A few Ohm s law questions in the second year of junior high school are very urgent!!!

1.Method 1: Connect a 1 resistor in series.

Method 2: Connect a 2V power supply in series with the positive and negative poles of the 10V power supply.

Method 3: After connecting two 2 resistors in parallel, it is equivalent to a 1 resistor and connecting the bulb in series!

2.Because, so the maximum current can only be otherwise, the B resistance will burn out) Now that you know the maximum current, you can calculate the corresponding maximum voltage! We know that i=u r, so ua=irAA i.e. uA= , and the same way uB=irB so.

UB = Because it is a series circuit u total = u A + u B so u total = 8v + 11v = 19v.

3.First of all, we should be aware that there may be two situations in this question, but after thinking about it for ourselves, we find that there is only one situation, why? We can calculate the rated current of lamp A according to the known conditions iA=u AA rAA i.e. iA=8v 32 = we can also calculate the rated current of lamp B i B=u B uA i B=12v 24 = If lamp B shines normally, then it means that the current flowing through lamp A and the current flowing through lamp B are equal, both are and the rated current of lamp A is, so at this time, lamp A will be burned out, which is not in line with the topic!

Then it can only be that lamp A emits light normally, when lamp A emits light normally, the current in the circuit is, which is less than the rated current of lamp B, which is in line with the dim luminescence in the title. At this time, we determine that the current in the circuit is At this time, the voltage of lamp A is equal to the rated voltage of 8V, and the voltage of lamp B is.

u B = IR B u B = So u total = u A + u B u total = 8v + 6v = 14v.

Note that the symbols above are the same as those below, and it is inconvenient for me to type on my computer! Thank you for your understanding!

4.Method 1:

Known: u=30v i=

Find: r=?Solution: r=u i=30v

When the voltage is 220V, i=U r, i=220V, 60 =3, 11A (11/3A, the fraction cannot be typed).

3 11A2A is not available.

Method 2: If it is located in a circuit of 220V, it can be used if the current = 2A, and the voltage = 220V

r hypothesis = u i r hypothesis = 220v 2a = 110

If you want to use a 220V circuit, the minimum resistance is 110, but its actual resistance is 60 60 110 can not be used.

I think my process is very good, you can look at it more, and realize that this kind of question is only average.

There is also a direct call if you don't understand!

Glad to answer for you!

The first question, I don't know how your circuit diagram is connected, I estimate that R1 and R2 may be connected in series, the voltmeter is connected to both ends of R1, U=6V is the power supply voltage, then the answer should be: the resistance value of R2 is at least 10 ohms.

The second question, because u2=2v, so u1=u-u1=8v-2v=6v.

i1=u1/r1=6v/12=

So the total resistance: r=u i=8v ohm.

In the first question, R2 should have a resistance value of at least 10 ohms. The total resistance of the second question is r=u, i=8v ohms.

1 question is not complete.

2, the answer is 4 ohms.

Because the light intensity is gradually increased, the resistance value of the resistance r decreases, and the resistance r0 is a fixed resistance, and the resistance value does not change.

The total resistance of the circuit becomes smaller, because the total voltage of the circuit does not change, then i=u r, and the total current of the circuit becomes larger.

Because the currents in the series circuit are equal everywhere, the number of current representations becomes larger.

And the voltage of the resistance r0 u1 = r0*i, because the resistance does not change, the current becomes larger, and the voltmeter also becomes larger.

So d answer.

When analyzing a circuit, the voltmeter does not exist.

In this way, the circuit is the power supply, the series connection of the fixed value resistor, the photoresistor and the ammeter.

When the light is enhanced, the resistance of the photoresistor decreases, the total resistance decreases, and the current increases, that is, the number of current representations increases.

U=r*i, when the resistance is constant, the larger the current, the greater the voltage, the voltage is on the side of the voltmeter, so the voltage representation increases.

I choose DThe resistance of the photoresistor decreases with the increase of the illumination intensity, so gradually increasing the illumination intensity of the photoresist will decrease the resistance, so that the current in the circuit increases, u=ir, r is certain, i increases, and the number of voltage representations also increases.

d The light intensity increases. The resistance of the photoresistor becomes smaller, and the supply voltage does not change, so the current increases.

Depending on the voltage division of the circuit, the ratio of RO to R becomes larger, so the RO load voltage measured by the voltage change also increases, so the voltmeter reading also increases.

The answer should be d.

Since it is a series circuit, when R decreases, the sum of the resistance values of R0 and R becomes smaller, and because the voltage does not change, the current value in the circuit becomes larger, that is, the number of A meters becomes larger. Since the current values of each resistor in the series circuit are equal, the value of the V-table is also larger.

i=u r0+r When the r value decreases, i must increase the current in all parts of the series circuit is the same, r0=u a r0 value remains the same, a becomes larger, then the voltage at both ends of r0 becomes larger, choose d

A indicates that the number has become larger, and V indicates that the number has become larger. Because it is a series circuit. As the photoresistor decreases as the light intensifies, the total resistance in the series circuit decreases, so the total current increases, and so does the voltage. u=ir

From the question, it can be seen that r (photosensitive) resistance becomes smaller, r=r (photosensitive) + ro becomes smaller, the power supply voltage is unchanged, that is, u, i=u r r is smaller, u does not change, so i becomes larger. Since the voltage is proportional to the resistance in series, the ratio of resistance to R (photosensitive) of RO becomes larger, so V also increases. d

1. The circuit is a series circuit, increasing the illumination, so the R resistance value increases, A indicator becomes smaller, because the series circuit voltage distribution is proportional to the resistance, so the voltage at both ends of R increases, so the indicator of V becomes smaller.

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.

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(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?

Since the relationship between current and resistance is to be studied, according to Ohm's law, the voltage should be set to a fixed value.

When the resistance becomes larger, the sliding rheostat should be adjusted so that the voltage at both ends of the resistor, that is, the number of the voltmeter, is the same as before the resistor is replaced.

Therefore, the resistance value of the rheostat connected circuit should be increased. So the sliding rheostat should slide to the right. Choose C

It can be seen from the question that the relationship between the current and the resistance is studied, so the voltage at both ends of the resistance should be controlled, and it can be seen from the figure that this is a series circuit, and the current at both ends of the resistance is proportional to the resistance, since the resistance is increased, the voltage at both ends of the resistance will increase. In order to keep the voltage at both ends constant, it is necessary to increase the resistance of the sliding rheostat to keep the voltage at both ends of the constant resistor.

To add) this method is the method of controlling variables

I don't know if there are any other components in the circuit base lead, just from the above Bohong good strip shouting parts, there are:

The solution is r=20;u=6v。

When it is not connected, the liquid line, when it is all connected, it is buried quietly.

To solve this system of asparagus equations, the supply voltage U = 6 V, and the constant resistance r = 20 ohms.