Physics in the second year of junior high school, about electricity, is really difficult to understa

Hello, I know your confusion, and the answer is as follows:

First of all, understand the difference between a short circuit and an open circuit

The "short" road means that the road is too "short" close to 0, and it can be reached immediately, and the resistance is 0, which translates to the resistance close to 0

The "broken" road means that the road is "broken", the road is not passable, the block is infinite, there is no way to pass, the translation is that the resistance is too large, and the current is 0

The resistance of the voltmeter itself is infinite, so adding the voltmeter to the circuit will not change the original circuit (the voltmeter is connected in parallel with the measuring appliance), and the resistance after parallel connection is equal to the original resistance, so the current through the original electrical appliance remains unchanged and the voltage remains unchanged.

When you say a short circuit in the voltmeter, the voltmeter is "short-circuited", that is, the two ends of the voltmeter are directly connected, and the voltage is 0.

It can also be understood as "a short circuit in a voltmeter is a situation where the voltage representation number is 0".

The more 0 the resistance of the ammeter itself, it is connected in series in the circuit, and the resistance added to the electrical appliance in series is still equal to the original resistance, so it will not cause harm to the circuit. However, once the ammeter is used incorrectly and connected in parallel with the electrical appliance, the electrical appliance will be "short-circuited".

A short circuit means that the current does not go from the appliance, but to another line with very little resistance. The voltmeter can be regarded as an open circuit, so it will not short-circuit the electrical appliance. Whether the voltmeter will be short-circuited depends on whether the two ends of the meter are connected to the line with low resistance.

Once an electrical appliance is short-circuited, it runs out of voltage. If this appliance is disconnected, the ammeter is also disconnected (the voltmeter can be seen as an open circuit).

1.The short circuit is to use the wire in parallel with the electrical appliances because the wire resistance is very small, so the current is also very smart, and the wire is good to walk, and the electrical appliances are short-circuited!! It happens at any time.

A short circuit is not a short circuit, of course the voltmeter will be short-circuited, even if the voltmeter is short, you use a wire to connect it, and it is turned on again.

2.There is a short circuit with the electrical appliance, and the value measured by the voltmeter is the voltage of the wire, which is close to 0In the case of a short circuit, if it is in series, when the current has no way to go, even if the resistance value of the voltmeter is large, the voltmeter will be walked, such as when there is only one electrical appliance left, close to the power supply voltage.

The ammeter can be broken, but the short circuit has nothing to do with the circuit, because the ammeter is equivalent to a wire, but it can't be measured! Got it?? Hee-hee.

(1) Method 1:

Solution: Because, resistors R1 and R2 are connected to the circuit in parallel, and the voltage at both ends is 24V, so the power supply voltage U=24V, and U1=U2=U=24V, because the current through R2 is i2=

So, r2=u2 i2=24

Answer: Slightly. 2) Method 2:

Solution: Because the resistor R1 and R2 are connected to the circuit in parallel, the voltage at both ends is 24V, the power supply voltage U=24V, and U1=U2=U=24V because, R1=80

So, i1=u1 r1=24 80=

Total circuit current: i=i1+i2=

Total resistance of the circuit: r=u i=24

1 r = 1 r1 + 1 r2

So, 1 r2 = 1 r - 1 r1 = 1 48-1 80 = 1 120 So, r2 = 120 ( ).

Answer: Slightly. I hope it helps you, and if you have any questions, you can ask them

I wish you progress in your studies and go to the next level! (*

First method:

The current through R1 is 80 ohms, then.

i1=24/80=

Because i2=, therefore.

Total resistance = 24 (i1+i2) = 24 ohms.

The two resistors are connected in parallel, and the total resistance is.

r1r2 (r1+r2) = 48 ohms.

The solution is r2 = 120 ohms.

The second party. i1=24 80=

According to r1 r2=i2 i1

Then r2 = 120 ohms.

Method 1: Find the R2 resistance directly.

r=u/i=24/

Method 2: First, find that the ratio of the two resistors of R1 current is equal to the ratio of the two currents r2=r1(i (u r1))=80 3 2=120 If you adopt it, you would be grateful.

I think this has to be explained by the essence of resistor generation:

The formation of electric current in the conductor by free electrons is the result of their directional movement under the action of voltage, and in the process of directional movement, the movement of electrons is affected by some other factors, such as the thermal motion of electrons, because any particle has to do thermal motion when the absolute temperature is not zero, and this thermal motion is irregular, and directional movement has an obstructive effect. Therefore, the resistance of the conductor varies with the time of energizing i.e., temperature. It can't just be about the material, the length, the cross-sectional area.

In addition, electrons are subject to the average hindrance of a large number of nuclei, the hindrance between electrons, the collision between electrons, the collision between electrons and nuclei, and so on. Of course, some of the various obstacles have a larger impact and some are smaller, and these microscopic impediments form a resistance at the macroscopic level.

However, there are countless electrons in the conductor, and the influence of each is large and small, and some unimportant factors such as temperature, cross-sectional shape of the conductor are often ignored for the sake of convenience in the calculation process. Therefore, within the scope of the landlord's involvement, there is no need to consider these factors, otherwise many things cannot be quantitatively calculated.

Of course, it's also very good for the landlord to recognize these o(o,I hope you can know some things in essence.,But don't get into the horns of the horns in learning.。

The same, no matter what kind of shape, as long as the area is the same, the resistance is the same.

In the opinion of the upstairs, resistance does not refer to the time it takes for an electron to travel through a certain section of wire, but rather a quantity like friction.

It's the same, this is calculated according to the formula, some things can't be based on your own feelings alone, but some things are still related to feelings, but it's not this problem

The conduction of metals is not determined by the length of the distance, but by the energy bands of the metal crystals. This involves the question of the energy band, valence band, and conduction band of the crystal. If the landlord is interested, you can go down.

In short, it has something to do with the transition of energy.

Well, as long as the area is the same, the resistance is the same, because his total area has not changed, but the shape has changed

The same. You can't feel it.

When a fixed resistor and a sliding rheostat are connected in series, why is the sliding rheostat the most powerful when the sliding rheostat is the same as the fixed resistance? To detail the steps.

Let the constant resistance value be R1, the maximum power of the sliding rheostat is P, and when the power of the sliding rheostat is maximum, the resistance value of the sliding rheostat is R2 and the supply voltage is U

Solution: obtained by p=i*i*r.

p=[u/（r1+r2）][u/（r1+r2）]*r2=u*u*r2/(r1+r2)

u*u [(r1*r1 r2)+(2r1+r2)] If p is to be maximum, [(r1*r1 r2)+(2r1+r2)] is the minimum and 2r1 is a fixed value, then [(r1*r1 r2)+r2] is the smallest because (r1*r1 r2)+r2 2 [(r1*r1 r2)*r2] (the mathematical proof of the equation is followed).

i.e. (r1*r1 r2)+r2 2r1, so it is calculated as r1-r2 0

Therefore, the minimum value of (r1-r2) is 0, i.e., when r1=r2, [(r1*r1 r2)+r2] is the smallest, so when r1=r2, p is maximum.

Mathematically proved: if a 0 b 0

There is a+b 2 (a*b).

a=b, the equation takes an equal sign.

Do you have a problem or a problem to solve? If you want to solve the problem, the following questions are for you to do: 1 There is a home light, the resistance of the normal light is set to r1, when the light is turned out, the filament resistance is set to r2, then the relationship between r1 and r2 is a r1 = r2 b r1 is greater than r2 c r1 is less than r2 d can not be determined Answer b2 r1, r2 and sliding rheostat r'In series in a circuit, r1, r'The voltages at both ends are U1 and U2 respectively, when the rheostat R. is sliding'When the slide p is in the other position, r1, r'The voltages at both ends are U1'、u2'Rule.

u1=┃u1-u1'┃，u2=┃u2-u2', then u1 u2(fill in》, =

Answer u1< u2 Note: If you want to know why, please ask! Thank you!!

Let's learn to draw circuit diagrams first, starting from the most basic electrical problems, if you don't learn electricity well, then your physical path will be difficult to follow.

Go buy contest questions to do. In fact, it is not difficult to study physics and electricity in junior high school.

15 (a homogeneous cuboid with a density of =500kg m3, length a, height b, and width c respectively, with a smooth surface, stationary on a horizontal plane, and held against a small wooden stake, 1) (What is the support force of the ground in the absence of wind?)

2) When there is wind blowing diagonally upwards at an angle of 45° horizontally to one of the faces of a long cube, it is shown. The pressure exerted by the wind on the smooth side of the cuboid is f, so how much force f is required to lift the cuboid up?

3) Experiments show that the wind produces pressure in the vertical plane on the smooth plane, and the pressure is proportional to the square of the wind speed, and the square of the angle between the wind and the smooth plane. Now let the wind blow from the upper left of the box, at an angle to the horizontal, as shown. When greater than a certain value, the box cannot be cocked, regardless of the wind speed.

Determine the value of by calculation.

It's hard enough.

Thinking about it from another perspective, in fact, the problem of junior high school is mainly a problem of thinking style, and the landlord can consider a topic like Lorentz transformation, which is very interesting.

Middle School Physics ??? Calm and calm、、

Wouldn't you know if you bought a counseling book?