I urgently ask the circuit teacher to help solve a circuit problem Thank you teacher 60

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<> solution: is (phasor) = 10 2 30° = 5 2 30° (a). ω=1000rad/s。

p1=u²/r1，u=√（p1×r1）=√（80×5）=20（v）。

i1=u/5=20/5=4（a），i3=20/20=1（a）。

is (phasor) = i1 (phasor) + i2 (phasor) + i3 (phasor).

Wherein: i1 (phasor), i3 (phasor) and u (phasor) are in phase, then: i1 (phasor) + i2 (phasor) are also in phase with u (phasor); I2 (phasor) lags behind u (phasor) by 90°.

So i1 (phasor) + i3 (phasor) and i2 (phasor) are perpendicular to each other, so according to the phasor diagram:

is=√[（i1+i3）²+i2²=√[（4+1）²+i2²]=√25+i2²=5√2。

i2=5（a）。

1）xl=u/i2=20/5=4（ω）l=xl/ω=4/1000=。

2) Total impedance of the circuit: Z=5 20 J4=4 J4=4 J4 (4+J4)=16 90° 4 2 45°=2 2 45°(

U (phasor) = is (phasor) z = 5 2 30° 2 2 45° = 20 75° (v).

u（t）=20√2sin（1000t+75°） v）。

3）u=20v，φu=75°；is=5√2a，φi=30°。φ=φu-φi=75°-30°=45°。

p=uiscosφ=20×5√2×cos45°=100（w）。

q=uissinφ=20×5√2×sin45°=100（var）。

Power factor: cos = cos 45 ° = 2 2 = .

<> solution: t=0-, the circuit is in a steady state, and the capacitance is equivalent to an open circuit, as shown in the figure above.

So: uc(0-)=us r3 (r1+r3)=100 10 (10+10)=50(v).

Commutation theorem: UC(0+)=UC(0-)=50V, that is, when T=0+, the capacitance phase is pinned in a 50V voltage source:

So: ic(0+)=us-uc) (r1+r2)=(100-50) (10+20)=5 3(a).

t= , the circuit enters the steady state again, and the capacitance is equivalent to an open circuit. Below:

Obviously: ic( )0,uc( pants balance)=100V.

The voltage source is short-circuited, looking at both ends of the capacitor, the equivalent resistance is: r=r2+r1=20+10=30 ( ) The circuit time constant is: =rc=30 50 1000000=.

Three-element method: uc(t) = uc( )uc(0+)-uc( )e (-t)100+(50-100)e (-t(v).

ic（t）=0+（5/3-0）e^（-t/ （a）。

The change curve is as follows:

<> solution: is (phasor) = 10 0°A, =10rad s. (Cosine phasor).

Disconnect the branch of the 5 series capacitor c from the circuit, and the breakpoint bucket sum is a and b.

z1=1∥j1=1×j1/（1+j1）=1/√2∠45°（ω

i1 (phasor) = is (phasor) z1 1 = 10 0° 1 2 45° 1 = 5 2 45° (a).

UOC (phasor) = UAB (phasor) = is J2 + 4I1 (phasor) = 10 0° 2 90° + 4 5 2 45° = J20 + 20 + J20 = 20 + J40 = 20 5 .

Disconnect the current source from the circuit. Apply voltage U0 (phase-free targeting) from ports A and B, and set the inflow current to I0 (phasor). Above.

Obviously i1 (phasor) = 0, 4i1 (phasor) = 0, zeq = u0 (phasor) i0 (phasor) = j2 ( ).

The Thevenin equivalent circuit of the circuit is as follows:

1) Obviously, when Z=R+J(XL-XC)=5+J(2-XC)=5, the RMS value of i (phasor) is the largest, that is, the RMS value of i is the largest. At this point, xc=2=1 (c)=1 (10c), c=1, 20=.

2) I (phasor) is 90° ahead of is (phasor) in the phasor, then: i (phasor) = i 90°(a).

i(phasor) = uoc (phasor) z=20 5 .

Then: 90°+arctan[(2-xc) 5]=.

arctan[（2-xc）/5]=，（2-xc）/5=tan（。

xc=2+5×，c=1/（10×。

<> 4 35°A current sources are connected in parallel with 50 resistors, equivalent to 4 50 35°=200 35°V voltage sources, connected with 50 resistors.

Disconnected from terminals A and B, then: UOC (phasor) = 200 35°V, Zeq = 50.

The net n in the dotted box has an equivalent impedance of :

According to the maximum power transfer theorem, when:

That is, when the conjugate complex of the impedance in the equivalent power supply is spring or , z can obtain the maximum power, so

Substitution: <>

xl=50ω。

Pmax=uoc (4req)=200 (4 50)=200 (w).

<> solution: t=0-, the circuit is in a steady state, and the capacitance is equivalent to an open circuit, so:

uc（0-）=us×r2/（r1+r2）=10×3/（2+3）=6（v）。

According to the commutation theorem, UC(0+)=UC(0-)=6V, i.e., when T=0+, the capacitance is quite worn to a 6V voltage source.

ic（0+）=10-6）/r1=4/2=2（ma）。

ur1（0+）=ic（0+）×r1=2×2=4（v）。Or: UR1(0+)=US-UC(0+)=10-6=4(V).

Solution: uc(0-)=0, according to the commutation theorem: uc(0+)=uc(0-)=0, so when t=0+, the capacitance is equivalent to a 0V voltage source, which is equivalent to a short circuit.

At this point, R1 is short-circuited, so: ic(0+)=us R2=10 100=.

UC(0+)=UC(0-)=US x R2 (R1+R2)=10 x 3 5=6V, IC(0+)=IC(0-)=0, Socks Touch UR(0+)=US-UC(0+)=4V. Search for dust.

S is UC=0, i.e. IC(0-)=IC(0+)=0.

When symmetrical loads are connected in a star shape, the voltage across each load is the phase voltage u (the voltage between the live and neutral wires). Then their total power p is equal to: p = 3 u r When symmetrical loads are connected in a triangle, the voltage across each load is the line voltage u' (the voltage between the live wires).

Then their total power p'It's equal to: p' =3× u'r because u'=3 u, so u' =3 u .So:

p ' 3×(3u)/r = 3×(3×u/r)= 3×p = 330 kw。Hope mine can help you!

Hello, I've got you covered! When symmetrical loads are connected in a star shape. The voltage at both ends of each load is the phase voltage, the voltage between the live and neutral wires, then their total power is equal to when the symmetrical loads are connected in a triangle fashion. Each load at both ends.

=2 f=2, then: xc=1 ( c)=1 (6280 .

i(phasor) = u(phasor) (r-jxc), uc(phasor) = i(phasor) (jxc).

So: uc (phasor) u (phasor) = (-jxc) (r-jxc) = xc (xc + jr) = [xc (xc +r )]arctan(-r xc).

UC (phasor) lag U (phasor) 60°, i.e.: arctan (-r xc) = -60°.

So: r xc = tan 60°, r = xctan 60° =.

Will you ask the teacher to solve it when you encounter the circuit, this should be a diagram, and you should find some electricians about the specific diagram.

Look at the picture, please see the picture for the detailed process. Problems 1-15 use the column mesh current equation to get r=, and the Kirchhoff method is used to get ux=.

Just unplug the net and you're good to go! Or copy this Weibo to open your phone**.