A topic about junior high school electricity

It's not just about the safety of the lab equipment, it's about making sure you can get the data you need.

For example, this question is to measure the rated power of the bulb, which is the goal, and we have to consider what the rated power will be calculated through.

Originally, it stands to reason that there is voltage and resistance, and the rated power can be calculated at once, isn't it painful! But he said that it was about 12 euros, which means that it was not accurate, it was only about 12 euros, it could be, it could be, it could be, the egg hurt +1.

Since the resistance is unreliable, it can only rely on the measured current, which is the purpose of Xiao Ming's design experiment.

The above is a casual thinking in your head, you may feel that it is not necessary, in fact, the above thinking is some ideas that will only be had when the topic is more than a lot of ideas. These thoughts are better than the cranky thoughts you see the topic and have no brains, right?

Next, we will officially enter the problem solving idea.

The bulb + rheostat is connected in series, and the current is equal. Roughly calculated, the rated current of the small bulb is about (directly obtained), and it is found that it does not exceed the maximum current of the two rheostats, then both can be considered.

The bulb + rheostat is connected in series, and the voltage has to be distributed according to the resistance. The rated voltage of the bulb is, the resistance is temporarily regarded as 12 ohms, the column formula, here is the voltage after 6V is separated, that is, the voltage of the rheostat, X is the resistance, calculate and find that X is greater than 5, then the "5 ohm 1 ampere" this rheostat is not qualified, then you can only choose the "50 ohm amp" of this rheostat.

If you think the other way around, it is okay to consider the resistance first and then the current, so that the "5 ohm 1 amp" rheostat is excluded first, then the remaining "50 ohm am" rheostat is naturally qualified. Your current question is okay, and there won't be any problems that the two rheostats can't work, so you can get the answer directly.

When the small bulb is working normally, its voltage is Then the current through it is about , and the voltage at both ends of the sliding rheostat in series with it is , so the resistance of the sliding rheostat connected to the circuit should be ohms, so we should choose a 50 ohm sliding rheostat.

Analysis: On the one hand, it is necessary to consider the safety of each electrical appliance, and the maximum current and maximum voltage that are allowed to pass through how the requirements work, especially the normal operation of the bulb or the special requirements of the circuit.

I don't know how to ask questions.

Find the lamp resistance r1: according to p=u2 r, r1=(220v)2 100w=484

The actual current of the circuit is measured according to the actual resistance of the bulb: according to P=I*IR, i=9 22A

According to U=IR, the actual voltage of the bulb U1=198V, the wire resistance can be regarded as the resistance R2 in series with the bulb, and the voltage of R2 can be obtained U2=220V-198V=22V, and according to the equal current in the series circuit, the power of the wire is P=UI=22V* 9 22A=9W

p rated = u square r

Find the r-lamp=484

Then use p actual = i square r lamp.

Find the current = 5/11.

Finally, use p total = i u

The total power obtained is 100 watts.

Subtract the power of the lamp and the power of the wire = 100-81 = 19

First calculate the resistance of the bulb, and then calculate the voltage at both ends of the bulb and the current flowing through according to the actual power, the total voltage is subtracted from the voltage at both ends of the bulb to obtain the voltage at both ends of the wire, and this voltage is multiplied by the current to obtain the power consumed on the wire.

You can think of a wire as a resistor, at 200V, the electrical power is 100W, p total = p bulb + p wire, and the power consumed on the wire is 19W

b right. It is obvious that the two appliances are connected in parallel, and option A is wrong;

When the switch is closed, point B is directly connected to the live wire, and the person will be electrocuted, while point C is connected to the zero line, and the voltage between the zero line and the ground is small (a few volts to 10 volts), and generally will not be electrocuted, option C is wrong;

When the switch is disconnected, point A is on the fire line, and point B is not on the fire line, so only at point A can make the neon tube of the voltage tester glow, option D is wrong.

1) As can be seen in the figure, when 40 degrees, it corresponds to 400 ohms, which is the same as the fixed resistance value, and the bipartite voltage is 12V

2) As can be seen from the figure, the higher the temperature, the smaller the resistance.

When the voltmeter reaches the extreme value of 18V i=18 400=> exceeds the extreme value of the ammeter, so the maximum current is.

U= U0=24-16=8V R0=8 ohms, and then you can see from the figure: 200 ohms corresponds to 60 degrees, so the maximum temperature is 60 degrees.

1) When the temperature in the incubator is 40, it can be seen from Figure B that the resistance value of R0 is 400 at this time, then the indication of the voltmeter U1=I total * R = U total (400 + 400) * 400 = U total 2 = 12V.

2) The range of the ammeter is 0 40mA, when the current in the circuit is maximum, the voltmeter U max = i max * r = 16V, then R0 = (24-16) ohm, known by the image, the maximum temperature is 60.

Circuit 2 resistors are connected in series, t = 40 rx = 400

rx=r u=uTotal 2=12v

Series current, etc., voltage division, resistance small voltage and large voltage, 18V 400 ohm = 45mA so it can not be calculated according to voltage.

According to the current calculation, (

First Light:

Maximum current: i1=p1 u1=6 12=

Resistance: r1= u1 i1=12 second lamp:

Maximum current: i2=p2 u2=2 8=

Resistance: r2 = u2 i2 = 8

Because of the series circuit, the current is equal everywhere, so the current cannot exceed the maximum current of the second lamp:.

At this point, r total = r1 + r2 = 24 + 32 = 56 so u = i2 * r total =

Let the maximum current of the first lamp be i

then there is UI 2=6 then i=

In the same way, the maximum current of the second lamp is.

Since it is in series, the maximum current is taken.

The maximum power is 6+

If the maximum voltage is U, the maximum voltage that can be calculated is 14

According to the formula r=u 2 p, the resistance of the two lamps is 24 ohms and 32 ohms respectively, when the voltage is applied at both ends of the circuit in series, the partial voltage of the resistive is larger, and the partial voltage of 32 ohms will be greater than that of 24 ohms.

And because the maximum partial voltage of the rear lamp is 8V, the voltage of the headlamp must be lower than 8V at this time, so the circuit is safe and should be calculated accordingly.

Because the resistance in a series circuit is proportional to the voltage across the resistor u r=u1 r1=u2 r2 where r=r1+r2=32+24=56

i.e. u 56=8 32 gives u=14

R2 is connected in parallel with R3, equivalent to R23

r23=30/2=15

R4 is connected in parallel with R5, equivalent to R45

r45=15*60/(15+60)=12

R23 is connected in series with R45, equivalent to R25

r25=15+12=27

R1 is connected in parallel with R25.

rab=27/2ω

Since R2 is connected in parallel with R3 and the resistance value is equal, the electrical power consumed by R2 is equal to the electrical power consumed by R3, and the power ratio is 1:1

R2 and R3 are connected in parallel with R4 and R5 in series, and finally R1 is connected in parallel with R1 rab=r1 (R2 R3+R4 R5)=27 (30 30+15 60)=27 (15+12)=

R2 is connected in parallel with R3, the terminal voltage is equal, and the power consumption ratio is:

p2/p3=u^2*r2/u^2*r3=r3/r2=30/30=1

I just graduated from junior high school, and I learned electricity very well, scoring 90 points on every test.

I also calculated, r(ab)=,p2:p3=1:1.

The practice is circuit simplification. Because R2 and R3 are connected in parallel, we can turn the two resistors into one, the formula is (R2 R3) (R2+R3)=15, then we will call this resistance R(2+3)!

Now there are only 4 resistors, and then, we look at R4 and R5, which are also in parallel, and the algorithm is the same as above, and the result is equal to 12.

Now there are only 3 resistors left, that is, r1, r(2+3), r(4+5), you see what circuit is now, it is a simple hybrid (there is a series connection, and there is a parallel connection), then now it is r(2+3) and r(4+5) in series, [r(2+3)+r(4+5)]=27, which is exactly equal to r1, then now the resistance of 27 is connected in parallel with r1, and finally r(ab) is equal to (27 +r1) 2=.

We know from p=iu and r=u i, p=u r, so p2:p3=(u r2):(u r3), because r2 and r3 are connected in parallel, so the voltage is the same, get the formula (u r2):

u²/r3）=（1/r2）:(1/r3)=1:1。

Hope it helps.

Answer: CA is wrong, because, the resistance value increases, the current becomes smaller i=u RB: The voltmeter measures the power supply voltage, the battery is not replaced, and the voltage is of course unchanged.

c;The power supply is connected in parallel with those two, so the voltage is equal, and the resistance of the lamp does not change, so the electric speed does not change the square r of p=u

d: The resistance of the whole circuit increases, the voltage does not change, and the electrical power of course decreases.

Of course, the power supply voltage is constant, because the power supply is not changed, and the voltage is of course stable.

1 A complete circuit should include four parts: wire and switch2 Outside the power supply of a closed circuit, the current flows from the pole to the pole of the power supply 3 The current is formed by the charge , and the direction of movement is defined as the direction of the current 4 The voltage is formed in the circuit is a device that provides voltage

Are these questions enough?

Yes, the supply voltage is always the same.

The power supply of junior high school is the ideal power supply, no matter what the circuit voltage source voltage is fixed at any time, the electricity meter of junior high school is also the ideal electric meter, the ammeter is the wire, and the voltmeter is disconnected.

In the figure, the bulb and the resistance are connected in parallel, the voltage at both ends is the power supply voltage, the ammeter measures the combined current of the resistance and the bulb, and the voltmeter measures the power supply voltage. So the magnetic field increases, the resistance increases, the current flowing through the resistance decreases, the bulb current does not change, the ammeter reading decreases, and the voltmeter reading does not change. According to the power p=u square r, it is known that the power consumed by the bulb is constant, the resistor power consumed decreases, and the total power consumed decreases.

So C is correct.

This problem is of relatively high quality, and it takes a few detours to solve it, which can be regarded as a difficult problem. In order to explain the idea of solving the problem, we use the method of backwards.

First of all, the problem is to find P2', assuming that the resistance values of A and B are R1 and R2 respectively, and the voltage at both ends of the original circuit power supply is U, and it is known that the voltage at both ends of the later circuit power supply is 2U, so.

p2' =2u)*(2u)/r2 = 4u×u/r2 ……1）

Now that both you and r2 are unknown, we have to find a way to calculate the power of the B resistor, so that we can connect p2 and p2' and avoid r2. Suppose that the voltages of A and B in the original circuit are u1 and u2 respectively, because A and B are connected in series, so u=u1+u2. Note that the power of the B resistor is.

p2 = u2×u2/r2………2）

Comparing (1) and (2), we find that simply finding the proportional relationship between you and u2 can solve the problem. In this case, it is necessary to flexibly apply the formulas and theorems that you have already mastered.

According to the formula p=i*i*r, the currents in the series circuit are equal everywhere, so p1:p2=r1:r2

According to Ohm's theorem u=i*r, the currents in a series circuit are equal everywhere, so u1:u2=r1:r2

This solves the problem of the proportional relationship between you and u2, because u1:u2=p1:p2, thus.

u:u2 = u1+u2):u2==(p1+p2):p2………3）

Substituting (3) into (1) obtains.

p2' =4×u×u / r2 = 4×(p1+p2)×(p1+p2) /p2×p2 * u2×u2/r2)

According to (2), p2 is in the last parenthesis of the above equation, and p2 in the denominator is eliminated to obtain it.

p2' =4×(p1+p2)×(p1+p2) /p2

In short, to solve this problem, you need to be very familiar with the relationship between voltage, current, resistance, and power of series circuits, which is very rare for a junior high school student.

A No, the switch can be placed on the trunk circuit to control all of it.

b No, the current of the series circuit is equal, and if it is not equal, it is in parallel.

c is in parallel, not in series.

d is the correct answer.

As for the series circuit you mentioned, there is also a situation where the voltage is equal, and it is only when the two resistors are equal (r u i, i equal, u equal).

Normal luminescence means that the actual power of the lamp is equal to the rated power.