Physics Electricity Problem Skills Urgent, Electricity Physics Problem Solving

There are diagrams for general electrical problems, and the most important thing is to learn to read diagrams. When encountering a circuit diagram, it is necessary to determine whether it is connected in series or in parallel. Use the voltmeter as an open circuit, remove it, and use the ammeter as a wire.

At this time, the judgment is much simpler, in what series circuit the current is equal everywhere, the voltage in the parallel circuit is equal everywhere, the series voltage is divided, and the parallel is shunted.

p=w/t=u^/r=i^r=ui

w=pt=uit

q=uit=i^rt

i=u r, these should not be much of a problem. The most important thing is to listen carefully in class, do homework seriously, and do some extracurricular improvement questions, and there should be no big problem in getting a high score.

In different regions, the regional characteristics of the high school entrance examination are very obvious, and there may be so-called 'skills', but they may not be practical in other regions. The general method is meticulous, rigorous, skillful, and not panicked. Before the assault on the high school entrance examination, experienced teachers will give an analysis of the usual way of asking questions in the local high school entrance examination, and even have pressure questions, in fact, this is the skill, at this time we must pay attention to the teacher's method of solving the problem.

I'm from the past, and I know it very well!

When you see a series circuit, you think that the current in the series circuit is equal everywhere, and when you see a parallel circuit, you should think that the voltage in the circuit is equal everywhere.

Generally, if you memorize the formula, you will know how to do it

Mark the conditions on the circuit diagram as you read.

Determine whether the series-parallel connection is equal to you or equal to I.

Sometimes it's a physical diagram, and you want it to be converted into a circuit diagram to do it!! 22！！@标准答案哦哦哦哦哦哦@@@

Understand the concepts of resistance, current, voltage, series, parallel, electric power, electric heating, and electromagnetism and the relationship between them.

If the ammeter and voltmeter are removed, the circuit is actually a single circuit in series between a fixed value resistor and a rheostat.

The ammeter measures the current (i) of the loop, which has:

i=u/(r+rₚ)

Because i cannot exceed the range of the ammeter, i.e., i.e., i, there is:

Solution: r

The voltmeter measures the voltage (U) across the rheostat and there are:

uₚ=u·rₚ/(r+rₚ)=

Since u cannot exceed the voltmeter range, i.e., u 3 (v), then there is:

Solution: r 10 ( ).

Synthesis getsThe value range of the rheostat is as follows:[Note 1] The above range does not exceed the maximum resistance value of 20 indicated by the rheostat, which is feasible.

Note 2] The maximum reading of the ammeter does not exceed the maximum current indicated by the rheostat 1A, which is feasible.

When r = , the ammeter obtains the maximum value and the voltmeter obtains the minimum value:

iₘₐₓuₚₘᵢ

When r = 10, the ammeter obtains the minimum value and the voltmeter obtains the maximum value:

iₘᵢₙuₚₘₐ

Namely:The ammeter reading range is:The voltmeter reading range is:

Summary. According to the title, when the sliding vane is located at the midpoint of the sliding rheostat rheostat, the electrical power of the sliding rheostat r is that after the sliding vane is moved for a certain distance, the electrical power of the sliding rheostat R becomes, so it can be seen that after the sliding vane moves, the electrical power of the sliding rheostat R increases, that is, the resistance decreases, so the sliding vane should move to the left. Because the resistance value of the resistance wire of the sliding rheostat r is proportional to its length, when the sliding vane is located at the midpoint of the sliding rheostat rheostat, the electrical power of the sliding rheostat r is, after moving the sliding vane for a certain distance, the electrical power of the sliding rheostat R becomes, it can be seen that the electrical power of the sliding rheostat R increases, but the total length of the sliding rheostat R decreases, so the distance of the sliding vane should be less than 9cm, so the sliding vane should move 6cm to the left.

Answer: The slide should be moved 6cm to the left, option b.

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According to the title, when the sliding vane is located at the midpoint of the sliding rheostat rheostat, the electrical power of the sliding rheostat r is that after moving the sliding vane for a certain distance, the electrical power of the sliding rheostat R becomes, so it can be seen that after the sliding vane moves, the electrical power of the sliding rheostat R increases, that is, the resistance decreases, so the sliding vane should move to the left. And because the resistance value of the resistance wire of the sliding rheostat r is proportional to its length, when the sliding vane is located at the midpoint of the sliding rheostat rheostat, the electrical power of the sliding rheostat r is, after moving the sliding vane for a certain distance, the electrical power of the sliding rheostat r becomes, it can be seen that the electrical power of the sliding rheostat r increases, but the total length of the sliding rheostat r decreases, so the distance of the sliding vane and the sliding vane should be less than 9cm, so the sliding vane should move 6cm to the left. Answer:

The slide should be moved 6cm to the left, option B.

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Ideas: 1. Remove the voltmeter first, observe the relationship between electrical appliances, L and Rab in series (that is, the movement of the sliding blade P has no effect on the size of the access resistance), so the current and voltage characteristics of the series circuit are connected i=i1=i2

u=u1+u2

2. Observe the object measured by the voltmeter, (whose voltage is measured by the voltmeter in parallel with whom) the voltmeter measures the sum of the voltages of L and Rap (the sliding rheostat can be divided into two parts: RAP and RPB).

3. Analyze the influence of the process of moving the slider P from A to B on the resistance of the entire series circuit (the total resistance to the access circuit remains unchanged at this time, but it has an impact on the range measured by the voltmeter). When the analysis is carried out, it is only necessary to grasp P, and when it is at point A and point B, the corresponding measurement objects are R lamp and Rap=0, respectively

R lamp and rapb = r max.

Current = Total Voltage Total Resistance.

i=u (r-lamp+.)

r max) corresponding to the voltmeter is indicated by ua=

IR lamp + rap) ubi

R light + rapb).

4. The actual power consumed by the lamp is p=

i*IR lamp.

There is a question to know, when the switch is closed. Only the resistor R1 works, so the power supply voltage is U=3A·R1 When the switch is disconnected, the two resistors are connected in series.

At this point there is p2 = (u total - u1)u1 r1

That is, 10W = 5V RI (3A·R1-5V).

The solution obtains r1=5, u=15v, u2=10vp2=u2 2 r2, i.e. 10w=1v 2 r2, and the solution gives r2=10

The rightmost A and C are connected in series with B in parallel to get R1, R1 is connected in series with C to get R2, and R2 is connected in parallel with A to get the equivalent resistance between A and B.

R equivalent = (A2+C2+2Ab+2BC+2AC)*R(Ba2+Ca2+AC2+2ABC).

r1*i1=u r1r2 (r1+r2)=r total r total *i2=u This is not two unknowns Two equations are well solved r1=200 3 u=40v

When S is disconnected, the circuit is connected in series:

r total = r 2 = 50

i=600ma=

u total = i * r total =

When S is closed, the circuit is connected in parallel

U1 = U2 = U Total = 30V

i2=u2/r2=30v/50ω=

Parallel i1+i2=i

i1=i-i2==

r1=u1/i1=30v/

That's it, thank you.

Is the A meter in this question an ideal ammeter?