Regarding physics, Ohm s law. Hurry!!!!!!!!!!!!!

It cannot be said that the resistance value is a fixed property of the resistance.

The voltage produces an electric current on the resistor, and the voltage is proportional to the current.

That is, the voltage is increased by 5 times, and the current is increased by 5 times.

Regardless of the principle, only the numerical consideration.

It can also be said that the current is increased by a factor of 5, and the voltage is also increased by a factor of 5.

It is correct that the resistor is larger and the more voltage is divided (in series), but you must remember that the current is generated by the voltage added to the resistor. The positive and negative ratios in numerical values are just one manifestation.

1. The resistance is not proportional to the voltage. You can control the voltage regularity. But you can't control how the resistance changes with temperature.

2. In the case you mentioned, the resistance is still not proportional to the voltage. Just think:

The voltage is fixed at 6V. The two voltages are 1 ohm and 2 ohms.

Resistor dividers: 2V and 4V. Change the resistance to 1 ohm vs 5 ohms. Resistor 2 has been increased by 3 ohms.

The partial voltage has been increased by 1V. If you make resistor 2 increase by 3 ohms. If the voltage divider is increased by another 1V, then the resistance is proportional to the voltage, but this is not possible.

Resistor 1 always detachs a portion of the voltage. So. Resistance 2 will never reach 6V, so it is not proportional.

3. The definition of proportionality is: One number increases regularly as the other increases. It doesn't have to be the same value. The ratio of the increased values remains the same.

1.Ohm's law can be used to calculate the magnitude of the resistance value, but the resistance is not proportional to the voltage, the resistance is a property of itself, and has nothing to do with the current or voltage. 2.

According to Ohm's law, u=i*resistance, when the current is constant, the greater the resistance, the greater the voltage, but it cannot be proved that the voltage is proportional to the resistance. 3.Regardless of the principle, only the numerical consideration.

It can also be said that the current is increased by a factor of 5, and the voltage is also increased by a factor of 5.

Resistance is a property of itself, which has nothing to do with the magnitude of voltage and current, but the magnitude of resistance can be calculated by voltage and current.

The proportional relationship can be illustrated by examples like the one you mentioned. A is 5 times larger, and B is also 5 times larger. That's right.

When the power supply voltage is constant, is the resistance proportional to the voltage across the resistor?

Answer: Resistance is a property of this appliance and does not change with voltage and current, i.e., it has nothing to do with them.

When the power supply voltage is constant or the current is constant: the greater the resistance, the more the partial voltage, can this statement mean that the resistance is proportional to the voltage?

A: It can only be said that the greater the resistance of Ann here, but the resistance has nothing to do with them.

Is it proportional or not: A is 5 times larger, and B is also 5 times larger.

A: Yes.

1. You're talking about the situation in a series circuit, right? No, assuming that the power supply voltage is U and there are two resistors in the circuit, R1 and R2 respectively, then the voltage at both ends of R1 is U1 (R1+R2), and the voltage at both ends of R2 is U2 (R1+R2), so in a series circuit, the voltage at both ends of a certain resistor is not proportional to the resistance.

2. Same as above, the voltage at both ends of the resistor R1 is UR1 (R1+R2), when R1 increases, the denominator also increases, so U is not proportional to R1.

3. Assuming that A B = K, then A is proportional to B, so A increases to several times of the original, and B also increases to several times of the original.

In the same circuit, the current in the conductor is directly proportional to the voltage at both ends of the conductor and inversely proportional to the resistance of the conductor, which is Ohm's law.

The value of the resistance is only related to the properties of the resistance itself, not to the voltage across the resistor.

Resistance doesn't change, as the teacher said. That's the definition.

In proportion, consider the question of squaring.

r=u i is.

r1:r2=u1:u2

The current is the same when connected in series.

i=u/r

The magnitude of the resistance is independent of any value.

Resistance is independent of voltage.

This is the derivation process of the ratio of partial pressure in series and shunt shunt.

Ohm's law r=u i is mainly used

The series currents are equal everywhere.

then i=u1 r1=u2 r2

Get u1 u2=r1 r2

The voltages at both ends of the parallel circuit are equal.

then u=i1r1=i2r2

Get i1 i2=r2 r1

Let's say a resistor and a sliding rheostat are connected in series. Electromotive force e, internal resistance r.

When the resistance of the sliding rheostat becomes smaller; The total resistance is getting smaller; The current through the slide is getting bigger... Mold starvation. That's right.

Aren't the series currents equal.?? Is the current going to change?

Equal series current means that in the same circuit, each component in the series circuit passes through the same current equally. It does not mean that the current of each component is equal in a circuit with different resistors. If the resistance of the circuit changes, the resistance of the entire circuit changes, but after the change, the current of each component is equal.

Voltage through the resistor = magnitude of current * magnitude of resistance.

The total resistance becomes smaller, the total current becomes larger, therefore, the voltage of the fixed resistance becomes larger (=IR), the total voltage remains the same, and the voltage of the rheostat naturally decreases.

The output power of the power supply is "maximum when the internal and external resistance are equal", so the resistance of the sliding rheostat decreases and the power increases. However, if the total external resistance is less than the internal resistance r, then the power becomes smaller again.

When the current is large, the resistor power becomes larger; The internal resistor power becomes larger (according to i 2*r).

The power of the sliding rheostat can be judged to be smaller by using the total power - the constant resistance power - the internal resistance power, and it can also be calculated using specific numbers.

d A 15cm resistive wire is connected in series. It must be in series to share the voltage, which goes without saying.

The bulb is, then the series resistance divider is . The voltage in series is proportional to the resistance, voltage 1; 3, the resistance is also 1; 3, bulb 5, then series resistance 15. Since the resistance is 1 cm, it is 15 cm.

A voltmeter can be thought of as a resistor. When there is a current passing, the pointer can be rotated and the degree appears, so the voltmeter has a degree, indicating that there must be a current passing.

Ohm's law: i=u has been unchanged, i wants to become 1 3, r must be expanded by 3 times, that is, from r to 3r, then 2r is increased, series connection can increase the resistance, so it is necessary to connect in series.

If the current is too high, the resistor will burn to death. Therefore, the maximum current to limit is r=18 ohms. The original resistance is 10 ohms, so at least a series of 8 ohm resistors can ensure that the original resistance is not burned to death due to excessive current.

Connect at both ends of the same power supply, the same power supply, of course, the voltage remains the same, and it is 1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:

Add a few more points. I'm so tired.

Your family I qq bar -m- It's too troublesome to say this 675854960

(1) The purpose of using resistance wires: to reduce the voltage at both ends of the bulb.

The parallel resistance cannot divide the voltage, so it must be connected in series to achieve the purpose of voltage division.

When the voltage at both ends of the small bulb is , it can emit light normally. The power supply voltage is 6V, so the series resistance is three times that of the bulb voltage divider resistance is required to ensure that the bulb emits light normally, that is, the voltage divider resistance is 15, and the resistance line of 15cm should be selected. Therefore, the treatment method [d] should be adopted.

What is the landlord's QQ? My 357304765. It's a long time coming.

The first option is D.

First of all, you should know Ohm's law, i=u r, and then you should know the difference between series and parallel circuits.

The first question is that the bulb voltage is rated, so it cannot be paralleled, because the voltage of the parallel circuit is the same and equal to 6V

Tandem should be used. From the series circuit, it can be seen that the circuit current is the same, and the sum of the voltages is equal to 6V, a small bulb, so the resistance line voltage is. Therefore, the total resistance of the resistance wire is three times that of the bulb, 15 ohms, so it should be 15cm.

You have to be clear about the working principle of the voltmeter, the voltmeter is actually connected in parallel with the electrical appliances at both ends, and the voltage at both ends is the same. The resistance of the voltmeter is very, very large, so the current through the voltmeter is very small, and the reading of the voltmeter is the opportunity of its current and resistance, which is done by the voltmeter when it was designed, and the voltmeter has a reading to indicate that the voltmeter has a current passing through, understand?

i wants to become 1 3, from i = u r, r becomes three times, and the resistance of the series circuit is additive, it turns out to be r, so it is necessary to connect 2r in series.

The rated current of the resistor is that it burns out if it exceeds this value. The maximum is that the rated voltage of the resistor is 5V, 10*, so the method of parallel connection cannot be used. If it is connected in series, the total voltage is 9V, current, so the total resistance is at least 18 ohms, so it is necessary to connect a resistor of 8 in series.

No matter what the two resistors are, they are connected to the same power supply at both ends, and the voltage at both ends of the resistor is the same as the power supply voltage, and the power supply voltage is the same, so the voltage ratio is 1:1. Of course, you can also find the current, the resistance multiplied by the current equals the voltage.

Hope it helps.

1. Select D, parallel shunt, series voltage division, and then the voltage of the bulb is equal to the ratio of the bulb resistance to the total resistance multiplied by the total voltage, and the resistance wire that needs 15 ohms can be calculated, which is 15cm.

2. The voltmeter has an indication that it does not indicate that there must be a current passing through, it is only the voltage difference of the measured circuit, just like a reservoir, the upstream water level is 30 meters, the downstream water level is 10 meters, and the water level difference is 20, but if the reservoir does not open the gate, there will be no water flow through.

Typing is too slow, and after looking at it, the basic solution is good.

This kind of problem is much easier to understand when it is related to the problem of water flow.

It can only be connected in series and parallel, which is equal to the power supply voltage, and the normal luminous current is 6 ohms, 20-5=15 ohms.

The voltmeter is modified from an ammeter.

It can be seen that the voltage is IR and the current is 1 3 i, then there is IR rx=1 3i, so there is rx=3r 3-1=2

The voltage of the fixed value resistance is 4V, R=4, ohms, and the current is 4a, 3a, U1=4x6=24, U2=3x8=24, and 1:1 can be obtained

It's so simple.

It's hard to say with a mobile phone.

You'll get to the bottom of it.

1 d series a 15cm resistance wire in series at a total voltage of 6 volts, the rated voltage of the bulb is volts, so the new wire should share volts, because the series voltage division, so series 15 cm resistance wire.

2 To put it bluntly, the voltmeter is an electrical appliance with a large resistance, and if there is an indicator, it means that there is a current, which is a matter of principle.

3 is still a problem of series voltage division, i = u r The voltage remains unchanged, the current becomes the original 1 3, and the resistance becomes 3 times the original, so it is series 2R

4 r total = u i = 9v = 18 r new = r total - r definite = 8).

5 Since it is the two ends of the same power supply, the voltage is of course the same, isn't this problem simple?

Solution: There is a problem to know that the LED, the sliding rheostat and the power supply are in series relationship, so there is V(LED)+V(Sliding Rheostat)=V(Power) is known V(LED)=3V, V(Power)=, V(Sliding Rheostat)=

Since the current = 20mA, the total resistance of the circuit can be calculated in units, and the resistance of the LED is twice that of the sliding rheostat according to V(LED)=3V(sliding rheostat)=.

Since the total resistance is 225, the resistance of the LED is 150, and the resistance of the sliding rheostat is 75

Solution: r=u i=3v

r total = r lamp + r resistance = u total i =

r-resistance = 750

This answer is correct, and the identification is complete.

LED worked normally when it was 3V.

(1) The dynamic friction factor between the wooden block and the conveyor belt.

Set the dynamic friction factor between the block and the conveyor belt

x=v0t+1/2at^2 36+1/2a*36=18 a=-1m/s^2=-μg μ=

2) The wooden block slides on the conveyor belt at a speed of 6m s, which is equivalent to the right movement of the conveyor belt, and the sliding friction force hinders the relative movement of the object and decelerates the movement.

After reaching 4m s, it moves at a constant speed.

3) Find the first question (2) and ask how long it takes for the block to reach B

The time for the wooden block to slow down from 6m s to 4m s t1=(4-6) -1=2s motion bit x1=(4+6) 2*2=10m constant speed time.

t2=(18-10) 4=2s The block reaches B with t=t1+t2=4s

Choose C, because the acceleration increases, the acceleration is provided by the frictional force, and f=ma, the friction force increases.

This question uses the knowledge of force analysis. You start with a ball that is subject to outward Coulomb force f=k q l (l=sin 1) because of the same charge, and because the ball is in equilibrium with 3 forces. Therefore, the Coulomb force f=mg tan 2 formulas can be used to find the amount of charge.

The rope is (mg cos).

It's a little messy. The symbols I have are not good to burn back. For such a topic, grasp the key word "balance".

Then the force cross is broken down to find the equal amount. In addition, you have to learn the holistic method and the decomposition method. Depending on the problem, consider the whole system or the parts of the system.

Questions like this have to grab a small ball to consider! It is estimated that you have just come into contact with this kind of problem, and you will be able to contact Duan Qing after more contact. There are also such problems, and the college entrance examination is basically not examined.

Because it's so simple.