex equation, and you say that you can just turn a complex number into a real number instead? (Complex numbers have real and imaginary parts), so it is not possible to do that, and honestly calculate complex numbers.

2. That's just a coincidence, even if you count correctly, the method is wrong.

Vector column equations represent vectors as complex numbers.

The following answers are in italics indicating phasors.

Solution: Let the voltage phasor of V1 be: U1=100 0°V, then the IC1 of A1=10 90°(A).

The impedance of the RL series loop is Z1=R+JXL=5+J5=5 2 45°(

Therefore, the current of the RL series loop is: I1=U1 Z1=100 0° 5 2 45°=10 2 -45°=10-J10(A).

The trunk current is: I=I1+IC1=10 90°+10-J10=10(A), that is, the reading of the ammeter A0 is: 10A.

uc=i×（-jxc）=10×（-j10）=-j100（v）。

The total voltage is: U=UC+U1=-J100+100 0°=100 2 -45°(V).

So the voltmeter v0 reads 100 2=.

<> solution: Let u (phasor) = u 0° and the angular frequency be .

xc1=1/（ωc1）=1/（1×ω/1000000）=1000000/ω（ωxc2=1/（ω×100/1000000）=10000/ω（ω

z1=2-jxc1=2-j1000000/ω，z2=8∥（-jxc2）=8×（-j10000/ω）/（8-j10000/ω）=-j80000/（8ω-j10000）。

u1(phasor)=u(phasor) z2 (z1+z2)=u(phasor) (1+z1 z2).

To make u1 (phasor) and u (phasor) in phase, the denominator of the above equation should be a real number, i.e., im(z1 z2)=0.

z1/z2=（2-j1000000/ω）/[-j80000/（8ω-j10000）]=。

So: 16 -10 10 =0, =10 10 16, =10 5 4=25000(rad s).