Ask the high school physics teacher to come in constant current problem .

The question is not very clear, and the corner mark is not clear.

In this case, r is varied.

When the parallel connection is started, the power of the three resistors is equal, it can be seen that the two fixed-value resistors are equal = r change at that time, when the hybrid connection, the voltage of r change becomes smaller, and r becomes larger, so continue the discussion.

You do it first, you won't resend the picture, it's clearer.

The equivalent power supply idea is to remove the resistance on the trunk circuit, and the internal resistance of the power supply increases, and the amount of increase is equal to the resistance value of the deleted resistor. In this way, the hybrid circuit can be simplified into a very simple series or parallel connection.

First, the part at the left end of the slider slide is connected in parallel with the resistor r on the left side above;

The part at the right end of the slider slider is connected in parallel with the resistor r on the right side above;

These two parallel parts are then connected in series.

Secondly, let's talk about the range of variation of the total resistance between CDs:

When the slide is at the "leftmost end", (the same is the case at the far right) the upper left resistor r is shorted, and the circuit is connected in parallel with the resistor r on the upper right.

It is not difficult to use the knowledge of the total resistance of parallel circuits.

1/r + 1/2r =3/2r

The reversal gives the total resistance = 2r 3

When the slide is at the "midpoint", the resistance of the parallel part on the left is two resistors with a resistance value of r in parallel, and the parallel resistance is r 2;

The resistance of the parallel part on the right is two resistors with a resistance value of r in parallel, and the parallel resistance is also r 2;

The total resistance of the two parts in series is R2 + R2 =R, so the range of variation of the total resistance between cd is:

Process from Left End to Right End 2r 3 r 2r 3 Description: The total resistance between the end and the midpoint can be determined according to.

The total resistance of the parallel circuit is less than the resistance of any one of the branches" and determine that its resistance is greater than 2r 3, which is less than r.

Finally, speaking of the direction of current flow, the slide vanes are at both ends needless to say, you get the idea.

The direction of the current in other positions is explained using the current law of Kirchhoff nodes. (I guess you don't understand this).

Based on the fact that you are a high school student, I will only give you a qualitative statement:

When the slide is in the range from the left end to the midpoint, the current flows from the slide to point B;

When the sliding vane is in the range from the midpoint to the right end, the current flows from point B to the sliding vane;

When the sliding vane is at the midpoint, the circuits on both sides are in equilibrium, and there is no current flowing between point B and the sliding vane, that is, the current is zero.

I guess there is a problem here, and I don't need to say the direction of the current in other places, you should know it yourself.

If a rheostat is considered to be two resistors connected in series, then the whole circuit is connected in series with two sets of resistors connected in parallel.

Range of variation 2 3r r

The variation range is 2 3r to r, with the resistance being maximum in the middle and the resistance being minimal at both ends.

The left half and the right half and then the left and right strings.

Because the full bias current of the ammeter is 300 microamps, the minimum resistance value of the loop is 5 thousand ohms. Considering the reading characteristics of the multimeter, only the pointer is more accurate when the pointer is in the middle of the range, that is to say, the resistance measurement value will be more accurate when the loop current is about 150 microamperes, so it is determined that the above answer is only in the range of b, and the pointer will stay near the middle of the range.

What you understand is this: positive and negative charges pass through a certain section in opposite directions, so the current is 4ne t Actually, this is not the case.

For the cathode, no electrons are lost per unit time, but a positive ion is obtained, which is 2ne t by the current definition

For a certain cross-section, 2n electrons reaching the anode per unit time come from both sides of the cross-section, that is, the side close to the anode does not pass through the electrons to reach the anode. So the number of electrons passing through this section is not 2n.

Let me start by asking you why you think it's 4?

First, the current is generated by the directional motion of charged particles.

In a metallic conductor is the motion of free electrons.

In the gas is the directional motion of free electrons and cations, this is to make the gas ionized equilibrium, there are n divalent positive ions moving, there must be 2n electrons moving in the opposite direction, so the generated current should be based on one of them.

This is similar to the movement of electrons and the movement of positive charges in a wire, which is actually a repetitive process, so the answer is correct.

It is indeed 2ne t, pay attention to the difference between the two types of questions, if there are n divalent positive ions per unit time through a certain section in the solution, it should be your answer, but if it reaches the cathode (anode), it is 2ne t. This is because, i=nqsv, note that this v, at the cathode, only positive ions reach (that is, there is velocity), and no anion passes through, v=0, that is, the current generated by the anion is zero, so i=2net

Well understood.

You imagine that at one point in space, a gas molecule (which is electrically neutral) is ionized, emitting positive ions and negative ions in two opposite directions. (The negative ions here are naturally electrons.) Do you understand?

So when calculating the amount of electricity passing through the cross-section, the gas ionizes to produce particles.