Electrical Experiments !! How is that resistance in series with the voltmeter determined?

The rated voltage of the electrical appliance to be tested is 10V, so according to the circuit diagram given, it should be ensured that the voltage at both ends of the electrical appliance can reach 10V, but not more than 10V, so according to the conditions given in the question, a DC power supply of 15V should be selected.

According to the rated voltage of the electrical appliance is 10V, the rated power is between 10 and 15, according to I=P U, the rated current can be known, from the figure in the title, it can be seen that the electrical appliance and the sliding rheostat are connected in series, so the current through the sliding rheostat is about 5V, and the voltage added to the sliding rheostat is about 5V, so the resistance value of the sliding line rheostat connected to the circuit should be about 5 ohms, and it may be slightly greater than 5 ohms, so it is more appropriate to choose 0-15 ohms and 2a.

From the title, it can be seen that the voltage added to the electrical appliance is 10V, but the range of the voltmeter is only 0-3V, so the voltage added to the resistance in series with the voltmeter is 0-7V, so its resistance value should be greater than the multiple of the voltmeter's resistance value, but not too much, (because it exceeds too much, the voltage added to the voltmeter will be small, and the pointer deflection is small, which is not conducive to reading) so it is more appropriate to choose 3 times, that is, 6k ohms.

I don't know if you can understand the above statement, if you don't understand anything, please ask questions again, I hope the above can help you, I wish you progress in learning.

The rated voltage of the application electrical appliance is greater than the range of the voltmeter, so it should be modified to increase its range. Calculate it with the series divider, and choose a resistor of 6K.

When voltmeters with different internal resistance are connected in series, the voltage with large internal resistance is higher than the voltage with small internal resistance. The sum of the measurements of the two voltmeters is equal to the voltage being measured.

A voltmeter is an instrument that measures voltage. It is composed of permanent magnets, coils, etc. The voltmeter is a fairly large resistor, ideally thought of as an open circuit. The voltmeters commonly used in laboratories at the junior high school level have ranges of 0 3 V and 0 15 V.

Principle:

Traditional pointer voltmeters and ammeters are based on one principle, which is the magnetic effect of electric current. The greater the current, the greater the magnetic force generated, the greater the swing of the pointer on the voltmeter, there is a magnet and a wire coil in the voltmeter, after passing the current, the coil will produce a magnetic field, and the coil will be deflected under the action of the magnet after being energized, which is the head part of the ammeter and voltmeter.

Because the voltmeter should be connected in parallel with the measured resistance, if the sensitive galvanometer is directly used as the voltmeter, the current in the meter is too large, and the meter will be burned out, so that a large resistance needs to be connected in series in the internal circuit of the voltmeter, so that after the transformation, when the voltmeter is connected in parallel in the circuit, due to the effect of resistance, most of the voltage added to both ends of the meter is shared by the resistance in series, so the current through the meter is actually very small, so it can be used normally.

The symbol of the DC voltmeter should be added with a " under v "The symbol of the AC voltmeter should be added with a wavy line under v ".

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Hello dear, I will answer this question for you, the solution to your problem is as follows: when the sliding rheostat is connected in series in the circuit, the resistance value of the resistance becomes larger, and the voltage at both ends of the sliding rheostat will also increase. Proportional premise is that the voltmeter must be at both ends of the resistor.

When a sliding rheostat is connected in parallel in a circuit, the resistance of the resistor becomes larger, and the voltage at both ends of the sliding rheostat is the supply voltage. There is no change in the number of voltage representations.

When the voltmeter measures the voltage of the series resistance, the highest accuracy is to report the large resistance or the small resistance.

Dear, hello, I saw your question, the teacher is very honored to judge the stool to be able to answer, is sorting out the relevant information for you, I hope to help you, please understand, please wait a while! <>

Dear, hello, I will answer this mu grinding problem for you, the solution to your problem is as follows: when the sliding rheostat is connected in series in the circuit, the resistance value of the resistance becomes larger, and the voltage at both ends of the sliding rheostat will also increase. Proportional premise is that the voltmeter must be at both ends of the resistor.

When a sliding rheostat is connected in parallel in a circuit, the resistance of the resistor becomes larger, and the voltage at both ends of the sliding rheostat is the supply voltage. There is no change in the number of voltage representations.

Hello dear, thank you very much for your inquiry. I hope the teacher's help to you, if there is anything you don't understand, please also rearrange the problem, and it is also convenient for the teacher to find relevant information for you, which is helpful to you, and please give the teacher a thumbs up to encourage you. Thank you very much for accompanying us, I hope you have a good life and good health!

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The resistance of the voltmeter is very large, and when the voltmeter is connected in parallel at both ends of the resistor, it has little effect on the total resistance of the circuit; So the answer is: big, and, small

Very simple, this question is mainly about the principle of voltage division, you are right, the resistance of V is very large, but it is not very large relative to R1, according to the principle of voltage division, the greater the resistance, the more voltage of the powder, the solution; Set the voltmeter resistor rx

3=rx (rx+3000)*, just solve rx.

There is definitely an impact.

A voltmeter is essentially a sensitive ammeter connected in series with a larger resistor.

When measuring, because the resistance r of the voltmeter is very large, much larger than the resistance to be measured, the current through the voltmeter is almost 0, which does not affect the original circuit.

The actual measurement principle is that the indication of the voltmeter (i.e. the measured voltage) is equal to the number of sensitive current representations (current value) in the voltmeter.

i) Multiply the organization of the resistance r in series, i.e.

U=IR Therefore, if you want to connect it with a resistor in series, it is equivalent to an increase in R.

If r is 1000 ohms and the resistance in series is only a few ohms, of course the effect can be ignored. However, if the resistance in series is more than a few hundred ohms, this effect cannot be ignored.

The range of the voltmeter is 3V, and the internal resistance is 3K, indicating that the maximum current he passes through is 3V 3K, first in series to it a 12K resistor, because the maximum current is unchanged, the total resistance is 12+3=15K, then the maximum voltage at both ends of the voltmeter is 15V, so the range becomes 15V. The answer A is false, B maximum current is unchanged, false.

When C indicates 2V, then it actually corresponds to 2V 3K, and the voltage is 3V 3K 15K = 10V, so C is wrong for D.

c;When a resistor is connected in series, the range of the original voltmeter is actually expanded. The voltage measured at this point should be the reading of the voltmeter plus the voltage on the series resistance. It is easy to calculate the actual voltage according to Ohm's law.

c。The voltage distributed according to the series circuit is proportional to the resistance. When the pointer indicates 2V, the voltage at both ends of the string resistor is 8V, so that the total voltage at both ends after modification is 2V 8V 10V.

The measuring range is doubled, and the numerals on the dial are doubled.

When the pointer reaches the maximum declination angle, the voltage distributed by the voltmeter reaches the maximum, and the resistor in series also receives the same voltage, and the overall voltage is twice the original range of the voltmeter.

Let me explain the answer:

A voltmeter measures the voltage of the mains supply.

Reasons:1The internal resistance of the voltmeter is quite large, and in physics, it can be considered infinite, so the current passing through the voltmeter can be regarded as zero.

2.According to Ohm's law, u=i*r, and at this time, since the voltmeter is in the circuit, the current is regarded as 0, so the current flowing through the resistance can also be identified as 0, since the current is 0, then according to Ohm's law, u=ir, u=0, since the voltage distributed on the resistor is 0, then the data measured on the voltmeter is of course the voltage of the power supply!

I hope this explanation can be understood.

According to the working principle of the voltmeter (the meter head is connected in series with a large resistance, this resistance is much larger than the resistance + power supply resistance in the circuit), when the voltmeter is connected in series with the resistance.

The voltmeter measures the voltage of the large resistance + resistance + power supply resistance in series, that is, the power supply voltage (there is an error, and the reason for the error is that there are many resistors and power supply resistances, which can be ignored). If the resistance is connected in parallel with the voltmeter, the voltage measured by the voltmeter is the parallel value of the large resistance and the measured resistance (there is a certain error, because the large resistance is much larger than the measured resistance, which can be ignored, so the voltage measured on the side is the voltage of the measured resistance) Because the voltage shared by the resistance is limited, the voltage through this resistance is measured. If the voltmeter and the power supply are directly connected in parallel, the principle is similar, and the power supply voltage is measured (the error is not counted).

The voltmeter works by means of an electric current, and as long as there is an electric current, it has an indicator. This is my speculation, and I hope that more professional people will explain it if it is incorrect!

The voltage at both ends of the voltmeter is also the voltage at both ends of the power supply.