# Measuring Current with a Voltmeter

## Warning

>Passing high currents through resistors can produce a hot of heat, don't touch them.

>Don't use this method with voltages above 12 volts unless you know what you're doing.

## Background

Most handheld digital multimeters have a knob that can be turned to different positions to select a mode of operation. Selecting the mode for current measurement (ammeter) allows the user to measure the current flowing through a chosen part of a circuit. However, some meters don't have this mode!!!

## A Meter without Ammeter Mode

Don't despair, if a meter cannot measure current with a built in ammeter feature, there is a solution.

A more expensive meter, pictured below, is fully capable of measuring current simply by selecting "20A" with the knob and moving the positive lead to a different socket. Still, even with this meter, the maximum current is limited to 20 amps. To measure higher currents the same trick can be applied.

## How to Do it

The way an ammeter (current meter / amp meter) works is by allowing a current to flow through a resistance of a known value. This is called a shunt resistor and is inside the meter. The voltage drop, or the voltage measured across the resistor, is proportional to the amount of current flowing through it. For example, if a voltage of 3 volts is measured across a resistor of 2 ohms, then the current flowing through it must be 1.5 amps. Voltage drop divided by resistance equals current (V/R = I).

## Simulation of the Circuit

## Detailed Example

The illustration above is from a circuit simulating software and already shows the current flowing the circuit. Had it not been so obvious, simply by knowing the value of the resistor and the voltage of the battery, the current can be calculated by Ohm's law.

Of course, if a meter doesn't have the ammeter function, there will be no internal shunt resistor so you will have to use an external resistor.

## Choosing a Shunt Resistor

A shunt resistor is just a regular resistor. The main idea is to choose a resistor that is a low value and accurate (plus or minus 1%). This resistor will need to be wired in series with the circuit that is to be measured. To measure the current flowing out from a car battery, for example, would require removing the negative cable from the battery and wiring a resistor between the cable and the negative terminal of the battery. The lower the value of the resistor, the less affect it will have on the circuit. In the case of the car battery, which could involve peak currents of 1000 amps, a very low resistance resistor is required. 0.012 ohm (12 milliOhms) would allow a maximum of 1000 amps at 12 volts (12 / 0.012 = 1000).

## Selecting the Correct Power Rating

In the previous example, a 12 milliohm resistor was selected for the shunt and installed in series with the load (the car) and the source (the battery).

The maximum current flowing through this shunt is 1000 amps and the maximum heat (power dissipated) through the resistor can be found by P = I^2 * R, where P is power dissipated, I = current in amps, and R = resistance in ohms. Solving the equation for the 12 milliohm resistor with 1000 amps flowing through it yields 12,000 watts, or 12KW of power dissipated! This is a lot of wasted power, and finding a resistor this size would be difficult.

To lower the power, and consequentially make the measurement more efficient. A lower resistance resistor is needed. Dividing the desired power dissipated (watts) by current (amps) squared gives us the resistor to use. 1 watt / (1000 amps*1000 amps) = 1uohm or 1 micro ohm. Remember the resistor must be a high precision type for accurate measurements. A resistor this low is surely to simply be a short fat piece of copper alloy wire, precision cut for accuracy.

In this automotive example, connecting this shunt resistor in series with the circuit will require the same strong terminals as the battery, to make good connection. A loose battery cable can prevent a car from starting because the resistance is raised to high at the poor junction for sufficient current to flow at the rated voltage of the battery.

## Performing the Measurement

Once an appropriately sized shunt resistor is installed in the circuit, attach the leads of the voltmeter across the shunt. Turn on the multimeter and select volts. Red lead to one size of the resistor and black lead to the other side.

Remember, the voltage measured by the multimeter (across the shunt resistor) is proportional to the current flowing through it. If the multimeter reads 1.5 mV (0.0015 volts) across a shunt resistor of 1 micro ohm... then 1500 amps is flowing through it! V / R = I. And power dissipated (by the resistor) is 1500 * 1500 * 0.000001 = 2.25 watts.

2.25 watts is a fairly large amount to be concentrated in a small resistor, it will get very hot unless the resistor is sufficiently large. The power rating on a resistor is what it can tolerate, not necessarily what your skin can! Be careful.

Tips:

This method is easy and standard for low current devices and medium sized DC loads. With AC circuits of high current, non contact methods can be used, such as a current clamp meter. A clamp meter measures the indirect effects through induction with a coil to determine current through a wire.

To keep the efficiency high and impact on the originally circuit as low as possible, the resistance should be chosen as low as possible. However, this may make voltage measurements more difficult since they might drop into the micro volt range which some volt meters cannot measure. Voltage amplifier circuits are typically used in this case to amplify the voltage to a detectable level.

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## 2 comments

Hey, that post leaves me feeling foilsoh. Kudos to you!

The reason your curenrt didn't work out is because you used the wrong values in your formula. Your right in saying the curenrt is the same in a series circuit, so that means if you want to find the Total curenrt, you take the Total voltage of the circuit divide by the Total resistance of the circuit. This will give you your 3.5 amps.If you place a meter anywhere in the circuit, you will get 3.5 amps.You can use the triangle method for? watts(goes up top)also.Ur series has been very informative