A shunt circuit is designed, which requires the shunt resistance to be calculated by reducing the ra

If you want to change the range of the multimeter from 500mA to 20mA. It is necessary to know or measure the full bias current ig, 500mA resistance value r ma resistance value r ma resistance value r2 of the meter head. The shunt resistor required for the 20mA range is then recalculated on the basis of the 10mA range.

For the Model 500 multimeter, the shunt resistance value is about 65 ohms. For the J0411 student multimeter, the shunt resistance value is about 27 ohms.

You are indeed asking a bit of a problem in this question, you should not look for shunt resistor, you should find series resistance, the essence of your question, it seems to be to modify the ammeter into a voltmeter. The physical quantity you give doesn't seem to be enough, you have to give what the voltage you add to the ammeter (know the resistance value of the ammeter before the modification, and you can also calculate the voltage you add to the ammeter), here the voltage added to the ammeter is assumed to be v, and the internal resistance of the ammeter before the modification is r, first turn the milliampere into ampere, divide by v, you can calculate the total resistance of the ammeter after modification, and then subtract r from this resistance, and calculate the resistance value of the series resistance!

I've only ever done to increase the range of the ammeter.

According to your requirements, the current flowing over the shunt resistor should be 4 times the current flowing over the internal resistance, so that the total current is 5 times the current flowing through the internal resistance, and the range is expanded to 5 times.

The shunt resistance is connected in parallel with the internal resistance, the voltage is equal, and the current is required to be 4 times, and the parallel resistance should be 1 4 of the internal resistance of the ammeter, that is, 250.

Before the modification, the internal resistance of the ammeter is r, you do ask a bit of a problem with this question, you should not find the shunt resistance, divide by v, you should find the series resistance, first turn the milliampere into ampere, you can calculate, the essence of your problem, as if you want to modify the ammeter into a voltmeter. The physical quantity you give doesn't seem to be enough, you have to give what the voltage you add to the ammeter (you can also calculate the voltage you add to the ammeter if you know the resistance value of the ammeter before the modification), and here the voltage added to the ammeter is assumed to be v

Assuming that the range of the ammeter is expanded to K times of the original, the internal resistance of the original ammeter is RO, and the amplification and parallel resistance are trapped in Huislip RX

Then there is a relationship between Wang La:

v/r = v/ro + v/rx

v/r = k*v/ro

Solve for rx = ro (k-1).

Classification of spring sources: Education Science >> Ascending Oak Pre-school Admission >> College Entrance Examination Analysis: The range of the ammeter is expanded to K times of the original, the internal resistance of the original ammeter is RO, and the parallel resistance of the current expansion is Rx

then there is a relationship: v r = v ro + v rxv r = k*v ro

Solve for rx = ro (k-1).

Summary. <>

Hello dear friends, it is our honor to serve you, the ampere ammeter needs to be expanded to 1ma, and 10 resistors can be connected in parallel, each resistor has a resistance value of 100 ohms, which are connected to the two terminals of the ammeter. Because the current is shunted in the shunt resistor, the current borne by the ammeter is only a part of the shunt resistor current, and the total resistance value of the shunt resistor is 10 ohms, so the current borne by the ammeter is 1 10 of the shunt resistor current, that is, 1mA 10=100ua, which meets the original range requirements of the ammeter. Therefore, the range of the ammeter can be expanded by shunting resistors in parallel, and a wider range of current measurements can be realized.

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Hello dear friends, it is a pleasure to serve you, I can't see your graphics clearly, you send me text form.

Dear friends, it is our honor to serve you, the ampere ammeter needs to be expanded to 1ma, and 10 resistors can be connected in parallel, each resistor has a resistance value of 100 ohms, which is connected to the two acres of terminal blocks of the ammeter. Because the current is shunted in the shunt resistor, the current borne by the ammeter is only a part of the shunt resistor current, and the total resistance value of the shunt resistor is 10 ohms, so the current borne by the ammeter is 1 10 of the shunt resistance current, that is, 1mA 10=100ua, which meets the original range requirements of the ammeter Xindan. Therefore, the range of the ammeter can be expanded by shunting resistors in parallel, and a wider range of current measurements can be realized.

ig=100ma=；r=1 ;

To make the ammeter range 1A, the current flowing through the shunt resistor is: i=;

The voltage of the parallel part is: u=igr=, then the resistor that needs to be paralleled is: r=ui==

A: Shunt resistors connected in parallel

Since it is a parallel resistor, then the internal resistance of the meter head can be equal to the voltage of the parallel resistor, and the meter head can only pass 100mA, then the current passed by the parallel resistance is 900mA (because the range is 1a=1000mA). Therefore, the equation can be obtained: according to Ohm's law, u=i*r, 1 ohm * 100 mA = r * 900 mA is obtained to r = 1 9 ohms.

It is equivalent to allowing 1A of current to pass through after parallel connection, and the flow rate of parallel connection is that in the same-parallel circuit, the resistance is inversely proportional to the current, that is, the resistance r r0 = 1 9, r = 1r0 9 = 1 9 ohms in parallel. r0 is the internal resistance.

10mA = 10000UA, because it is in parallel, so there is r1 * (10000-50) = 2000 * 50 to get r1 =.

1000*

In other words, the direct voltage at both ends of the ammeter is up to 1V without series resistance.

This is talking about parallel connection, not series connection, so the voltage at both ends of the parallel circuit is only 1V.

1V 5A=, which is the total resistance of the parallel circuit.

If we know that in a parallel circuit 1 r1 + 1 r2 = 1 r, then we may know that the shunt resistor has a resistance of 1

In practice, there is no need to be so troublesome in the calculation, and the voltage is calculated directly, and the shunt is used directly by using the parallel circuit. Then use the voltage at both ends, which is 1V, and divide it.

When the ammeter is full of deviation, its current is 1mA, then the current of 5A-1mA should flow through the shunt resistance when the range is 5A, because the ammeter head and the shunt resistance are connected in parallel, so the voltage is the same, that is to say, the current of the same voltage on 1k is 1mA, the voltage is 1V, the current on the shunt resistor is 5A-1mA, and the shunt resistance is almost equal.