Learn physics, physics masters come in and take a look

1. The internal resistance of the ammeter is small, so it can be used as a wire. The internal resistance of the voltmeter is large, so it can be "wiped off".

2. The voltmeter measures the electrical appliances that are "connected in parallel" with it (the electrical appliances refer to resistive bulbs, etc.) The ammeter measures the electrical appliances that are connected in series with it.

3. The key to doing this kind of problem is to be careful, first determine what type of circuit (parallel or series), and then think about the characteristics of parallel or series circuits. If there are many circuits, you can use the ammeter as a wire voltmeter, and the idea can be erased and simplified. The last point can only be said so much, you have to read more books and do more questions to exercise your basic knowledge.

1.The ammeter is made as a wire, and the voltmeter is wiped off.

2.The ammeter measures the current of its branch, and the voltmeter looks at the part surrounded by the two terminals and measures the voltage of the part.

3.The most important thing is to collect the knowledge of the circuit, and clearly distinguish the value of the part of the meter. Then list all the conditions in the problem as formulas and solve the equations.

There are also more questions, and the most important thing is to understand why you do this

1. The voltmeter resistance is very large, and the current passing through it is very small, so it is considered that the current does not pass through it, which is what you call quite "wipe off". The resistance of the ammeter is very small, which is equivalent to a wire, and it can be processed according to the wire.

2. The current of which electrical appliance is measured in series between the ammeter and which electrical appliance, and the voltage of which electrical appliance is measured in parallel between the voltmeter and which electrical appliance. The rheostat does not measure the physical quantity at both ends of the device, it has the function of protecting the circuit and changing the current in the circuit.

3. To do the problem of circuits, we must first analyze and understand how the current goes, and then analyze whether the relationship between each electrical appliance is in series or parallel, and then solve it according to the relationship between current, voltage and resistance in the series-parallel circuit, as well as Ohm's law formula.

I won't talk about the answer to the question, the few people upstairs have already explained it thoroughly.

I just have a little bit of experience learning physics when I was a kid. I was good at physics when I was a kid, and I won a prize in the province, hehe.

In terms of electricity, I generally think of an electric current as a stream of water. Ammeter = Water Meter, Voltmeter = Dam. Rheostat = pipe, so you'll be a little more intuitive.

Just imagine the current as a water flow, and everything is done, first think about a simple circuit, fully understand, and then develop into a complex one.

1pv=nrt due to constant pressure so outside p = p = we (work done) = p outside δv = pδv = nrδt = 1 * 2qp, m = integral number ncp, mdt = cp, m*72 = units? ）cp,m=

This should be solved with calculus, I haven't touched physics for a long time.

w=∫(0-4)f*x=∫(0-4)[(10+6x^2)*x]=∫(0-4)[10x+6x^3]5x^2+

w=ekinetic energy=

v = root number 464 (m s).

It should be that the speed of the hammer handle when it touches the ground immediately becomes 0, and the hammer head will move downward due to inertia to squeeze the hammer handle, so that the hammer head is tightly attached to the hammer handle.

Answer: Solve the system of equations s=1 2*at

v=at substituting the numbers of the antelope and the cheetah into the system of equations, respectively, can be obtained:

Antelope. a=t=4(s)

Cheetah. a=t=4(s)

The hood is known to:

1) Stupid antelope displacement 1 2

Cheetah displacement 60m

x = i.e., the value of x is not greater than.

2) Antelope displacement 50 + 25 (4-1) = 125m, cheetah displacement 60 + 30 4 = 180m

x=180-125=55m, that is, the value of x is not greater than 55m

I can give you some advice, physics is such a thing, in fact, the amount of knowledge is very small, and there are very few things that you need to master, but many students always feel that they don't know how to start when they learn physics or do physics problems, in fact, this is exactly what you need to use knowledge and exercise this ability. First of all, you must be familiar with the textbook, how to be very familiar with the textbook, let's put it this way, a physics class will talk about one or two pages, you have nothing to do for five minutes to read it, it doesn't take time, so you have time to flip through more, turn a few times to be very familiar. Next, it is the use of knowledge, some students choose the sea of questions tactic, in fact, to be honest, this tactic is really ineffective, because physics problems are always ever-changing, and a slight change becomes a problem you have not seen, then you will not do it.

So I suggest here that physics should be learned by heart, how to learn by heart, first of all, the basic knowledge learned, and the concepts in the textbook (there are not many, only a few pages for several days, read a few more times, just remember the formula) and formulas, the emphasis here is to remember, because only by remembering, can you effectively carry out the next step, the next step, that is, to do the problem, but I very much disagree with the sea of questions tactics, but training tactics, that is, first of all, you can do the knowledge of physics after class, from simple to complexIn the process of doing it, don't look at the concepts and formulas, including the previous examples, even if you don't know where to start, even if you don't have a clue, hold back, don't look at the front, this time is the time to train you, you just read the questions over and over again, think over and over again, try to take the knowledge you have learned over and over again, believe me, you must be able to do it, because that is what you have learned, but you don't know how to use it, the price of doing so is that the time to pay may be longer, Sometimes it may take one to two hours to do a problem, but you can rest assured, this is not without learning efficiency, physics questions focus on independent thinking, and although there are many questions, there are few question types, and there are a few in the past, when you learn the ability to think and make problems, then you can easily use the knowledge you have learned to solve all kinds of problems, at this time you will also find that physics is actually a little thing, too little too little, of course, there is another point I say here is very important, It's the usual homework and the practice questions in your reference book, don't look at the answer when you do it, even if you can't do it, you put it down first, and when you have time, take it out and think, take it out and think, and resolutely don't look at the answer, when you make this question through a few more thoughts, what you have mastered is not only this question, you have mastered a very powerful method, which will make you feel that physics is getting simpler and simpler, this is the independent thinking ability that teachers tell students all day long, However, most teachers just told their classmates all day long and didn't tell them what to do, so many students didn't know why and didn't know how to exercise their independent thinking skills!

Hehe Physics requires strong logical thinking ability, force, unlike liberal arts things that will be done if you remember them, so learning physics needs to be energetic, active thinking, and can not fight fatigue battles, if the results are not ideal, you can start with the simplest, pay attention to the summary, physics is not all difficult problems, and the score is not low to do the simple ones correctly.

According to the general review, first review by chapter, then review by topic, then comprehensive review, and finally do a set of questions.

Is it a diagram or a diagram B.

U refers to the potential difference, so that the node between R2 and R3 is B, there is an electric potential UB, and then the node on the right of R4 is C, and there is a potential UC here, and the voltmeter measures the potential difference between the two points of BC, so U=UB-UC, and the node on the left of R4 is A, then there is, UR3=UB-UA, UR4=UA-UC, then UR3+UR4=UB-UA+UA-UC=UB-UC=U

So u=ur3+ur4

Is the rightmost one R1 or R4, and does U refer to the voltmeter voltage or the total voltage?

In short, the voltage of the parallel circuit is equal everywhere, and the total voltage of the series circuit is the sum of the subvoltages. It's a series and combined circuit... Landlord, please make the resistance clear. Can not see.