Please ask the physics teacher for advice, hello physics teacher, how to do it

Of course, it doesn't conform to Ohm's law.

As far as the transmission circuit is concerned, it is also connected to the user, just like a motor, and it has to output power outward, not a pure resistance circuit.

So the power loss can only be calculated as i 2*r and not u 2 r

Voltage boost is used to reduce losses by reducing current.

Addendum: It seems that the landlord thinks very deeply. Good phenomenon ...

I can't use Ohm's law because in transmission lines:

u=u1+u2

U is the voltage on the secondary side of the factory booster, U1 is the voltage drop caused by the resistance of the transmission line, and U2 refers to the self-inductance voltage on the primary side of the user's step-down transformer.

u1=i r That's right.

But u=i r is obviously not right, understand? Do you know what's wrong?

The voltmeter can be used, but the key is which voltage you use to set Ohm's law, increase u, i drop, use Ohm's law to calculate u1 drop, this is not a contradiction...

1: U2 is the self-inductance voltage of the coil, which is a voltage that can be measured in real terms, and it is not designed by me to solve the problem.

2: When there is only a circuit with electricity and power supply (as you U2=0, U=i r analogy) if the coil changes, the output power must be different, and the constant power transmission line of the power plant is different, the power plant circuit must have 3 parts. The calculation method is completely different.

3: Regarding the question behind you, I can't understand it after reading it 5 times, I really want to help you, I'm sorry...

There is a question of total power involved

The total power output of the power plant is certain, we know that when the voltage increases, the current will naturally be smaller, because the total power is certain, and the electrical energy lost by the wire is mainly in the form of thermal energy, which is only related to the current, resistance and time.

Summary. The fifth problem solution: (1) when the motor does not rotate, all the electric energy consumed is converted into internal energy, then the resistance of the motor coil r=u i (2) when the motor is working normally, the voltage at both ends of the motor u1 = 2v, when it is suddenly stuck, the voltage at both ends of the motor is still 2v, at this time the current = i1 u1 r = a

3) When the motor is working normally, the electric energy consumed by rotating for 60s w=ui't = 2v 1 a 60 s = 120 j, the heat generated by the current through the coil during normal operation q=i 2rt=(1a) j, the mechanical energy obtained by the motor when it is working normally w machine city = w - q = 120j-30j = 90 j

Hello physics teacher, how do you do it?

Okay, are you getting both of these questions right?

Well. 8) Close switch S2 and disconnect S1

8）③rx=u1r0/u2

The fifth problem solution: (1) when the motor does not rotate, all the electric energy consumed is converted into internal energy, then the resistance of the motor line auspicious cherry dust circle r=u i (2) when the motor is working normally, the voltage at both ends of the motor u1 = 2v, when it is suddenly stuck, the voltage at both ends of the motor is still 2v, at this time the current = i1 u1 r = a(3) When the electric meditation motor is working normally, the power consumed by rotating for 60s w=ui't = 2v 1 a 60 s = 120 j, the heat generated by the current through the coil during normal operation q=i 2rt=(1a) j, the mechanical energy obtained by the motor when it is working normally w machine city = w - q = 120j-30j = 90 j

To measure the resistance, the component must be disconnected from the circuit, so the first blank fill: S2 small bulb resistance is generally a few ohms to tens of ohms, for example, 8 ohms, so the selector switch is 1, and the reading is 8 ohms.

The multimeter is used as a voltmeter, which is connected in parallel with the small beads, so the voltage of the small beads is measured.

To measure the small bead current, the ammeter must be connected in series in the circuit, so S1 is closed and S2 is disconnected

S1 is disconnected because the ohmmeter cannot measure live.

1 ohmmeter median value is 15 The scale around the median value is relatively uniform, the measurement is more accurate, and the resistance value of the small bead is about 10 ohms.

8 ohms reading is fine, and we chose a magnification of 1

l Parallel to the light.

S1 S2 Let the multimeter be connected in series with the bulb.

Question 9, power is equal to 220 times 3 = 660 watts.

Question 10: Work is equal to power multiplied by time One kilowatt-hour of electricity is equal to 3,600,000 joules. The last question is divided by 25 by dividing the work calculated in the previous question by 25, and the time can be calculated.

Left 27-10

Then use the spring scale to pull up the metal block vertically and put it into the water, due to the existence of buoyancy, the indication of the spring scale will change, let the indicator of the spring scale at this time be f1, f-f1 is the buoyancy f float. f float = water vg

Water f-f1

33'Incorrect, the bulb resistance increase is nonlinear.