The generator has a circuit voltage of 220V, an output power of 4400W, and an output resistance of 2

Sophomore physics haha.

Line = p u = 4400 220 = 20a

So U loss = i line r = 20x2 = 40V

So u user = u-u loss = 220-40 = 180 vp loss = i line r = 20 x2 = 800w

Therefore, p = p-p loss = 4400-800 = 3600W: 10, then the output voltage of the transformer becomes the original 10 U loss'=1 10 loss = 4v, so u user = 220-4 = 216v

P-loss'=1 100p loss = 36w

So p user = 4400-36 = 4364w

Is your load purely resistive? If not, what is the nature?

If it is a purely resistive load, the load resistance is 11 ohms, the total resistance is 13 ohms, the total power is, the output power obtained by the load is 186V.

If the voltage is increased from 220V to 2200V with the same resistance of the load (still assuming a purely resistive load), the voltage rises proportionally to 1860V and the power is proportional to 315kW.

In fact, not all loads have linear power consumption. If it is a constant power load, the power is still the same, and the voltage applied to the load is 2197V.

This problem is too theoretical, and the conversion efficiency of the transformer is not considered at all.

(1) According to p=ui, i=p u

220a=200a

Then the voltage loss on the transmission line u loss = IR = 200 , power loss p

loss = i2 r = 40000

So the voltage obtained by the user u user = u-u loss = 220-80v = 140v, and the power p obtained by the user = p-p loss = 44000-16000w = 28000w

2）u u1

Get u1 = 2200vi i

i1 = 20a

Then the voltage loss on the transmission line u loss = i1 r = 20 power loss p

loss = i2 r = 400

Therefore, the input voltage of the step-down transformer U2 = U1 -U loss = 2200-8V = 2192V

The input power of the step-down transformer P2 = P-P loss = 44000-160W = 43840WU2U user

Then the voltage that the user gets u user =

The input and output power of the transformer are equal.

So the power that the user gets p user = p2 = 43840w

The output voltage of the generator is 220V, it can be seen that this voltage is single-phase alternating current, and two wires are used for single-phase alternating current transmission, one is the live wire and the other is the neutral wire, and the two wires are connected in series from the circuit point of view, so the total resistance is multiplied by 2

Because the power line doesn't just have to go into the user, it has to come out of the user's home and back to the power plant, so it's multiplied by two.

(1) According to p=ui, i=p u

a＝200a

then the voltage loss on the transmission line is u loss.

IR=200, power loss P loss. i

r40000×

So the user gets the voltage u user.

u-u loss. 220-80V = 140V, the power that the user gets p the user.

P-P loss. 44000-16000w=28000w 2)uu12200vii

i1 = 20a

then the voltage loss on the transmission line is u loss.

i1r=20×

Power loss P loss.

ir＝400×

Therefore, the input voltage of the step-down transformer u2u1u loss.

2200-8v=2192v

The input power of the step-down transformer is p2

P-P loss. = 44000-160w = 43840wuu users.

Then the voltage that the user gets u user =

The input and output power of the transformer are equal.

So the user gets the power p user.

p243840w．

p ui, the voltage ratio of the secondary coil is equal to the ratio of the number of turns of the coil, the voltage of the secondary coil is 2200V, the current is equal to 44000 2200 20A, P i 2R, P 80W, and the user gets 44000-80 43920W. You don't need to change the pressure to ask directly.,Digging fingers is about the same as above.,I'm so tired of playing my phone to give one!

1）25a；（2）1：16；（3）190：11。

Test question analysis: due to the voltage of the original coil of the step-up transformer U1250V, the current passed: <

a, power loss of power transmission line: <>

According to the power lost on the transmission line in relation to the current and resistance, it can be obtained:

Current on the transmission line (current through the secondary coil of the step-up transformer, the primary coil of the step-down transformer):

A 2) Since the current of the original coil at the power supply is i1

400A, the current passing through the secondary coil is i2

25A, so the ratio of turns of the primary and secondary coils of the step-up transformer is <>3) because the current in the original coil of the step-down transformer is i3

I225A, the step-down transformer sub-coil current (user current) is:

Therefore, the ratio of turns of the primary and secondary coils of the step-down transformer is <>

Active power p=

Rated current i=p

i=u r refers to the pure resistance single-phase circuit of DC circuit or alternating current, and the generator belongs to a three-phase AC circuit, which includes capacitive reactance, inductive reactance and impedance, etc., with apparent power, active power, reactive power, etc., and the calculation method is different.

In three-phase electricity, there are three types of power, active power p, reactive power q and apparent power s.

The cosine of the phase difference ( ) between voltage and current is called the power factor, which is represented by the symbol cos, and numerically, the power factor is the ratio of the active power to the apparent power, i.e., cos = p s

The three power and power factor COS is a right-angled power triangle: the two right-angled sides are the active power p, the reactive power q, and the hypotenuse side is the apparent power s.

In a three-phase load, all three types of power are always present at the same time at any time

s = p +q s= (p + q ) apparent power s =

Active power p=

Reactive power q=

Power factor cos = p s

The current can be 0!

If you follow the ideal model.

p=i2 * r (p=1000kw, i, r are the total resistance of the total current).

i*(r1+r2)=5kV (r1=10,r2=load) r2=u i (u=5kv, i is the total current) The next is manual work, calculate it yourself.

p=ui ==Tongqing 44000=220*i == Wu Tsai i=200a;

Line resistance loss: ur=i*r=200*;

Therefore, the voltage obtained by using the cavity wheel is 200-40=180V

Generator output rated current:

i＝p/u＝44/

Voltage drop due to transmission conductor loss sensitivity:

U-descending ri-aggressive.

The voltage that the user gets:

u＝220－40＝180(v)

The power that the user gets:

p ui bridge lead large 180 200 36000 (w) 36kw

It's basically still 220V, is the line loss considered?

Loss: i6%PR

30A 10A, the output voltage of the step-up transformer is: U Pi

10V 5000V, the turn-to-turn ratio of the step-up transformer is: NN UU 240

The voltage lost on the transmission line is: U=I2R=10 30V=300V, then the input voltage of the step-down transformer is: U3=U2- U=5000-300V=4700V, and the turns ratio of the step-down transformer: nn U

u＝4700

3) The power that the user gets is: P = U3I2 = 4700 10W = 47000W, then: N = P P light.

2) The turns ratio of step-up transformer and step-down transformer are 6:125 and 235:11 respectively (3) It can make 470 "220V100W" lamps shine normally