Help with an electrical calculation problem

The resistance of V 60W is 110 * 110 40 according to the principle of voltage division 110V 60W should be divided into 88V voltage, 110V 40W should be divided into 132V, 2, the current in the circuit is: 220 (110 * 110 60 + 110 * 110 40 ) =, power consumption: 88 * of 110V 60W, 132 * of 110V 40W

3. This series connection cannot be used, 4. If the two lamps are 60W, the power consumption is 60W, the voltage is 110V, and the current is; Both are 40W, and the current is.

1.When 110V 60W, 110V 40W are connected in series, the 60W resistance is 110 2 60=200. The 40W resistance is 110 2 40 = 300

The total resistance is 200 + 300 = 500, and the total current is 220 500 = so the withstand voltage of 60W is, and the voltage of 40W is. 2. It is higher than the rated voltage, so it cannot be connected in series like this.

3. All are 60W, the voltage is 110V, the current is 60W, and the power rate is 60W. Both are 40w is also available. The voltage is 110V and the current is. The power consumption is 40W.

The withstand voltage is 88V and 132V respectively, the power consumption is and the current is respectively.

<> "Solution: The formula for calculating transformer capacity: s = 3ui - where rolling you and i are the line voltages of the transformer.

Line current. For the high voltage side, U1L=35000V=35KV, SE=8000KVA, so line current:

i1l=8000/（√3×35）=。

The primary side adopts the Y-shaped wiring method, so the line current = phase current =.

Secondary phase voltage.

10500v=, so keep an eye on the filial line current:

i2l=8000 ( 3 凯稿.)

The secondary side adopts the connection method, so: I2P (phase current) = I2L (line current) 3 = 440.

The secondary side is the connection, and the line voltage = phase voltage = 10500v =.

Explain which reputation is pure such as the back of the plum trousers.

The high-voltage side is connected Y, and the line electric car is current=phase current=8000 35 The low-voltage side roll front is wired, and the line current = 8000

Phase current = 440

Low-voltage side wiring, phase voltage sail disadvantage = line voltage =

According to the formula of p=u*i, the formula can be calculated as v3= 4=6v, 2=12 volts-v4=6v, i2=1a, p2=6w.

Firstly, through the known positive link of element 4, the missing voltage value v4 can be obtained, and then the current of element 3 is obtained;

Then, element 3 acts as the sum of the currents of Sun 4, which is the current flowing through element 2, which is also the output current of the power supply;

The power emitted by the power supply is equal to the sum of the absorbed power of each component, so that the power of element 2 can be obtained;

The key points are the direction of the current and the polarity of the voltage: the current flowing from the positive pole of the voltage of the component is flowing from the component that absorbs power;

According to KVL's law, the voltage in the circuit in Figure 5 is U=2 6-2=10V.

I don't understand your calculation question, which probably means to supply power to a load with a power of 670kw and a power factor of 10kv, and the ambient temperature is 30 degrees, and the allowable voltage drop is required to be 5%, right? If you don't find the cross-section of the wire first, where can you get the current carrying capacity?

[There is a 10 thousand volt overhead line at the place with a calculated load of 670kwcos, which is equal to a calculated load at 2km; The ambient temperature is 30 degrees; The allowable voltage loss is 5 percent; Try to find the allowable current carrying capacity of the cable. 】

Try to find: 1. Calculate the total current of 0-2km first (the influence of power factor needs to be considered);

2. Calculate the current (also need to consider the influence of power factor);

3. According to the allowable voltage loss is 5%; The ambient temperature is 30 degrees; overhead lines; Select the cross-section and core of the cable.

4. Find out what is the allowable current carrying capacity.

Do you think the conditions are complete? How to find the allowable current without wire parameters? It seems that the total current of the load is required, and it is difficult to understand.

(All Authorities referred to below are International Authorities).

Known: r1=2, r2=6, r3=3, e1=4 (positive and negative), e2=9 (positive left, negative right. If it is the opposite, please calculate it yourself according to the gourd painting scoop).

Solution: Use Thevenin's theorem twice, the first time for the auxiliary and the second time for the requirements of the problem.

1) Remove r3 temporarily. In this case, the voltage at the AB terminal is the equivalent electromotive force E'=e2-r2*[e1/(r1+r2)]=9-6*[4/(2+6)]=6；Equivalent resistance r'=r1*r2/(r1+r2)=。

2) Put the R3 back in its place. In this case, the voltage at the AB terminal is the equivalent electromotive force e=r3*[e'/(r'+r3)]=3*[6/(；Equivalent resistance r=r'*r3/(r'+r3)=。

The equivalent is a 1 ohm resistor and an 8V voltage source in series (if the 9V voltage source is left-positive, right-negative).

If it is right-positive and left-negative, the voltage source is changed to 4V.

Obviously, the AB port is connected to a 1 ohm resistor for maximum power. I don't need to calculate this.

This seems to be a junior high school physics problem!

z=280+20+j*2*

i=u/z=220/

The voltage at both ends of the lamp is ua=i*r=

The voltage on the rectifier UB=I*ZL=

The phase angle of the physical quantity is indicated after the sign.

R-L series AC circuit, phasor diagram by phasor method, and then calculated. It's not difficult, I believe you have your own ideas.