High 2 Physics Constant Current, High 2 Physical Constant Current Problem

1. Between 1000 and 1500 ohms.

2. There is a current in the cross-section, and there is no current on the arrival surface, which is the difference.

It can be seen from the questions raised by the landlord that the first question is a physics question, and the second question is a chemistry question.

First of all, take a look at the physics that the landlord mentions always.

The ohmmeter is different from the measuring instrument of Pu Energy, and the biggest difference is the particularity of its uneven scale - the resistance measurement of the meter in its range is smaller and more accurate, and the larger the more inferior. As a result, ohmics have rarer scales for small values and denser scales for high resistances. From this point, it can be seen that the landlord's "hand is in the middle of the 100 and 200 scales when measuring" means that the value of the dial reading will definitely be less than 150, and then multiplied by the magnification "*10", the measurement result must be:

r<1500 ohms.

Answer the second question: There must be two ions (positive ions and negative ions) in the electrolyte, the positive ions have a positive charge, and according to the theory of electrode attraction of opposites, it will move to the negative electrode (chemical name is cathode); In the same way, negative ions have a negative charge, they move towards the positive electrode, and according to the theory of electric neutralization, the total charge they each carry is the same, but the electricity is opposite. So, if you only ask about the difference between these two ions, it is that they carry opposite electrical properties.

r=1500

The positive and negative ions of a certain section are the sum of the negative ions and the negative ions reaching the anode and the positive ions reaching the anit.

Agree with the opinion upstairs, there is really a problem with education now, the digital multimeter has long been used, and everyone has to learn the old things, isn't this to let people learn the old technology??

Education is like this, what else is there to talk about educational innovation?

r=1500? I don't think so. Because the scale of the ohmmeter is uneven, I forgot about this, so you better ask your physics teacher.

r=150.

The positive and negative ions passing through the cross-section are the solution ions in the battery, but the ions reaching the cathode and anode are the ions in the electrolytic solution.

The upstairs can't say that, now that the war is fought with guns, isn't it true that the troops don't fight boxing when they are training?

A bulb L, marked with the words "6V, 12W", R lamp = 3 When connected in parallel, the motor works normally, the motor U is rated = 6VL in series with D, the bulb P actual = 9W, I = 3A when the motor is connected in series, P mechanical = P total - p heat = u motor I-I 2r = (6 3-6) w =

Mechanical efficiency pmechanical pTotal =

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.

Select C, first exclude R1, because the internal resistance of the power supply is not considered, so there is no need to consider R1 in parallel in the circuit

When R3 is broken, the total resistance of that group of resistors becomes larger, and the dry circuit current becomes smaller, so the voltage division of the parallel circuit where the A lamp is located becomes smaller, and the voltage division of the B lamp becomes larger, so it will be brighter.

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.

R1 remains the same, R2 becomes larger, and the brightness of the small bulb L becomes relatively brighter, because the total resistance of R1 and R2 is connected in parallel with the bulb, no matter which R1 or R2 becomes larger, it will make its shunt smaller, so that the bulb will get more current and become brighter.

The ammeter reading will be larger because the greater the resistance in parallel, the smaller its total resistance, and the greater the current!

R1 thermal power will become smaller, because R1 and R2 are connected in series, the current is equal, the voltage is proportional to the resistance value, R2 becomes larger, R1 is relatively smaller, the voltage drop becomes smaller, and the power is naturally smaller!

R2 becomes larger, the total resistance becomes larger, the total current becomes smaller, and the ammeter reading becomes smaller.

R2 becomes larger, the voltage at both ends of the small bulb L does not change, and the brightness of the small bulb L does not change.

R2 becomes larger, the current flowing through R1 becomes smaller, and R1 generates less heat.

Elect Dansen B. From the meaning of the title: e=including r+5+7) and e=i(r+5+15). The solution gives i=, and when r=0, i=. When r is infinite. i=。Since r is not equal to 0. Therefore, I is late to change to (,

B when the resistance of the resistance box increases, the total spring rubber resistance of the parallel circuit increases, the trunk current decreases, the voltage at the end of the road increases, and the current flowing through the voltmeter increases, so the indication of the ammeter decreases, only possible, if the influence of the positive auspicious resistance and the resistance value of the voltmeter in the power supply is not considered, the current representation number can be so 1 excluded, select b

Answer: a

Circuit analysis: the voltmeter is connected to both ends of the power supply, and the voltage at the end of the circuit is measured (the indication changes with the "external resistance r") The ammeter measures the current (dry circuit current) through the fixed resistance on the left, and when it is set to R1 when the dicing p is at the A end, the rheostat is short-circuited, and the external resistance r = R1 When the dicing is between A and B, the rheostat "left half BP" and "right half AP" are connected in parallel, and then connected in series with R1. And when the dicing is at the "midpoint" between A and B, the "parallel resistance" r reaches the maximum value. The external resistance r = r1 + r and maximum.

When the dicing p is at the B terminus, the rheostat is also shorted, and the external resistance r = R1 slider p slides from A to B

From a midpoint:

r and gradually increase , the external resistance r = r1 + r and.

The external resistance r = r1 + r and r, the external resistance r increases gradually.

Then: the voltage at the end of the road should be increased, and the "voltage representation number should be increased"; As the total resistance increases, the trunk current decreases, and the "current representation number decreases".

From the midpoint b:

r and gradually decrease, the external resistance r = r1 + r and.

The external resistance r = r1 + r and , the external resistance r decreases gradually.

Then: the voltage at the end of the road should be reduced, and the "voltage representation number should be reduced"; The total resistance decreases, the trunk current increases, and the "current representation number increases".

That is, the number of current representations decreases first and then increases; The number of voltage representations increases first and then decreases.

PS: The shunt resistance is smaller than that of any one branch.

The current is 10 milliamperes, that is, through the Coulomb charge per second, 240 meters of orbital limbs, 1 electron chain flows at the speed of 1 10 light, and will pass through the orbital section 125000 times per second, that is, 1 electron can produce 125000* Coulomb charge, and the Coulomb charge needs the number of electrons to be B

The first side of this question must first clarify the definition of current, the amount of electricity through the conductor cross-section (not the unit cross-section) in the unit time of fortune, 10mA is the amount of electricity through the cross-section of the circular track per unit time is 10*10 -3 coulombs, and then divide by the amount of each electron to get the number of electrons passing through the cross-section per unit time, and then multiply the result by the time of the electron motion for a week is the final answer, choose b, time is divided by the circumference by the speed to pull the side ball.