Physics in the second year of junior high school, simple circuits, physics in the second year of jun

According to what you said, it should be: r short circuit, which is equivalent to turning into an ideal wire with a resistance of 0, according to Ohm's law deformation formula U=IR, the voltage at both ends is 0, and the voltmeter is connected in parallel with it, so the indication is 0, in addition: the voltmeter is equivalent to an infinite resistance, and of course it is short-circuited when connected in parallel with the wire.

If r is broken, it is equivalent to becoming an infinite resistance, and simply according to Ohm's law i=u r, the current is of course equal to 0. As for the voltmeter, when R is broken, the voltmeter is actually connected in series into the original circuit, because the resistance of the voltmeter is large, according to the series voltage division, the power supply voltage at this time is basically given to the voltmeter, and the indication is close to the power supply voltage.

I hope it's clear.

R short circuit, the voltmeter basically has no current, and the current is left from R because the voltmeter resistance is too large.

r short circuit The actual circuit has current, but the default voltmeter resistance is infinite, which is regarded as an open circuit, so the current representation is zero, and the voltage representation is the power supply voltage.

1.Resistor short circuit voltmeter will not short circuit 2Short-circuited ammeter will become zero voltmeter string like in the circuit.

Being short-circuited does not have this concept.

None of the circuit diagrams don't know how you.

It's late, and the answer is already below.

The ammeter is not zero, and the reading is the current through a small bulb. The voltmeter will also not be short-circuited, and the reading is the voltage at both ends (of the bulb series power supply).

1 No other resistor voltmeter will be short-circuited.

2 The ammeter goes to zero.

Solution: Estimate the maximum current rating of a small bulb:

i max = p u min =

AI Amount = Au Slip =

v If the supply voltage u total = 2

u light = v = vu forehead =

v If the rated voltage is taken as.

v, then the resistance value of the rheostat deviates from the midpoint is nearly 1, and the maximum resistance of the rheostat is only 5, which does not conform to the title "near the midpoint").

If the power supply voltage u total = 3

v=ulamp=v=vu=

v (Since the supply voltage is.)

V, so it is not possible to confirm that the rated voltage of the bulb must be V when the small bulb U =

v, p = i u =

a w when small bulb u forehead =

v, p = i u = a w

Solution: (1) The product of the voltage on both sides and the current passing through the normal light emission p=ui=(2) Because the resistance of the small bulb increases with the heat, it can be used r=u i=2v approximately equal to)

3) The more power the bulb, the greater the brightness.

4) Because this circuit is connected in series, the current in which the bulb is obtained from (2), r=. Therefore p=u i=i r=approximately equals, which is actually.

These are my own hard work, you have to look at the pictures more, do more of this type of questions, not as difficult as you imagined.

(1) Because it emits light normally, p=ui=

2) From Ohm's law, r=u i=2v

3) The greater the actual power of the bulb, the greater the brightness of the bulb Be sure to write the actual power otherwise it is wrong.

4) In a series circuit, r total = (10 +

p=u/i=i²r=

1 p=ui=

2 r=u/i=2v/

3) The more power the bulb, the greater the brightness.

4 u=ir=10ω×,p=ui=

(1) The normal power is p=u*i=

2) r=u i=ohm.

3) The higher the power, the brighter the bulb.

4) The voltage in the circuit is u=i*(r+rlamp)=, then the power p=u*i=watts.

(1) (3) Bulb brightness Bulb power.

4) p=i2(r+rl)=

The bulb resistor in the fourth question cannot be used because the power supply electromotive force (i.e. the supply voltage) is not told. You have to consider whether the voltage is 2V when the bulb is passing current

Such a simple question needs to be asked.

The current flows from the negative electrode to the positive electrode, starting from the negative pole of the battery, connecting the -pole of the ammeter, and then connecting other electrical appliances from its positive electrode. When entering, it is the negative electrode to the negative electrode, and when it comes out, the positive electrode can be to the negative electrode of the other table.

Well,Actually, you didn't understand the circuit diagram for so many of your problems.,I ignored the problem and explained it directly.。。。

This question is actually very interesting, even I thought wrong at first, this kind of problem is better to draw an equivalent circuit...

The picture below is on.,Excuse me.。。。

In the end, it was clear that this is a parallel circuit of L1, L2, L3 in parallel...

Select C When the switch S is closed, and the sliding blade P of the sliding rheostat moves to the A terminal, the lamp L emits light normally.

The power supply voltage is 18V, the current through the lamp i=p u=18W, 18V=1A, and the indication of the ammeter is.

Slide rheostat resistance r=u i=18v (ohms.

When the switch S is disconnected and the slide plate P of the sliding rheostat moves to 3/4AB length at terminal A, the voltmeter shows 15V

So at this time, i=u r=15v (12*3 4)=5 3a, because the voltage at both ends is the smallest at this time, so the power of the lamp is the smallest at this time.

So lamp p=ui=(18v-15v)*5 3a=5w The correct answer should be 5w and choose c

First of all, there should be the following two circuits for junior high school level, I don't know what the teacher will say:

Electricity meters in junior high schools are generally regarded as ideal meters.

The ideal ammeter is treated as a wire with no resistance;

An ideal voltmeter is considered an open circuit where the current does not pass through.

The switch is closed, and when the slide P moves to the A terminus, the sliding rheostat and the bulb L are connected in parallel.

The voltage of each branch of the parallel circuit is equal, and the bulb l emits light normally, and the current through the bulb l: i = p u = 1a, and the current i through the sliding rheostat is obtained by differential'=( 。

It has already been said that the voltages are equal, and the resistance r of the sliding rheostat'=u/i'=12ω 。

The switch is disconnected when the slider P moves to a distance of 3 4 lengths, that is, the 3 4 length of the sliding rheostat is connected in series with the bulb.

The voltmeter measures the resistance of the 3 4 length sliding rheostat (i.e. the resistance of 9).

Here's the key:

The topic says that to find the minimum power of the bulb, as long as the bulb voltage is small and the current is small.

If p now slides to the right, the total resistance increases, then the bulb current decreases and the voltage decreases (u=ir, r definitely).

Then the bulb power p=ui is small.

However, if the number of voltage representations exceeds 15V, it will burn out.

At most, you can only slide right to the distance A 3 4, which is the place mentioned in the title. (Critical).

Based on the analysis and the corresponding conditions, the minimum power of the bulb can be found.

The total current i at this point''= 15 9 A = 5 3 A (bulb voltage u.)''=（18-15）v=3v

p=ui=3*5/3 w=5w

I also think that choosing C, the upstairs pro should be right.

I suggest you check the answer again.

1. All bulbs are on. 2. Same. 3. The circuit of the voltmeter is a branch circuit, and the others are trunk circuits.

1. The closing switch resistance is ohms, and the disconnecting switch is 2 ohms.

2. The closing switch is L1 on, and the disconnect switch is the same as bright.

3. The current of the closing switch is large in L1, and the current of the disconnection switch is the same.

4. The disconnect switch R3 is not in the circuit, and the other circuits are trunk circuits. Closed switch, R1, L1, ammeter, power supply are trunk circuits, and the other 2 are branch circuits.

All bright is no...

Closed 3 ohms disconnected 4

Close L1 (L2 is not lit, - disconnected as bright.

Close L1 and disconnect as well.

There is no branch to disconnect, and there is no ---to close

What the fuck are you talking about, I've never seen it before...

1. Both bulbs are on.

2. The value of A1 ammeter is the same as that of A2 ammeter.

3. There is only one line (L1, L2 in series), and the voltmeter is regarded as an open circuit.

1. I don't know the resistance of L1 and L2, and I can't calculate the total resistance of the circuit.

2. If the resistance of L1 and L2 is the same, L1 is brighter.

3. When the switch is turned on,The current through L2 is the same as the current through L1. When the switch is closed, .The current through L2 is less than the current through L1.

You think of an ammeter as a wire and a voltmeter as a short circuit. The first picture, although not very clear, is that the light bulbs are all on.

Figure 1: Both bulbs are on.

The value of the A1 ammeter is the same as that of the A2 ammeter.

The series circuit current has only one voltmeter and is considered an open circuit.

The resistances of L1 and L2 in Figure 2 are not known, and the total resistance of the circuit cannot be calculated.

It is impossible to judge. When the switch is on, ...The current through L2 is the same as the current through L1. When the switch is closed, .The current through L2 is less than the current through L1.

Figure 1, 1, all bright.

2，a1=a2

3, V1 is a branch road, other trunk roads.

Figure 1: 1: Fully bright.

2: Yes 3: The voltmeter is a branch and other trunk circuits.

Figure 2: 1: Europe.

2：l1>l2

3: l1>l2

4: Except for the ampere meter and L1 are trunk roads, all other branch roads.

Because the resistance is constant, the higher the voltage, the greater the current in the circuit, and the current in the series circuit is equal everywhere. In order to protect the circuit, only low currents can be considered.

Solution: r total = 16 euros + 22 euros = 38 euros.

The maximum current after series connection is i=

According to Ohm's law:

U=IR=Euro=19V

Therefore: the answer is d

Note: In this question, it is important to clarify the meaning of 16 euros and 22 euros. The maximum resistance value is 16 ohms and 22 ohms, respectively; If the maximum permissible current is exceeded, there is a risk of burning out the fixed-value resistor.

u1=16*

u2=22*

So the maximum voltage is u=u1

The answer is: c

I'll give you an equation and I'll do the calculations yourself

S2 is closed, R2 is shorted, S1 is closed, and R1 and R3 are connected in parallel.

Then the ratio of the current flowing through R3 to the current flowing through R1 is equal to the inverse ratio of resistance, i.e.

i1/i3=r3/r1 （1）

i1+i3= （2）

Substituting the resistance values of R1 and R3 can find the size of I3.

What is the relationship between r1 r2 r3? Otherwise, no way!

Classmate, your conditions for this question are not enough, and the question is not clear, you can take a closer look at the question.

The second floor is very nice. You look at the resistance in the question. Substitution is fine. The calculation is also not difficult.

The conclusion is that, according to the principle of current distribution in parallel circuits, the current through the resistor R3 is: a

Children's shoes, I think you are talking less r1=r2=r3.

If this is the above, then it can be solved