Physics and electricity in the third year of junior high school, as long as the last is empty, ask f

Answer: p=

Analysis: When the switch S is pulled to position 1, the ammeter measures the current i1 flowing through the resistor, and the bulb and the resistor are connected in parallel, so u=ul=ur=i1*r. When the switch S is pulled to position 2, the ammeter measures the current i2 in the trunk circuit; The bulb and resistor are still connected in parallel, and u=ul=ur=i1*r.

Then the current on the resistor is still i1, and the current flowing through the bulb is: i2-i1, so the resistance of the bulb rl=ul (i2-i1)=i1*r (i2-i1). The rated power of the bulb p = the rated voltage of the bulb u amount x the rated current of the bulb i amount.

The rated current of a bulb is equal to its rated voltage divided by the resistance, i.e.: i = u [i1*r (i2-i1)]. Bring in the formula p=u amount xi amount to get:

p = (u forehead) 2 * (i2-i1) amount =, and finally get: p =.

Use the square of the rated voltage of the small bulb, that is, the square divided by the resistance of the small bulb.

p=u2 r=(.

wo I don't understand. You started that picture in**Ahh Brothers.

The power is 880W when blowing hot air, and the motor and heating wire work at the same time, and the total power of the two is 880W

Blowing cold air is 80w power. Only the motor works, and the power of the motor is 80W

Then the electric power of the heating wire is (880W-80W

880W is not thermal power, Q is thermal power, and the mechanical power of the hair dryer is 80W when blowing cold air, so subtracting 80W is Q

Because it's a hair dryer that works normally, so I don't think it's a cold wind or a hot wind, so < 880w--80w>.

r1 = u1 i1 = 2v ohms.

If the voltage is expressed to the full scale, the range of the voltmeter is selected as 3V, and the indication of the voltmeter is 3VI2 U2 R1=3V 10 ohms=

The maximum current in the circuit is: i=U R1=6V 10 ohms If the current is expressed to the full scale, the range of the ammeter is selected, and the number of the ammeter is 6V

Solution: When the current representation number is ampere, it is obtained by i=u r, u1=i1*r1=ohm=2vu2=i2*r2=ohm=4v

u total = u1 + u2 = 2v + 4v = 6v = u power supply means that the internal resistance of the power supply is 0

When the rheostat is moved to a new position, if the current is expressed to full-scale amperes, then u1'=i1'*R1 = Euro = 6V>3V, so if the voltmeter reaches the full scale of 3V first, then the current representation number is:

i"=u1'r1 = 3v 10 ohms =

If the voltmeter comes first, it is 3V, then the R2 resistance is 10 ohms, the current is, and the ammeter is not full, so it can.

If the ammeter arrives, then the ammeter shows the least, the maximum 3A, 3A is obviously impossible, then the ammeter display must be 6V, that is, the sliding rheostat access resistance = = 0 ohms, which is already exceeded.

Analysis:1Understanding of commonly used electricity meters. The voltmeter has two ranges, 3V and 15V; The ammeter has two ranges and 3A

2.According to the circuit diagram shown in the figure, the range is selected. The power supply voltage is 6V, and when P is moved to the leftmost end, the voltage is 6V on R1, and the maximum current is at this time, so the voltmeter range must be 15V.

As for the range of the ammeter, since the maximum current of the circuit is, any range can be, but if it is 3A, the ammeter and voltmeter cannot be full, which is not in line with the topic, so the range that should be selected (can be full).

3.Calculation and analysis: There is only one that represents the full article, which is the ammeter, since it is a full article. According to Ohm's law, the number of voltage representations should be 6V.

4.Based on the above analysis, the position of the slide p should be at the far left, as there is no partial pressure.

There is a question to know: the range used by the ammeter and the voltmeter is respectively.

When the voltage is expressed to 3V on the full scale, the current is expressed as the number is, which is in line with the topic.

When the current is expressed at a full scale, the voltage at the R1 terminal is 6V, which is not in line with the topic (rounded).

The ammeter is full scale amps (500 mamps), and the voltage is expressed in the number of 5 volts.

The charge moves in the electric field, perpendicular to the direction of the electric field, and does not do work. Therefore, the effective displacement of ball A is 2l and the effective displacement of ball b is l. Because the sum of the potential energies of the two spheres A and B does not change, the electric potential energy increases and decreases.

As for who increases and who decreases, there is no requirement, so A and B have heterogeneous charges, and there is no requirement for who is positive and who is negative. Options A and C are incorrect. The amount of change in electric potential energy = eql Because of the uniform electric field, e is the same, and the displacement ratio is 2:

1, so the ratio of power is 1:2, and option b is correct. Because the electric potential energy does the work, that is, the electric field force does the work, the d option is wrong.

Analysis: When both switches are disconnected, R1, R2, L are connected in series.

i＝i1＝i'2 u1 r2, i.e. r2 u1 i1 ......and: p1 i1 2r1, i.e. r1;

When only switch S1 is closed, R1 and L are short-circuited.

r2＝u2/i2……and i2

From and obtain: u1 u2 i1 i2......By the title: u1 u2 1 2 ......Substituting and substituting the value of i1 gives r1 10( ).

When only switch S2 is closed, R2 and L are disconnected.

p'1＝i'1 2r1 , substituting the r1 value into , gets: i'1. At this time, the power supply voltage: u u'1＝i'1r1＝

It is not difficult to find: r2 u i2 6 ;

The voltage at both ends of the power supply is unchanged, i1 (r1 r2 r lamp) i'1R1, i.e. (R1 R2 R lamp) R lamp 60 R1 R2 60 10 30 20 ( ) Since S1 and S2 are closed, R1, R2, L are connected in parallel, 1 R total 1 R1 1 R2 1 R lamp 1 10 1 30 1 20, i.e. r total (60 11) (

At this time, the lamp emits normally, and the total electrical power of the circuit when the lamp l emits normal light: p total u 2 r total 6 2 (60 11).

When both switches are disconnected, the three conductors are connected in series.

When switch S1 is closed only, R1 and L are short-circuited.

When switch S2 is closed only, R2 and L are short-circuited.

When both switches are closed, R1 and L are connected in parallel, R2 is shorted in series with three conductors, and the electrical power consumed by resistor R1 is. When the switch S2 is only closed, R2 and L are short-circuited, and the electrical power consumed by R1 alone is pushed out by P=U2 R: when the three conductors are connected in series, R1 distributes voltage to 1 6 of the total voltage

u1:u2=1:2, which is proportional to the resistance of the series voltage distribution r1:rl:r2=1:2:3

When the lamp l emits light normally, R1 consumes P=, Rl consumes P by P=U2 R, parallel R1:Rl:R2=1:2, and Rl consumes P as a quarter of R1).

Then the total power consumed by the circuit is .

According to Ohm's law, the voltage of the power supply is calculated in combination with Figure B, and then according to p=u2

R combined with Figure C to find the electrical power of the bulb is the rated power of the bulb U=U2=I2R2=, in Figure D, the rated power of the lamp L pl=

u2rl(6v)2

I agree with you, this blank filling of 5 euros is already the main point of the question.

1. The fixed-value resistance r1 = 5 ohms is the known condition given by the question, and the value of r1 is also calculated in the calculation, and the known condition is overturned, which is not allowed in this question.

2. The calculated average value is in Europe, which is actually the average value of (R1+RG), that is to say, the error is caused by the ammeter.

3. Compared with 5 ohms, this error is negligible in general calculation problems.

As a ** question, it is not required to be so precise.

You have a point. However, the title says that we should also measure a data that u i is equal to ohms, that is, move the sliding rheostat, read out the current and voltage indicators, and find the set of data with a ratio of ohms, which is the data that takes into account the error, because it is the average value. When doing this, we only need to read the current and voltage, and the ratio must have a decimal.

Agree with your idea, fill in 5 euros for the third question.

This is very simple, according to the resistance rod (similar to the sliding rheostat) access resistance to calculate the height, the higher the height of the access resistance, the larger the voltage representation, the sliding rheostat should understand, the resistance series will be calculated, then it is very simple.

Solution: (1) The indication of the light change voltmeter is 6V, and the voltmeter is connected to both ends of the power supply, indicating that the circuit fault may be the light bulb open circuit or the resistance value of the sliding rheostat connected to the circuit is 0, and the light bulb is not lit and the ammeter has no indicator, and the circuit fault should be the light bulb open circuit.

Because the indication of the voltmeter is, the bulb emits light normally, and the current indicates the number i=, so the rated power of the bulb p = u i =

2) When the voltage at both ends of the bulb is obtained U=I1R, and when the switch S is pulled from position 1 to position 2, the connection mode of the three resistors in the circuit does not change, so the voltage at both ends of the bulb is still I1R, and L is connected in parallel with R, the bulb current is obtained by i=i2-i1, and the bulb resistance rl=i1r i2-i1 is obtained by Ohm's law

Due to the thermal effect of the current, when the bulb emits light, the resistance of the filament will increase with the increase of temperature, so when the voltage at both ends of the bulb is not equal to its rated voltage, the resistance obtained is not equal to the resistance when it emits normal light, so that the power obtained is not the rated power;

Therefore, if you want to find the rated power of the hidden bulb with the bright key, you must make the voltage at both ends of the bulb equal to the rated voltage.

Therefore, the answer is: (1) the filament is broken;

i1r i2-i1, the bulb does not shine at the rated voltage, the temperature change of the filament causes the resistance of the filament to change, the switch S is pulled to position 1, and the slide p is moved, so that the current representation number is i1=u0r

According to the title, there are two cases where r2 is 0 and 15, and the following is to calculate that when r2=0, the power supply voltage u=, r1 power p1=(when r2=15, the current in the circuit is i2=u (r1+15) r2 power is p2=(i2) 2r2

Based on the power ratio of 9:4 on R1, P1:P2=9:4 can be calculated, and the result: R1=30

Power supply u=9V

Show digital 3V to draw a circuit diagram, depending on the situation, in fact, it is very simple.

When R2=0, the power supply voltage U=, R1 Power, P1=(When R2=15, the current in the circuit is I2=U (R1+15), and the power of R2 is P2=(I2) 2R2

Based on the power ratio of 9:4 on R1, P1:P2=9:4 can be calculated, and the result: R1=30

Power supply u=9V

Indicates digital digit 3V