How to talk about the electrical power part of junior high school physics class

There are not many formulas, only with your heart, you can remember:

1. Formulas of current, voltage and resistance of series circuits.

i=i1=i2；u=u1+u2；r=r1+r22, parallel circuit current, voltage, resistance formula.

i=i1+i2；u=u1=u2；1 r=1 r1+1 r23, Ohm's law expression.

i=u r4, electrical power definition.

p=w t electric power measurement calculation formula.

p=ui5, Joule's law expression.

q=i^2rt

Actually, it's not much.

Partial pressure, shunt.

Ohm's Law. Electric work, electric power.

That's all there is to the electrical.

There is also an infrequently used i=q t

Different formulas are used and functional, remember when you have to understand the memory, especially to see how the example questions are used, it is better to remember the example questions, so that you can use the formula like a fish in water, don't wait until you finish the formula to memorize the formula. 1. Formulas of current, voltage and resistance of series circuits.

i=i1=i2；u=u1+u2；r=r1+r22, parallel circuit current, voltage, resistance formula.

i=i1+i2；u=u1=u2；1 r=1 r1+1 r23, Ohm's law expression.

i=u r4, electrical power definition.

p=w t electric power measurement calculation formula.

p=ui5, Joule's law expression.

q=i 2rt -- physical quantity unit meter.

Name Symbol Name Symbol.

Mass m kg kg

Temperature t degrees Celsius m=pv

Velocity v m s v=s t

Density kg m3 kg m3 = m v force (gravity) f Newton (N) n g = mg

Pressure p Pascal (Pa) Pa Pa P=f S

Work w joules (joules) j w=fs

Power p Watts (Watts) w p=w t

Current i amps (amps) a i = u r

Voltage U volts (volts) v u=IR

Resistance r ohm (ohm) r=u i

Electric work w joules (coul) j w=uit

Electrical power p Watts (watts) w p=w t=ui heat q joules (joules) j q=cm(t-t0)--study o( o

It's just a matter of personal feelings

I feel that the problem is difficult, and it is difficult to be difficult.

I'm also confused, I'm tutoring people, and the child's foundation is poor, so I should understand how the formula came about first.

The representation symbol w of a little energy, the relationship between electrical energy and electric work, their representation symbols are w, the SI units are (joules), and the commonly used units have (degrees, kilowatt-hours) The relationship is 1 degree = 1 kilowatt-hour = 6 joules, and then there is electrical work, the work done by current, the formula is w = uit

1. B The electric energy consumed by electric fans, incandescent lamps and electric heaters is the same.

W=pt, because the electric power p is the same, and the energizing time t is the same, so the electric energy consumed is the same.

The electrical energy consumed by the fan w is converted into mechanical energy and thermal energy.

The electrical energy consumed by an incandescent lamp w is converted into heat energy and light energy, q w.

All the electric energy W consumed by the electric heater is converted into heat energy Q, Q W.

2. When the filament is broken and then put on, the resistance decreases, the voltage at both ends of the lamp remains unchanged, and the actual electrical power of the lamp increases according to P=UU R.

The brightness of the lamp is determined by the actual electrical power of the lamp, and the increase of the actual electrical power means that the conversion of electrical energy into light energy increases every second.

So the lights are brighter than they are.

3. When connected in series, the electric squeeze is the same, and in the same time, the amount of electricity passing through is equal. The relationship between the amount of electricity and the flow and time of electricity: q=it. Brother Chun.

The resistance of iron wire is greater than that of copper wire. When the current is connected in series, the energizing time is equal, and according to Q=IIRT, it can be seen that the heat generated by the current through the iron wire is more than the heat generated by the current through the copper wire.

The formula is not much, only with heart, you can remember:

1. Formulas of current, voltage and resistance of series circuits.

i=i1=i2；u=u1+u2；r=r1+r22, parallel circuit current, voltage yield, resistance formula slippery.

i=i1+i2；u=u1=u2；1 r=1 r1+1 r23, Ohm's law expression.

i=u r4, electrical power definition.

p=w t electric power measurement calculation formula.

p=ui5, Joule's law expression.

q=i^2rt

Voltage, resistance formula.

i=i1=i2；u=u1+u2；r=r1+r22；u=u1=u2, you can remember: finger touch.

1. The series opening and voltaining circuit quietly carries the current and resistance formulas.

There are not many formulas for i=i1+i2, only intentions, parallel circuit current, voltage; r=1/；1/r1+1/r2

3. Ohm's law expression.

i=u/

Choose Kai Li D, use the power formula p=u r so p1:p2=2w r:3w r about r, and 2:3 a is right.

Looking at the diagram obtained by p=u r, utotal=u2=u3 utotal=3w*r1, the first figure obtained by the method of a stove is obtained u1=2w8r1, then r1:r2=1:2 b is right.

P1:P2=U1*I:U2*i is correct to get 2:1 c by removing i.

So choose D

The power of R1 in the diagram is p=i2r=[U (r1+r2)]2r1=4w---

The power of R1 by Figure B2 R1 = 9W --

The r1 r2=2 option is correct.

In Figure A, the two lights in series can be seen from P=I2R, the power ratio is equal to the resistance ratio, and the width is correct with the C option.

In Figure A, the ratio of the series voltage is equal to the resistance ratio. Therefore, the voltage ratio of the two lamps is u1 u2=2 1, i.e. u1 = 2 3 * u and the voltage of diagram b l1 is u so, option a is correct, and option d can be based on p = u2 r, figure a p = u2 r1 + r2).

Figure B p=u2 r1 +u2 r2

Then be cautious about the potatoes on r1 r2=2 1

Three-way coupling, can be solved A and B two diagram circuit total power consumption ratio is 2 9 I wish you progress!

Set the filament resistance of L1 to be R1

The filament resistance of lamp L2 is R2

In Figure B, the power of lamp L1 pB=U2 R1=9 (1).

The power of L1 in the nail diagram is PA=U'Counting spring 2 r1=4w (2).

u'/u=2/3

A is correct u'Envy does u=2 3,u'=2U3, then the brother balance voltage U2=U-U at both ends of the lamp L2 in Figure A'=u/3

B Error In Figure A (set the current iA), the power of lamp L1 p A = i A U', the power of L2 P2 = i A (U-U')

p2=u'/(u-u')=2u/3)/(u/3)=2/1

c is correct and p a p2 = i a 2r1 i a 2r2 = 2 1, then r1 r2 = 2 1

In Figure B, the ratio of the power of the lamp (U 2 R1) (U 2 R2) = R2 R1 = 1 2

D False Correct answer ac

Content from the user: Immortal Directions.

Power" Lesson Plan [Teaching Objectives].

1. Understand the physical meaning of power.

2. Understand the concept of power.

3. Know the formula of power, and will make simple calculations about power.

4. Know the unit of power "watts (w)" and be able to apply it correctly in calculations.

New Lesson Introduced:Student Activities:

Pick up the textbook vertically from the table, then pan it, and finally stand still.

Ponder: What are the two necessary factors to do work?

And analyze under what circumstances did the students do their work on the book? Under what circumstances do you not work on books?

Multi** display: two necessary factors to do work:

1. The force acting on the object;

2. The distance that the object passes in the direction of the force.

Student Activity: Computing.

1. Xiaohong weighs 500N, and when she runs to a 10m-high building within 5s, Xiaohong does J's work.

2. Xiaoli weighs 400N, and when she runs to a 10m-high building within 5s, Xiaohong does J's work.

3. Xiao Ming weighs 500N, and when he walks to a 10m high building within 10s, Xiao Ming does J's work.

4. Xiaoqiang weighs 600N, and when he runs to a 12m high building within 10s, Xiaoqiang does J's work.

Multi** display: w=fs,w—j; f—n;s---m；1j=Student Discussion:

Among the three classmates above, Xiaohong and Xiaoli are quick to work, how do you compare?

Xiao Hong and Xiao Ming, do fast work, how do you compare?

Xiaohong and Xiaoqiang, do fast work, how do you compare?

In physics, a new physical quantity: is used to express the speed of work done by an object.

Teacher's Board Book: Section 3 Power --p).

Lesson**: Teacher's question: How do we compare the speed and slowness of object motion?

The student will answer:

^r=ui=u^2/p=（10v）^2/1w=100ωi=p/u=1w/10v=

The rated voltage is 10V

p fixed value resistance = u 2 r = (10v) 2 16 = this is the basic question of junior high school physics and electricity, a certain version to understand.

If you still don't understand, please ask.

Hello landlord:

Solution: by Ohm's law u

ir, also has its own power p i * r

bai i*(dur 16) zhi10vi *r 1w

Solving the system of equations yields r 4 dao ,i

or r 64 i

The rated voltage is 2V or 8V

The power of the resistor is 4W or I*R.

It's a tandem, right? I'm going to have a tandem answer to this.

The bulb is connected to a 10V power source, and the power is du

for 1w. The bulb happens to shine normally.

Explain the rating of the bulb DAO.

The return voltage is A10V, and the rated power is 1W

The resistance of the bulb r1 = 100 1 = 100

The current through the bulb is i1 = 10 100a =

The power of the resistor p=100 16w=

p=u^2/r

So r=u2 p=100 ohms.

i=u/r=

Because the bulb emits light normally, u rated = 10V

Presistance=U2R=

From the question p=u*i=u r

R1 = U1 P = 10 ohms, I1 = 1A, R2 = 15 ohms, I2 = maximum current in series, U = (R1 + R2)*, P = U * I=9W The maximum voltage is 9V when connected in parallel, I = (1 R1 + 1 R2) * U=, P = U * I = current maximum power is.

Connect them in series, then the maximum voltage allowed to be added at both ends is 15 V. If they are connected in parallel, the maximum current allowed through the trunk circuit is A. The maximum power of the current consumption is.

Two lamps in series to see the maximum current (the total current can not exceed the maximum rated current of any one), two lights in parallel to see the maximum voltage (same as above).

Concatenation: i1=1(a),i2=, so i(actual)=actual)=(10+15)*

p=i(actual)*u=9(w).

Parallel: u1=10(v), u2=9(v), so u(actual)=9 (v) i=u(actual) r1+u(actual) r2=

p=u(actual)*i=1(w).

The maximum electrical power is.

Solution: (1) Only close S1, R1 and the sliding rheostat are connected to the circuit in series, and the voltmeter measures the voltage at both ends of the sliding rheostat, U1 = IR1 = 1 A 5 ohm = 5 volts.

Supply voltage U = 5 volts 4 volts = 9 volts.

2) Close only S2, the lamp is connected in series with the sliding rheostat, the voltmeter measures the voltage at both ends of the sliding rheostat, and the voltage at both ends of the sliding rheostat U2=IR2=A2 ohm = volts.

Resistance of the lamp r lamp = u lamp i = (9 volt ampere=8 ohms, disconnect s1, close s, s2, only the lamp works, the rated power p of the lamp = the square of u r lamp = 9 volts squared 8 ohms = watts.

3) Only close S2, the lamp is connected in series with the sliding rheostat, the maximum current allowed by the sliding rheostat is i = 1 amp, at this time the sliding rheostat access circuit resistance is the smallest, the voltage at both ends of the lamp U lamp = IR lamp = 1 A 8 ohm = 8 volts.

At this time, the power of the sliding rheostat is small, p2 = u2i = (9 8) volts 1 amp = 1 watt.

When the resistance of the sliding rheostat reaches a maximum value of 5 ohms, the minimum current i passing through the circuit'= u r total = 9 volts (8 5) ohm = amp.

At this time, the power of the sliding rheostat is maximum, p2' = i' squared, r2 = ohm squared, 5 ohms = watts.

The variable range of the sliding rheostat power is (1 watt.