The main points of physics and electricity The teacher of electricity basically did not understand w

In a circuit, if the potential at both points is high, the voltage at the two points may not be much different, so "if the potential is high, the voltage is also high." "It's wrong. （f）

Note: This question is a true/false question, and the one with (r) after it is right, correct; There is (f) is FLSE, false. Therefore, this question is wrong.

Knowing the direction of gravity and the direction of motion, make a triangle, g is the hypotenuse so the electric field force is the minimum, and do the perpendicular line in the direction of motion, so e is the minimum (root number 2 2*mg) q

Direction; Direction of vertical movement.

mg q vertically upwards Since the particles move in a uniform linear velocity, the resultant external force on the particles is zero, and the particles are only affected by gravity and the electric field force, so the gravity is equal to the electric field force, and the direction is opposite, so e=mg q, and because the gravity force is vertically downward, the electric field force is vertically upward!

Upstairs, you are wrong, the electric field force is perpendicular to the direction of velocity, the magnitude is [mg cos45 degrees], the particles are moving in a straight line, and it is not said to be moving in a straight line at a uniform speed!

1: Current Definition: The movement of an electric charge in a medium forms an electric current.

The positive and negative poles of the power supply gather equal amounts of positive and negative charges respectively, so there is a potential difference between the positive and negative electrodes, that is, the voltage.

At the beginning of the capacitor, there is no charge on the two plates, so the potential difference between the two plates is zero, that is, the potential of the two plates is equal, and the potential is uo, the positive potential of the power supply is u+, and the negative potential is u-, then u+ is greater than uo and greater than u-.

When the capacitor is connected with the power supply, the potential of the positive pole of the power supply is higher than that of the capacitor, so a part of the positive charge (the direction of the positive charge is from the high potential to the low potential, and the direction of the negative charge is opposite) is transferred from the positive pole of the power supply to the positive plate of the capacitor along the wire, and at the same time, a part of the negative charge is transferred from the negative pole of the power supply to the negative plate of the capacitor, and the positive charge of the positive plate gradually increases and the electric potential gradually increases until it is equal to the positive potential of the power supply; The situation of the negative plate and the positive plate is similar, so I will not go into details. This transfer process is very short, so only instantaneous currents can be formed.

2: Because the amount of power stored in a fixed capacitor is proportional to the potential difference between its poles, and has nothing to do with any other factors, this ratio should be used to measure the power storage capacity.

3: The current directly forms a magnetic field, not an electric field, the landlord may have made a typo, right? Alternating currents form an unconstant magnetic field.

4: Constant current and direct current are two different things, the former is to show a state, and the latter is mainly to examine the nature of the power supply. Direct current is a direct current power supply, and it does not necessarily produce a constant current, for example

In a DC power supply, the two poles of a series of an ammeter and a sliding rheostat, the experimenter adjusts the resistance of the rheostat from the smallest to the maximum, and then to the minimum, and repeats this action constantly, the pointer of the ammeter will keep changing, obviously this is not a constant current, but this is a DC circuit, that is, the positive and negative poles of the power supply are always the same, and the voltage between the two poles is always the same.

It's all caused by voltage, such as high-voltage wires and the ground, which is also a big capacitor. In fact, a lot of things carry electricity, but they don't show it. In winter, the bed quilt will flash, that is, there must be certain conditions, and the capacitor will be charged and discharged under the condition of voltage.

The reason for the formation of electric difference is due to the presence of positive and negative charges, some of which are positive and negative ions. To measure the power storage capacity of the capacitor with the capacitor (charge amount and potential difference), this is stipulated by the predecessors, there is no need to pay attention to it, and there is a study at the university. The electric field formed by the alternating current is certainly not constant.

The direction of its current is constantly changing. Constant current is not direct current, they are not the same concept, constant current indicates that the magnitude does not change, the direction does not change, direct current is.

The direction remains the same, straight, just like a cow. Silly only one main direction.

1.The question to ask is how much resistance should be connected to the AB terminal at least when the rated current is 2A (which cannot be exceeded). In other words, it is an external resistor when the rated current is full.

You will definitely understand the calculation of the parallel resistance, and I will directly explain your problem, when the slider changes in the middle, the change of the total resistance.

2.In the simplest terms, the parallel resistance is smaller than the series resistance.

So the case of the slide rheostat slider at both ends is:

When the slider is connected below, the total resistance of the circuit is the largest, which is R0+R(24), which cannot reach the rated current.

The rated current can only reach the maximum when the slider slides to the top, and the higher it slides, the smaller it becomes. (Proven later by two methods).

a.Simple reasoning: Considering that it is a problem, in a more practical case, it can actually be assumed that an infinitesimal resistor can be connected in parallel at ab, which is equivalent to a resistor that is almost zero in parallel. Therefore, the resistance is minimal when the slider gradually slides upward.

b.Computationally derived:

I set the rheostat resistance to rb in parallel, then the one that is not connected in parallel is (22-rb), the internal resistance has been calculated before is 2, and the minimum resistance is set to rx.

r total = 2 (internal resistance) + (22-rb) + rb * rx (rb+rx).

Since we are talking about r always being the largest when the slider is in what position, we can set rx to n*rb first (to simplify the calculation, I set this constant to n times rb, n can be any constant greater than or equal to 0):

rtotal = 24-rb+rb*rx (rb+rx); Simplified to rtotal = 24-rb+rb*n*rb (rb+n*rb) (where rx=n*rb).

rTotal = 24 - (1-n (n+1))*rb

Obviously, even if I have an external parallel resistor infinity (n= ), it only makes r always get closer and closer to 24, and the larger rb, the smaller r always becomes; If (n=1) is only to bring r always closer and closer to 24-rb, the larger rb is, the smaller r always becomes.

So in any case, the larger the RB involved in parallel connection in a slip rheostat, the smaller the total R.

The meaning of the third question is to investigate: the maximum value of the trunk current is when the total resistance is the smallest If the rheostat slides to the top end, and then a small resistor is connected, the total resistance will be r+r0 r, that is, the resistance value of r0 and the external parallel resistance is required to be the smallest because the total resistance value after parallel connection is less than any of the resistances Think about the dynamic process The dicing from the top to the bottom The small resistance is connected to both ends of the ab There is an extreme assumption that the idea is equivalent to r at the bottom is about equal to r0 Then there will be r"r0 at any position above It's going to turn off the lights, so much to talk about, in fact, it can be proven, if you need QQ, send you a 383493998

The sliding rheostat is adjusted to the lowest end, which is called no-load, not load.

It is recommended that you take a good look at the relevant example problems before doing each problem, understand them thoroughly, and find out the idea of solving the problem. When you do the problem independently, follow the steps of the example problem solving, develop the habit of solving the problem systematically, and don't solve the problem with a hammer in the east and a stick in the west.

Make sure to get the problem in the future, and immediately have a rough way to solve the problem.

You should develop the habit of solving problems independently, even with the help of other students, you must do it again independently.

Don't engage in sea tactics.

Learn physics and electricity to see more, ask more, more hands-on, physics is a very experimental discipline, you not only have to watch the teacher do a variety of demonstration experiments, but also do a lot of experiments by yourself, some experiments can be done at home because of the simplicity, it is recommended that you find some popular science physics books to look at first, improve your understanding of physics, interest, do not rigidly memorize definitions, memorize formulas, set of formulas to solve problems.

Physical electricity is mainly divided into the common sense memorization part, which is summarized and memorized.

The law of series-parallel connection of motion Ohm's law Calculation of electrical power, Joule's law Perform the problem solving part. Find a universal disintegration pattern.

Experimental part: follow the teacher's ideas to conduct experiments, understand the experimental procedures, and precautions. Selective memory.

To memorize and understand all the formulas first, which is necessary to do physics problems, you can try to use formulas at the beginning, do more to know which formulas should be used for that kind of questions, in short, you still have to do more practice.