Physical electrical power in the second year of junior high school

Oh, then let me say, when we only connect the power supply in the circuit, this electrical appliance, switch such a simple series circuit, the power supply voltage remains the same, the actual voltage is the rated voltage, you can know according to P is equal to the square of U divided by R, the resistance of the electrical appliance is determined on the appearance, will not change, the resistance is certain, and according to P=UI, it can be known that I=U R is Ohm's law, which is a known and invariant condition.

You don't have to jump in the analysis of the problem, you will know it step by step.

First of all, electrical appliances will not change, so r is certain, then the actual voltage is equal to the rated voltage, then the voltage u is unchanged, according to i = u r, the current is fixed and constant, so p = u * i must be a fixed value, the following formula can naturally explain the problem.

。。。Assuming that the electrical resistance does not change with temperature, then as long as the voltage does not change, the current will not change, and p=U2 R is the same.

As for the rated power, it refers to the electrical power of the electrical appliance at the rated voltage, and the rated voltage in China is 220V. . .

The definition needs to be clarified.

Abandon Hello.

The formulas of physics are different from the formulas of mathematics, and have a scope of application and certain limitations.

For example, in this question, you can only look at p=u 2 r, r must be constant (junior high school range), u=u amount, p is equal to p amount.

p=ui has two variables and cannot be applied.

To do physics problems, choose the formula that is most beneficial to you.

Hope mine is helpful to you.

I wish the landlord an unlimited future and everything is powerful!

Since the electrical resistance is certain, when the electrical appliance is a pure resistance, when the voltage is the rated voltage, according to Ohm's law, the current is a fixed value, no matter what formula is calculated, the actual power is equal to the rated power. But if it's not a pure resistor, the first formula can only be used. Besides, there is a definition of rated power in the book, isn't it the power of the electrical appliance under the rated voltage?

Solution: According to the meaning of the question.

The supply voltage is u=4

When the sliding rheostat is moved to the maximum resistance.

The total resistance in the circuit is r=r, lamp + rresistance=40 +20=60, i=u, r=6v, 60=

p light min=i r=(

p rheostat max=i r=(

When the sliding rheostat is moved to a resistance value of 0.

i=u/r=6v/40ω=

P light max=UI=6V

Therefore, the range of change is.

0w=I'm sorry.

Just read wrong.

+q 384413722 I sue you · reason + answer ·

The answer on the 4th floor is not correct, he counts it as the total power

The answer on the 1st floor is yes, but I can give you the reason

When the current in the protection circuit is 5A, it is a critical state, and the resistance value of the rheostat is maximum.

I didn't calculate the specifics, but I figured out that the rheostat resistance is 0 to that value, and then I calculated the lamp power in two states.

The range of power variations of the lamp.

The range of power variation of the rheostat.

Because it glows normally at a and because p=u*i so i=p u= r lamp=u i = 12 ohms at b the lamp consumes 1 4 p,= i= i = u r r=u i = 48 ohms so slip the maximum resistance is 36 ohms power = i 2*r=

1) By P=UI, I=P=60W=220V=3, 11(A)2)W=PT=11W*2000H=22 kWh.

W white = pt = 60w * 2000h = 120 kWh.

Power saving: w white - w section = 98 kWh.