Physical electricity and heat under the second year of junior high school, physical electricity and

For a non-pure resistive circuit, the total power is not equal to the thermal power.

As an example, for electric motors, there are:

The total power consumed by the electric motor =

The heating power of the motor + the power of the motor to do external work.

The total power consumed by the motor is calculated as p=ui.

The heating power of the motor is calculated as p=i square r.

You may ask, don't p=ui and p=i squared r want to wait? Actually, these two formulas are not equal. If you take physics notes, you should remember:

p=ui calculates the total power (regardless of what power is included in the total power), and in a purely resistive circuit, p=i squared r because u=ir.

In non-pure resistive circuits, U=IR is not true at all! (You can read the textbook on this, Ohm's law does not hold true in impure resistive circuits), so p=ui but u is not equal to IR, so the results obtained by the square of w=i rt and w=uit must be different.

If you don't understand, look at the following points: (Be sure to understand every sentence carefully).

1.Power is divided into total power, thermal power, and other power. In any one circuit, there is: total power = thermal power + other power.

2.In any one circuit, total power = ui and thermal power = i square r. Therefore, there is: other power = other power ui - i square r.

3.It's just that in a pure resistive circuit, all the energy is used to generate heat, and the other power = 0, so the total power = thermal power = ui = i square r, from which Ohm's law is derived (because ui = i square r, so u = ir).

4.In a non-pure resistive circuit, the total power is greater than the thermal power, so UI is not equal to i square r, and the meaning of the two formulas is different, and the number is of course different. It can also be seen that in non-pure resistive circuits, U is not equal to IR.

Here's an example:

An electric motor for lifting heavy loads, connected in series with an ammeter and in parallel with a voltmeter. The ammeter is 10 A, the voltmeter is 100 V, and the resistance of the motor is known to be 2 ohms. Seeking:

1) The heating power of the motor. (2) The power of the motor to lift the heavy object.

Analysis: If you look at the data in the question, you will know that U is not equal to IR. (That's for sure, because it's the electric motor).

The first question is good to ask, and the formula is set directly:

p = i square r = 200w.

The second question: It can be seen from the total power = thermal power + other power:

UI = i square r + the power of the motor to lift the heavy load.

Gain: The electric motor lifts the power of heavy loads.

100x10-200

800w。I'm exhausted. Hope you understand.

UIT does not calculate the electrical work of a pure resistor, it is applicable to everything. The square rt of i is only used to find the electrical work of a pure resistor, and p=ui is also a formula that applies to everything. If you understand what I just said, there will be no contradictions.

You may have made a mistake in class or the teacher made a mistake. It is also possible to calculate the pure resistance by UIT!!

p=i squared r is the power consumed on the resistor (pure thermal energy) p=ui is the power of the entire electrical appliance, where there can be a lot of energy consumption, such as heat, light, kinetic energy, etc. If it is a pure resistive circuit, the square rt of w=uit and w=i is equal, I wonder if this can be explained.

You are mistaken, the actual electrical power of an electrical appliance will change, but the rated electric power will not change.

The square rt of w=i is obtained from the square r of w= and p=i.

w=uit is derived from w= and p=.

Therefore, the physical quantities sought are definitely different.

No, W=i squared RT is mainly used in series circuits because the currents in series circuits are equal and the total resistance can be added. w=uit is used to know the current and voltage of the desired appliance, not because the physical quantity is different.

Just like p=ui, p=i squared r, p=u squared r

It is not that the physical quantities sought are different, but that different formulas are used in different circuits.

Q=W=I RT is mainly suitable for series circuits, and parallel circuits are generally Q=W=(U R)T, and after the resistance is reduced, the heat increases.

Both of these formulas can be reduced to q=w=uit, assuming that the voltage is constant, then when a resistor is connected in parallel, the resistance r will decrease, and the current i will increase, and finally the heat will increase.

The mistake in the analysis of the problem is that when the resistance r decreases, the current i will increase, which will make the formula q=w=i rt a little confusing, so choose the formula that can avoid using i, that is, q=w=(u r)t.

The total resistance becomes smaller after parallel circuits, but you have to look at the prerequisites of this circuit, w=i2rt is only suitable for pure resistive circuits.

The key point is: in the parallel circuit, when the voltage is constant, it is necessary to follow p=U2 r, the smaller the resistance, the greater the rated power, and the more heat generated per unit time.

And the situation you said later is only true when the current is constant, according to p=i 2r, the current is constant, the greater the power, the more heat is generated per unit time.

To put it simply, one is for parallel and one is for series.

Read more physics books for yourself and understand more concepts.

The key point is that if a resistance wire is connected in parallel, then the total resistance decreases, and the current decreases when the voltage does not change, so the formula q=i*i*r*t (the square of the current is multiplied by the resistance, and then multiplied by the time).

In this case, use the formula: q=(u*u r)*t (the square of the voltage multiplied by the time, divided by the resistance).

Got it? The voltage does not change, and the smaller the resistance, the more heat is generated with equal time.

1) When the switch is disconnected, the electric blanket is in a state of insulation.

2)1min=60s. p0"=w/t=60j/60s=1w.

r0 = u 2 p0 = (220) 2 100 = 484 euros. p0"=u0"2 r0=1w , the solution is u"=22v

Because the electric blanket heats up when the switch is closed. So the power supply voltage = 220V so at this time r0 voltage: r voltage = 1:9Because the series circuit currents are equal everywhere, r0:r=1:9

Since r0 = 484 ohms, r = 4356 ohms.

After answering, if you have any questions, please feel free to ask

Calculate which of the two currents is greater to know which power is larger, because Joule's law Heating power p=i ro

Therefore, when it is disconnected, the current is smaller, and it is in a state of heat preservation.

2、q=i²ro*60=60j

The data of the nameplate for resistor RO can be obtained as RO=U P=484 ohms.

Then we get i= 1 484=1 22 a

Whereas Ohm's law i=u (ro+r)=1 224840=484+r

r = 4356 euros.

You check it, you don't know how to ask.

(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.

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?

A false. Because it can also be based on the formula of specific heat capacity: q=cm&t, the specific heat capacity and mass of the ground are not small.

B yes, I almost read it wrong, it's the same object. According to the formula q=cm&t, so it is true.

c wrong. It's like an ice-water mixture, he doesn't absorb enough heat, his temperature is still 0 degrees, and then he absorbs enough to rise.

D false. Because it's also the formula above. Think about ice-water mixtures and you're fine.

I choose d d that the temperature is high and does not necessarily release heat, and the temperature drop may be external work, and the absorption of heat may also be used to do external work.

If there is a problem with the topic, if you look at it with the knowledge of high school.

It is difficult for junior high school students.

The rise and fall of temperature depends on the work done and the heat transfer" is sufficient.

A: False, it is also related to the temperature difference and the mass and specific heat.

B: Yes, it should be for an object.

C: False, if the external work is more at the same time as absorbing heat, the internal energy will be reduced D: False, the higher the temperature, the greater the heat, but it is not proportional.

c. Use the method of elimination.

Because if A is right, then D is right, and this is a multiple-choice question, then it is impossible...

b.Wrong, the conditions are insufficient, if two objects are in contact, it must be the conduction of heat and high temperature to low temperature, you don't say the conditions, this problem can't be said.

A. Wrong from Q = cm (T0-T) can be seen that the heat released is related to the specific heat capacity C, mass M, and the change in temperature (T0-T), and the conclusion can not be drawn only based on the original temperature T0 is high, but also look at C, M, T.

b. The implicit condition of this sentence refers to the same object, that is to say, c and m are the same, and it can be seen from q=cm(t0-t) that the mass m is the same as the specific heat capacity c, and the more the temperature drops [that is, the change in temperature (t0-t) is larger], so q is of course more.

c. The wrong object absorbs heat, and the temperature does not necessarily rise, such as ice in the process of melting, constantly absorbing heat, but the temperature has been 0 degrees Celsius, and it does not heat up until it is melted. The crystals are all the same when they melt, absorbing heat but not heating up.

d. False"Heat" cannot be said to "have", it must be in the process of heat transfer before this word can be used. This is defined by definition.

B is not necessarily related to the amount of heat emitted by the object and the temperature A is wrong.

Crystals (e.g., ice) absorb heat when melted, but the temperature does not change.

Heat is the amount of process, and it can only be said that it releases or absorbs heat, and it cannot be said that it has heat D.

b is correct, and it can be considered water and ice. b is the same object. A is the specific heat capacity of the body.

Haha, the ...... of the second year of junior high schoolI miss it so much.

So much nonsense to choose A.

Answer C Question Hint Blocks indicate that the density of the object does not change in water (generally speaking, the less dense the object is, the more pronounced it is with temperature, and the main influence here is, of course, water).

According to the thermal expansion and contraction, 90°-10°, the volume of water will shrink and the mass will remain the same, then the density will increase, and the buoyancy generated by the same volume of water will become larger. The block density does not change, so it floats.

Children don't understand and ask, it's very promising.