1. Physics and electricity problems in junior high school. Physics and electricity questions from th

The resistor power p is 27W, P=U2 R, substituting R=3 ohms, you can find U, U=9V.

According to the voltage distribution law of the series circuit, the voltage of the lamp is 15V-9V=6V.

The voltage distribution is proportional to the resistance, then r'= 6 9r = 2 euros.

The second question is that if it is the rated power, you don't need to calculate it, there is one in front, and it is estimated that it is the rated voltage or current that you ask.

The algorithm for the rated voltage is to use p=U2 R again, substituting the rated power and the calculated resistance value.

The algorithm for the rated current is p=i 2r.

Thirdly, the actual power is p=U2 r, substituting the actual voltage 6V and resistance 2 ohms calculated before. p'=36/2=18w

By: Q=PT=I 2RT

Get: i 2=q rt=27 3*1=9

i=3a by: i=u r

Get: total resistance r total = u i = 15 3 = 5 ohms.

The resistance of the small bulb: rl=r, total - r=5-3=2 ohms.

The actual power of the small bulb in the circuit PL = I 2RL = 3 2 * 2 = 18W has been given in the text The rated power of the lamp L is 32W. Also find (2) the rated power of the small bulb???

I'll give you an example question first.

The cars are equipped with an indicator light to remind the driver whether the door is closed or not.

As soon as one of the four doors is not closed properly (equivalent to a switch disconnected), the indicator light will glow.

Please click on the original image to enlarge it and select the correct one.

Choose B, because if all 4 doors are closed and equal to a wire (short circuit), then the current will not pass through the bulb.

If one door is not closed, then it is equivalent to an open circuit, and the current will not pass through.

to analyze Figure B.

At this point, you will find that if all four switches are turned off, then it is a short circuit!

And in this diagram the resistor r is on the trunk and the bulb is on the branch!

I can't find the picture, I hope you can understand!!

I don't know how to give a QQ yet, I'll teach you.

The power of the electric lamp becomes the original one, and it can be seen from P=i r that for the bulb, the resistance does not change, so the current becomes the original, the ammeter is connected in series with the bulb, and the current is equal, and the current representation number becomes one half of the original.

For electric lamps, p=i r, the resistance is unchanged, the electric power is proportional to the square of the current, the power becomes 1 4, the current becomes 1 2, choose c

Let the original current be i and the later current be k, then (1) i square r=p(2) k squared r= can get k=