You will need

- — ammeter;
- tester;
- the conductor of the cross sections of the known material;
- — table of specific resistances.

Instruction

1

Connect a shunt in parallel with the ammeter to expand the possibilities of its measurement. In this case, a shunt current passes through the primary, and that part which is to be measured, passes through the ammeter. Rated current in the network calculate the formula.

2

To calculate the shunt, find out the maximum amperage that will have to be measured by the device. To do this, measure the voltage at the current source U in volts and divide it by the total resistance of the circuit R in ohms. All measurements performed by tester if the current is constant in this case, consider the polarity of the device. Find the rated current in the circuit by dividing voltage resistance I=U/R. Look at the scale of the ammeter and find out the maximum current it can measure.

3

Find the resistance of the shunt. To do this, measure your own resistance-ammeter R1 in ohms, and find the resistance of the shunt by dividing the product of the maximum current that can be measured by the device I1 and its resistance R1 to the rated current in the network I (R=(I1∙R1)/I).

4

Example. Need to measure the current in the circuit, where the maximum value can reach up to 20 A. it is proposed to use the ammeter with the maximum possible measurement current of 100 mA and a resistance of 200 Ohms. The resistance of the shunt in this case is R=(0,1∙200)/20= 1 Om.

5

As the graft use the standard resistors. If those are not available, make a shunt yourself. For the manufacture of shunts it is best to take conductors of copper or other material with high conductivity. To calculate the desired length of the conductor shunt l, take the wire the cross sections S and find out the resistivity ρ of the material from which made this device. Then, the resistance R, times the cross section of the conductor, measured in mm2, and divide by its resistivity, expressed in Ohm∙ mm2/m, taken from a special table l=R∙S/ρ.

6

To manufacture the shunt to the ammeter from the above example of a copper wire 0.2 mm2, take its length, which calculate according to the formula l=1∙0,2/0,0175=11,43 m. the same principle use and bypass any other part of the circuit.

# Advice 2: How to choose a shunt

In the practice of electrical measurements is often necessary to measure current, the value of which exceeds the upper limit of the available ammeter. The output of such a situation is application

**of the shunt**to the ammeter. The shunt allows you to change the amount of current allowable for the instrument.You will need

- - copper or nichrome wire;
- - power supply with adjustable output voltage;
- - ammeter;
- - ohmmeter;
- the load, for example, tungsten.

Instruction

1

To calculate the resistance of

Rш = (Ra*Ia)/(I-Ia),

where Rш – the required resistance

**the shunt,**use the following formula:Rш = (Ra*Ia)/(I-Ia),

where Rш – the required resistance

**of the shunt**; Ra – the resistance of the winding of the ammeter; I – upper value of the measured current; Ia is the current full deflection of the ammeter.2

Determine the scale of the existing device maximum measuring current Ia. For example, its value is 100 µa and you need to measure current up to 25 A.

3

Find the magnitude of the winding resistance RA of the ammeter. It can be taken from the passport of the device or to measure with a ohmmeter (with acceptable error). Let this is equal to 1750 Ohms.

4

Substitute the values into the formula and get the result:

Rш = (1750*0,0001)/(25-0,0001)=0,007 Om

Rш = (1750*0,0001)/(25-0,0001)=0,007 Om

5

Now empirically using a model of the ohmmeter to find the length of the wire. The obtained value is small, and for the manufacture

**of the shunt**will need to cut copper wire. It is more correct to use a solicitor with the appropriate shunt resistance value.6

If in the future it is necessary to conduct measurements with a given error, the device with the installed shunt must be approved in metrological laboratory since the installation

**of the shunt**lowers the measurement accuracy.7

If for any reason it is impossible to find out the data of the measuring device, pick up shunt experimentally. It is desirable to confine small currents up to 5 A.

8

Sequentially connect the load, monitoring the ammeter and the test ammeter. Terminal of the test ammeter will short-circuit a piece of nichrome or copper of minimum length.

9

Set the power supply voltage to the control device showed a current value for which the selected shunt. Arrow test ammeter should not exceed the limit.

10

Slightly releasing the clamp one of the terminals of the ammeter, increase the length of wire between the terminals as long as the needle reaches the maximum scale. Lock the clamps on the test device, trim the excess wire. Turn off the power supply. Current meter range you need ready to use.

# Advice 3: How to choose a shunt resistor for ammeter

The sensitivity of modern dial indicators are so high that many of them are current full deflection does not exceed one hundred microamperes. In practice, it often is necessary to measure currents of hundreds of milliamperes, and even amperes. Comes to the aid of so-called

**shunt**.Instruction

1

Before you start making

**the shunt**and it is necessary to measure the internal resistance of a dial indicator. Use ordinary tester or multimeter (either dial or digital). In this case, it is necessary that the current through the test device was not too large, otherwise the arrow may be deformed.2

Now calculate the voltage that should be provided to the indicator that the arrow has deviated completely. For this, set the full deflection current in amperes and the measured resistance of the device in ohms. Then substitute them into the standard formula of Ohm's law:U=IR where U is the voltage required for full deflection, and I is the current full deflection, R is the measured resistance of the frames.Do not be surprised that calculated according to this formula, the voltage will be very small.

3

Now you need to calculate the resistance of the

**shunt**. It will be so small compared to the resistance of frames that the latter can be neglected. Resistance**shunt**should be such that when passing current through it, which is**of the ammeter**limit,**the shunt**e fell a voltage equal to the calculated in the previous formula. Thus, this resistance may also be calculated by the standard formula of Ohm's law, but modified in the following way:R=U/I, where R is the required resistance of**the shunt**and, U - naprijenie full deflection of the indicator calculated by the previous formula, I is the limiting current, the measurement of which will be calculated your ammeter (if it is expressed in milliamperes, put it in pre-amps).4

Correctly connect between the indicator and

**the shunt**. Namely himself**, the shunt**will engage directly in the circuit, the current in which is to be measured, and indicator connect the wires to it. Thou to do the opposite, turning on an led indicator in the circuit, and**a shunt**connecting the wires to the indicator, the latter will explode or even burn. Think about why.5

Manufacture for micro

**ammeter**new scale having graduations in milliamperes or amps and the corresponding scale.Note

Connecting and disconnecting the implement when de-energized circuit. Observe the correct polarity. When measuring AC, use the indicator in conjunction with the detection circuit.