Physical electrical work and electrical power problem steps just talk about ideas .

You have too many topics....I won't make ...... for you

These problems are generally done through the rated voltage and rated current of the bulb, and a current can be calculated through the data given in the title with the formula p=ui, which is the so-called maximum current allowed to pass, if a bulb is in the circuit, because the rated voltage is less than the power supply voltage, it is necessary to connect a resistance voltage divider, and the specific resistance value of the voltage divider resistor should be done through the resistance of the bulb and the current of the series circuit equally. And to not exceed the rated current, the actual power, to use the actual voltage and actual current of the small bulb to make up, to the actual power of the minimum, that is, to the small bulb voltage divider, to the rheostat resistance value is large enough.

When calculating this kind of problem, we should pay attention to the meaning of each voltage and current, it is recommended to use a more obvious subscript, otherwise it is easier to confuse the rated and the actual one, to draw a good circuit diagram, don't be too troublesome, know the amount of nearly two marks on the diagram, and then pay attention to ", this kind of indication is the resistance of the bulb, the rated voltage and the rated power do not mean anything.

That's pretty much it, I don't know if it will help you......

General method: Calculate the resistance from the rated power, current, and voltage.

Then look at the connection method given by the question, and use the resistance law to find the overall resistance related to the problem.

Look at whether the question is given a closed circuit or a constant voltage circuit, and find the total current of the circuit according to the corresponding law.

Find the branch current according to the current distribution law or the corresponding power according to the power distribution law.

Sometimes, light and dark, heat, are the limitation.

I'll tell you the answer!!

One "220v

The 1100W" electric heating wire R0 burned out.

The resistance of the original resistance wire is.

r=u p=(220v) 1100=44 Your answer is to choose R2 resistance wire 220V

2200w two in series! The resistance of the R2 resistance wire is R=U P=(220V) 2200=22

When two are connected in series, it is 22 x2 = 44

R2 resistance wire 220V is selected

2200w two in series!

Pure hand fighting!

The first problem is not clear by typing, in short, it is similar to the second question, the rated voltage is the same, the power is small, the internal resistance is large, the voltage is naturally divided, and the power consumed is large.

The first problem is Ohm's law for closed circuits. The method is to calculate the resistance value of each heating wire, and connect it in series and parallel according to the option, and the resistance value is the same as that of the original original. Question 2:

At the same voltage, the power with small resistance is large, and the power with large resistance is small. Just calculate the resistance values for comparison.

The slider is at the A end, the lamp is short-circuited, U power supply = IR =

The slide is at the B terminal, the lamp and the resistor AB are connected in parallel, i resistance = u r = 5v 10 = i lamp = i - i resistance =

P light = ui light = 5V * 2A = 10W

Solution:"8v16w"The resistance is 8 2 16 = 64 16 = 4 ohms"12v36w"The resistance of the bulb is 12 2 36 = 144 36 = 4 ohms then the bulb l1 p = i 2r1 = 16w

So: i 2 = 16 4 = 4

So: i=2a

Maximum voltage: U=IR=2*(4+4)=16VThe total power of the two bulbs is: I 2(R1+R2)=4*8=32W

1. When the slider is in the first position.

The voltage of the sliding rheostat is 8V, the current is, and the resistance value can be 20 ohms when the sliding vane is in the second position.

The voltage of the sliding rheostat is 6V, the current is, the resistance value can be obtained as 10 ohms, the power supply voltage is U0, and the sliding rheostat and the fixed value resistor are connected in series.

The series electrical appliances share the voltage according to the size of the tissue.

First position: 20u0 (20+r)=8

Second position. 10u0/(10+r)=6

When the two equations are combined, it can be found that r = 10 ohms, and the supply voltage is 12V, then the first position:

The electrical power of r is ui = 4v times watts.

Position 2. The electrical power of R is 6V times watts.

So the change in electrical power is 2 watts.

Supply voltage u fixed-value resistance r

The voltage is expressed in the number u'The current is the number i

ir+u'=u

Then there is a solution r=10 ohms.

The electrical power of the constant resistance is p=i 2r

The variable is (

1. The turntable rotates 20 times in one minute, which means that the power consumption in one minute is 20 600 = 1 30 kWh 1000 * 3600 30 joules.

The actual power is p (1* 60) 2000 watts.

2. The same principle as the previous question, the electric energy meter is marked with the words 3000 revolutions per kilowatt hour, the washing machine works for 20 minutes, and the turntable rotates 300 revolutions, and the electric energy consumed during this time is kilowatt hours joules joules.

So the actual power of the washing machine is p (20*60) 300 watts.

3. The lamp is marked with the words pz220-100, which means (the rated voltage is 220 volts, the rated power is 100 watts), and the home circuit voltage usually refers to 220 volts, so it emits light normally, the current is i p u 100 220 amperes, the filament resistance is r u 2 p 220 2 100 484 ohms, and the power consumption in 10 seconds is p*t 100*10 1000 joules.

1 kWh of electricity 1 kWh joule, which can make the lamp shine normally The time is.

t e p 100 seconds 10 hours.

4. The bulb resistance is r u 2 p 3 2 ohms, so a 15 2 ohm resistor should be connected in series.

Reason: In series circuits, the voltage is proportional to the resistance, and the ratio of the lamp voltage to the voltage on the resistor to be stringed is 3:(:1, so the resistance to be stringed is half of the lamp resistance.

1. Use a kilowatt-hour electric meter to turn 600 rpm, that is: electric energy consumption e=1kwh to 600 rpm, one minute to 20 rpm, power consumption e1=20 600kwh=120000j, actual power p=e1 t=120000 60w=2000w2As with one, the key is to put the larger unit of energy:

kWh is converted to: joule calculation, i.e., watt-second calculation, kilowatt to watt 1000, hour to second and then 3600, i.e. 1kWh = 3600000J

The power consumption of 3000 rpm is 1kWh = 3600000J, then the power consumption of 300 rpm is E1 = 300 3000 3600000J, and the power is P=E1 T=360000 (20 60) = 300W3The working voltage is 220V, the power is 100W, the current is 100 seconds, the power consumption is 1000J, and the 1 degree point can emit light for 10 hours, which is too simple and has no process.

1...20 rpm is 1 30 kWh, 1 minute electricity is 1 30 kWh, and the power is 2000 watts.

2...3000 rpm is kilowatt hour 300 rpm is of course kilowatt hour 3....Power 100 watts Voltage 220 volts Current is 100 220 = amps Resistance is 220 * 220 100 = 484 ohms 10 seconds power consumption 10 * 100 = 1000 joules 1 kWh is 3600000 joules Can emit light normally 3600000 100 = 36000 seconds Both 10 hours.

4...Tandem (ohms.

1.Solution: It is known that 600r, t=1min=60s, according to p=w t, get:

p=20/600/60/3600=20w

20 minutes = 1 3h

p=w/t=

3 pz – general lighting bulb, 220 – rated voltage 220 volts, 100 – rated power: 100 watts.

This is a 100W bulb.

i=p amount, u=100w 220u=

r=u 2 p=(220v) 2 100w484 ohm w=pt=100wx10s=1000j 1kwh=3600000j t=w p=3600000j 100w=36000s=10h

4 i=p u= uresistance=u-total-ulamp=

r=uresistance, iresistance=ohm.

I hope it is useful to write it all according to the format of the examination.

1.The kettle turns 20 in one minute, and the kettle turns 60 * 20 = 1200 in one hour, so the power of the kettle = 1200 600 = 2 kw

2.The washing machine turns 300 in 20 minutes and 900 in an hour, so the power = 900 3000 = 300w

3.Voltage: 220V. The power is 100W. p=ui。。i=p u=100 220=5 11a Joules forgot,, glow for 10 hours.

4.A resistor in series with a ohm... In order to make the small lamp shine normally, it is necessary to use the voltage divided by the resistor, so use series connection.

When the small lamp is shining normally, the current = p u = resistance = u i = 3 ohms.

Because the series currents are the same, the resistance = u i = (ohm.

1 solution: w = 20r 600r kw * h = 1 30 kw * h = 120000j 1min = 60s

p=w t=120000j 60s=2000w2 solution: w=300r 3000r kw*h=1 10kw*h=360000j 20min=1200s

p=w t=360000j 1200s=300w3 solution: p amount=100w

U = 220V

w=pt=100w*10s=1000j, t2=w2 p2 p2= 1kw*h 4 solution: i1=u1*i1 from p1=u1 r1 r1=u1 i1=3v ohms from i total = u total r total r total = u total i total i = ohm r2 = r total -r1

Ohm - 15 ohms = ohms.

1.Solution: It is known that 600r, t=1min=60s, according to p=w t, get:

p=20 600 60 3600=20wA: Omitted.

Give you a hint and you'll be done.

2) Because the power of the two bulbs is the same and the resistance is the same, the power supply voltage is equally divided between the two bulbs after series connection. That is, the voltage on each bulb is 110 volts.

3) According to the above, so in the case of two bulbs in series, the power consumption of each bulb is 10 watts, and the total power consumption is 20 watts.