Physics and electricity calculation questions in junior high school, and physics and electricity cal

R1 voltage at both ends of the 3V sliding rheostat access resistance 10 ohms.

r2=u2÷i=2÷

2.From the inscription, it can be seen that the current after moving is: i=

u1=u2=

Resistance after movement: r2=

1、i=;u=6v

u1=i*r1=4v

r2 = u i - r1 = 5 ohms.

2. If the sliding rheostat moves to the right and R2 becomes larger, the current indication number becomes smaller and the voltage variable number remains unchanged.

and i=; u=6v

According to the problem, it can be obtained: let i2 be the current after the change.

i-i2) u=ammeter range, voltage range.

1) The range of the ammeter is; Voltmeter range is 0-15V;

i-i2)/u=1/25;

i=;u=6v

i2=Valid.

After moving. u1=

u2=r2=

2) Through the calculation, it can be seen that the combination of other meter ranges is not valid.

1)u1=i*r1=4v;u2=6-u1=2v;r2=u2/i=0/5a

2) I don't understand what it means, so it's easier to solve the problem in detail.

Solution: When EF is connected to 36V voltage at two points, S1 and S2 are disconnected P total = P = 36 2 R2

When AE is connected to 36V closure total = p'=36 2 (r1+r2) it is known that p:p'=9:5 so:

p：p’=（36^2/r2）：=9：5

When EF is connected to 36V voltage at two points, S1 and S2 are closed P Total = , Total Resistance: R1*R2 (R1+R2).

P total = 36 2 (r1 + r2) r1 * r2 = put 5r1 = 4r2

r1=4r2 5 substituting 36 2(r1+r2) r1*r2= get:

r2 = 40 ohms.

r1==4r2 5=4*40 5=32 euros.

So i=36 (40+32)=

p1=i^2*r=

Answer: The electrical power consumed by R1 is P1=8W

Solution: When the EF is connected to 36V voltage at two points, S1 and S2 are closed, the equivalent resistance at both ends of EF ref=R1R2 (R1+R2)....①

p total = 36 ref ....②

S1, S2 disconnect P total = P, there is P total = P = 36 R2 When AE is connected to 36V to close S2, the total resistance at both ends of AE Rae = R1 + R2P total = P' = 36 Rae

From the inscription, p:p'=9:5

It can be found from the above ratio.

1) The normal working current is 60 divided by 220, which is equal to how much, how much, how much. [Dear,Press your ** calculator.。。 I know the most annoying thing about the high school entrance examination physics 220 what...

2) This... Isn't it just 1) that baa = =

3) This is a little more complicated, in 1) is not the calculation of the rated current, and then calculate his resistance, the resistance is negligible change... That's a fixed value.。。 Then use this constant resistance, and 3) the 200V you tell here can be calculated.

That is... 60 220 = rated current, 220 rated current = resistance, 200*200 resistance = actual power

4) [To be honest... I already want to hit the wall = = junior high school physics is particularly annoying = =].

Ahem. Let's move on. 4) means that .

How much extra resistance do you need to add to make 220V the same as 200V? Other words.. As a light bulb now he is normal!

It's just a 2b that adds an extra resistor to her. [Flip the table].

200V Rated Resistance = Actual Current, 220V Actual Current = Actual Resistance.

What I call actual resistance is... Bulb resistance + that extra resistance.

Mmmmmm I even know how to do junior high school questions... It means that I am not completely degraded = = child, you have to study hard, come on

First of all, you should know that the bulb is marked with 220V 60W to tell you the following data, the rated power of the bulb is 60W, the rated voltage is 220V, the rated current = p u = 60 220 =, the resistance r = u 2 p = 220 * 220 60 = 807 ohms, so, 1).i=p/u=, 2).i1=u/r=220/807=, 3).

p=u^2/r=200*200/807=50w, 4).r1 = u 2 50-807 = 220 * 220 50-807 = 968-807 = 161 (ohm).

In the circuit shown in Figure 16-25, LL and L2 are marked with the words "6V 3W" and "6V 6W" respectively, and the words L3 rated voltage are not legible, while the rated power is 4W.

1) Find the resistance r1 and r2 when L1 and L2 emit light normally;

2) When the switch S1 is disconnected and S2 is closed, there is just one bulb that emits light normally, at this time, what are the numbers of the ammeter and voltmeter?

3) When the switch S2 is disconnected and S1 is closed, if the actual power of the light-emitting bulb is 1W, what is the indication of the ammeter at this time? What is the rated voltage of lamp L3? [

First, calculate the resistance r=u2 p=5 according to the nameplate of the bulb; Only S3 is closed, the sliding rheostat slides to the A terminal, and the circuit only has R1, ammeter, V2, so the supply voltage U=, R1=U I=5

1) All closed, the bulb and the sliding rheostat are connected in parallel, R1 is short-circuited according to P=U2 R, you can calculate the power of the bulb P=, the total power so the power of the sliding rheostat P=, according to R=U2 P=15 (2) Only close S3, the sliding rheostat and R1 are connected in series, V1 measures the sliding rheostat, V2 measures R1 to protect the ammeter first, R total = U i=, that is, the minimum resistance of the sliding rheostat must have a protection voltmeter, the total voltage is, V1 is up to 3V, It is proportional to the voltage distribution and resistance of the series circuit.

The maximum rx:r1=2:1,rx=10 of the sliding rheostat is the same as rx:r1=1:2,rx=

Therefore, the sliding rheostat is the smallest and the maximum is 10