Electrical exercises for junior high school, especially in relation to electrical power

1.When the slider P is in the midpoint position, the R lamp and half of the R change are connected in series, and the bulb is normal at this time, then the voltage at both ends of the bulb is 6V

Available at this time. u total = 6 + (6 r lamp) * 20 2 When the slider p slides to the maximum resistance position of the rheostat, the r lamp and r are connected in series, which can be obtained.

u total = 4 + (4 r lamp) * 20

The two-way solution is solved.

Supply voltage U total = 12V

The resistance of the bulb r lamp = 10 ohms.

The rated power of the bulb.

p = u r light = (6v) 10 ohm =

2.The current of the circuit when the resistance of the rheostat connected to the circuit is the maximum value.

i=4 rlamp=4 10=

The heat generated by the passage of an electric current through a rheostat.

q=i *r becomes t=(oh * 180s = 576j

For the first question, fill in 20w. The voltage obtained by the two electric soldering irons is half of their respective rated voltages, so the actual power is 1 4 of the rated power, which is 10W, so the total power is 20W.

The second question is C. The resistance of the two lamps is 40W, so when it is connected in series, it divides more voltage and the actual power is larger, but the actual power of both is less than their respective rated power, because the voltage obtained by each is less than their respective rated voltage.

The electrical resistance of electrical appliances considered in junior high and high schools is basically unchanged, but it will be found that it changes with temperature later.

Question 1: Two identical soldering irons have the same partial voltage, each is 110V, the voltage is halved, and the power is reduced by 1 4, so the actual total power is 20W

Question 2: Choose C

It can be calculated that the resistance of 110W is larger than that of 40W, so the 40W partial voltage will be a little larger, the voltage is high, and the bulb will be brighter, because the voltage does not reach the rated value, so it is less than the rated power.

Summary: In a series circuit, the current is the same, so the voltage division is mainly considered.

Solution: (1) 3000r kWh means that the meter consumes 3000 turns, and the kWh of electricity is, so it can be calculated that the power consumed by each turn is w0;

Now the dial of the whole process of a laundry has been installed 84 times, so the total power consumed in the whole process of a laundry is 1200J 80R 96000J

2) The electrical energy consumed during dehydration is: w uit 220v;

3) Then the electric energy during washing is w total w 96000j 21120j 74880j

4) by w uit, so washing motor working current i w ut 74880j (220v 12 60s);

Hope it helps, thank you!

1.Because the color TV turned 2 times in 5 minutes.

So 1 minute turn 2 5 = turn.

And because the light turned 5 times in 1 minute.

So the power of the electric light is 5 times the standby power of the color TV: 5 times.

Because of the 100w bulb.

So color TV standby power: 100W

Day = 4800j

The work of 100w1 minute is 100*60=6000J, then the work of the meter per revolution is 6000 5=1200J, the standby power of the refrigerator is 1200*2 (5*60)=8W, the monthly standby power consumption is 8*60*60*20*30=17280000J, and the monthly standby power consumption is 17280000 60 60 1000=kWh.

1. W lamp = pt = 100w 1min = 100w 60s = 6000j

Then the turntable of the electric energy meter turns to indicate that 6000J 5 turns = 1200J electric energy is consumed, then P color TV = w t = 2 1200J 300s = 8W2, W = PT = 8W 20h 30 =

Don't know if the answer satisfies you? ^_

It's very simple, power p = ui, the voltage of household appliances can be regarded as 220V, there is a current flow of 4A, then its power p = 880W, it should be an electrical appliance such as a refrigerator. The power of air conditioners is generally in the order of kilowatts, while fluorescent lamps are generally tens of watts, and electric fans are about 100 watts.

So choose C, refrigerator.

This depends on the rated current of the two resistors, p=ui=i r, i= (p r), from which it can be calculated that i1 = (45 5) = 3a, i2 = (60 15) = 2a, if the two resistors are connected in series, it is obvious that the maximum allowable current i=i2=2a, because if it is 3a, then one of the resistors will be burned. None of the options are correct.

Obviously, if the two bulbs are the same, then they receive the same voltage, which means that each bulb is assigned a voltage of 110V (series divider). P=U R, so it is certain that after adding 110V to both ends of the bulb with a rated voltage of 220V, the actual power of the bulb is only 1 4 of the rated power, so the actual power of the 60W bulb is only 15W, and the answer is C.

Look at the resistance, r=u p, the rated voltage is the same, the resistance with a large rated power is small, that is to say, the lamp resistance of 60w(r is smaller, then the voltage it is distributed in series should be smaller. If you want to quantitatively analyze, you can calculate the voltage on the 60W bulb U=220V [(220 60) (220 60+220 40)] 88V, and then substitute it to calculate the actual power of the two bulbs. The actual electrical work p'=u' r=[u (u p) (u p+u p)] u p=p p (p+p) ,p' =p p (p+p), obviously, the actual power of the 40w bulb is larger.

Pick B. PZ6-2 means "6V, 2W for general lighting (PZ means 'general illumination'), which is the value under rated working conditions. From this, the rated current i1 = 2 6 = 1 3 (a), i2 = 8 12 = 2 3 (a), then it can be known that the series current is 1 3 (a), and the pz6-2 emits light normally. The resistance of PZ12-8 r=u p=18, the actual voltage at both ends u'=ir=18 (1 3)a=6v, so the total voltage is 6v+6v=12v.

Pick B. When it glows normally (it is only effective for incandescent lamps, the principle of fluorescent lamps is different from that of incandescent lamps, so it cannot be calculated in this way.) )①r=u²/p=(220v)²÷100w=484ω；

Because p=u r, when the actual power of the bulb is 81W, the voltage on the bulb U'=U (81 100)=198V, that is to say, the voltage drop on the wire is δU=220V-198V=22V. The actual current on the bulb i=u'r=198V 484, and the work consumption on the wire is δP=δUI=9W.

1.In the circuit shown in Figure 1, the power supply voltage is 16V, and although the specifications are marked on the small bulb L, it is not clear except for the word "12V". After the switch S is closed, adjust the sliding plate P of the rheostat so that the small bulb can emit light normally, and the indication of the ammeter is 0 8A at this time, assuming that the resistance of the small bulb remains unchanged, find:

1) The resistance value of the small bulb when it emits light normally.

2) Re-adjust the sliding blade p of the rheostat, when the indication of the ammeter is 0 6A, what is the electrical power of the small bulb?

Answer: 1(1) The resistance value of the small bulb when it emits light normally is 15;

2) The electric power of the small bulb is 0 54W.

Whether you want an example problem, or ask for the solution of a certain example problem, you should write the question on the solution of the example problem.

Do you need sample questions, or do you need to explain them?