Lesson Explainer: Design of the Voltmeter | Nagwa Lesson Explainer: Design of the Voltmeter | Nagwa

Lesson Explainer: Design of the Voltmeter Physics • Third Year of Secondary School

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In this explainer, we will learn how to describe the combining of a galvanometer with a multiplier resistor to design a DC voltmeter.

A galvanometer is used to measure the current of a circuit by placing it within it.

When there is a current in the circuit, it passes through the galvanometer, which measures it. The galvanometer also has a resistance 𝑅G. Together, these values are be related to the voltage using Ohm’s law, 𝑉=𝐼𝑅, where 𝑅, in this case, is the resistance of the galvanometer, 𝑅G. This relationship cannot be used with most galvanometers, however, as they are typically very sensitive and only measure small amounts of current, on the micro- and milliscales. We can say that the maximum current on either end of the scale a galvanometer can measure is the galvanometer current, 𝐼G.

This makes their use by themselves impractical as voltmeters, as they would only show up to the maximum current 𝐼G even if the actual current is higher. There is a way this can be solved, though, and it is by adding another resistor in series with the galvanometer that has a much higher resistance.

The additional resistor in this circuit causes the total current in the circuit to go down, as the voltage stays the same in 𝑉=𝐼𝑅. The resistor is called a multiplier resistor, and if its resistance is large enough, it can allow the current in the circuit to fall down to a region that the galvanometer can measure, making it a suitable instrument for measuring voltage.

Let’s look at an example.

Example 1: Multiple Resistor Size in a Voltmeter

A voltmeter is used to measure the voltage of a direct-current source that is estimated to have a voltage of several volts. The galvanometer in the voltmeter has a resistance of a few milliohms. Which of the following correctly explains why the multiplier resistor in such a voltmeter must have a resistance much greater than the resistance of the galvanometer that the multiplier is connected in series with?

  1. If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the current through the galvanometer will be greater than the current that would produce a full-scale deflection of the galvanometer arm.
  2. If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the resistor will generate a magnetic field that significantly changes the deflection of the galvanometer arm.
  3. If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the voltage of the source will be significantly increased.
  4. If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the direction of the deflection of the galvanometer arm will reverse and no reading will be displayed on the voltmeter.

Answer

The direction of the galvanometer arm will not switch if the multiplier resistor is too small; that would require the current to be going in the opposite direction. So, it is not D.

A galvanometer uses a magnetic field to deflect its needle, but the resistor will not be generating one, so it is not B.

The voltage of a source does not increase or decrease depending on resistors in the circuit either. It is not C.

Ideally, the current going through the multiplier resistor will be much higher than the one going through the galvanometer, so the correct answer is A.

This added resistor is called a multiplier resistor, as it effectively multiplies the amount of voltage a galvanometer can measure on its own and is of a value much larger than the galvanometer resistance. Its resistance value is shown as 𝑅M, and this value helps determine how much voltage can be measured. This only works when the multiplier resistor is in series with the galvanometer, not in parallel.

It must be noted that this is for setting up a galvanometer and multiplier resistor circuit for use as a voltmeter, not the connection of a voltmeter to a circuit. A correctly connected voltmeter measuring the voltage across a resistor 𝑅 looks like the diagram below.

For constructing the voltmeter in the first place, though, the multiplier resistor must be in series with the galvanometer. Deconstructing the voltmeter in the diagram above into its galvanometer and multiplier resistor, 𝑅M, reveals this:

Let’s look at an example.

Example 2: Correctly Placing a Resistor for a Galvanometer-Based Voltmeter

Which of the following circuit diagrams most correctly represents a galvanometer combined with a multiplier resistor being used as a voltmeter to measure the voltage of a direct-current source?

Answer

Circuit A has its resistor in parallel, not in series. This would cause the effective total resistance of the circuit to actually decrease, which is not what we want when trying to increase the range we can measure! While a voltmeter must be connected in parallel to the circuit to properly measure voltage, when constructing a voltmeter using a galvanometer, it must be in series with the multiplier circuit.

Circuit B has a resistor in series, but also one in parallel with the galvanometer. Just like circuit A, this would cause the total resistance to decrease, decreasing the range that the voltmeter can measure.

Circuit C has a resistor in series with the galvanometer and nothing else. This is the correct setup for combining a resistor with a galvanometer to use it as a voltmeter.

The correct answer is circuit C.

The maximum voltage that can be measured by the galvanometer is equal to the product of the maximum measurable current of the galvanometer and the total resistance of the circuit, 𝑅t. Expressed using Ohm’s law, it would look like 𝑉=𝐼𝑅.maxGt

The total resistance 𝑅t for a series circuit is just all the resistances added together, so it looks like 𝑅=𝑅+𝑅.tGM

The equation for the maximum voltage a galvanometer can measure is thus 𝑉=𝐼(𝑅+𝑅).maxGGM

Let’s look at an example using this equation.

Example 3: Voltage across Multiplier and Galvanometer Resistors

The voltage 𝑉 in the circuit shown is 12 V, which is the greatest voltage that can be measured using the circuit as a voltmeter. The resistance of the galvanometer is one-hundredth of the resistance of the multiplier.

  1. Find 𝑉G, the voltage across the galvanometer. Answer to the nearest millivolt.
  2. Find 𝑉M, the voltage across the multiplier. Answer to one decimal place.

Answer

Part 1

We are given the maximum voltage that can be measured, 12 V, and the ratio of galvanometer to multiplier resistance: 𝑅100=𝑅.MG

The voltage across the galvanometer, 𝑉G is found using Ohm’s law: 𝑉=𝐼𝑅.GGG

We know the resistance ratio, but we do not know 𝐼G. We can find it by relating it in the equation for maximum galvanometer voltage: 𝑉=𝐼(𝑅+𝑅).maxGGM

By dividing both sides by (𝑅+𝑅)GM, we can isolate 𝐼G: 𝑉(𝑅+𝑅)=𝐼(𝑅+𝑅)(𝑅+𝑅).maxGMGGMGM

They cancel out on the right side to give 𝑉(𝑅+𝑅)=𝐼.maxGMG

Since 𝑅G is one-hundredth of 𝑅M, or 0.01𝑅M, we can substitute it in the equation to get 𝑉(0.01𝑅+𝑅)=𝐼𝑉1.01𝑅=𝐼.maxMMGmaxMG

With this relation, we can substitute it back into the voltage across the galvanometer: 𝑉=𝑉1.01𝑅𝑅.GmaxMG

𝑅G is, once again, 0.01𝑅M, so we can substitute 𝑉=𝑉1.01𝑅0.01𝑅.GmaxMM

𝑅M cancels to give a relation of 𝑉=0.0099𝑉.Gmax

𝑉max is 12 V, so the voltage across the galvanometer 𝑉G is 0.0099(12)=0.1188.VV

To get this answer in millivolts, we multiply it by 1‎ ‎000, since there are 1‎ ‎000 mV in 1 volt: 10001×0.1188=118.8.mVVVmV

To the nearest millivolt, 𝑉max is 119 mV.

Part 2

Now, to find 𝑉M, we can just use the relation of total voltage in a circuit, understanding that the maximum voltage is also the total voltage in the circuit: 𝑉=𝑉+𝑉.maxGM

So, 𝑉M is 𝑉𝑉=𝑉+𝑉𝑉𝑉𝑉=𝑉.maxGGMGmaxGM

Putting in the values gives (12)(0.1188)=11.8812.VVV

To one decimal point, the multiplier voltage is thus 11.9 V.

Sometimes it is helpful to determine the exact value of resistance we need in the multiplier resistor to measure up to a specific voltage. We can do this by solving for 𝑅M in this equation. To begin, let’s start by distributing 𝐼G: 𝑉=𝐼𝑅+𝐼𝑅.maxGGGM

We can then subtract 𝐼𝑅GG from both sides 𝑉𝐼𝑅=𝐼𝑅+𝐼𝑅𝐼𝑅𝑉𝐼𝑅=𝐼𝑅.maxGGGGGMGGmaxGGGM

Now, we can divide both sides by 𝐼G to isolate 𝑅M𝑉𝐼𝑅𝐼=𝐼𝑅𝐼.maxGGGGMG

𝐼G cancels out on the right to give 𝑉𝐼𝑅𝐼=𝑅.maxGGGM

The terms on the left can be broken up since they share the same denominator: 𝑉𝐼𝐼𝑅𝐼=𝑅.maxGGGGM

The 𝐼G’s cancel out, simplifying to 𝑅=𝑉𝐼𝑅.MmaxGG

Let’s look at an example using this equation.

Example 4: Required Multiplier Resistor Value for Specific Maximum Voltage Range

A galvanometer has a resistance of 175 mΩ. A current of 20 mA produces a full-scale deflection of the galvanometer. Find the resistance of a multiplier resistor that, when connected in series with the galvanometer, allows it to be used as a voltmeter that can measure a maximum voltage of 15 V. Answer to the nearest ohm.

Answer

Let’s look at the equation we have set up for finding the multiplier resistance: 𝑅=𝑉𝐼𝑅.MmaxGG

We are given the value of maximum voltage, 15 V, but we need to put the other values in their base units first. Let’s start with the current of 20 mA.

There are 1‎ ‎000 mA in 1 A: 11000.AmA

Multiplying this relation by 20 mA gives 11000×20=0.02.AmAmAA

Similarly, for the resistance of the galvanometer, there are 1‎ ‎000 milliohms in 1 ohm: 11000.ΩmΩ

Multiplying this by 175 mΩ gives 11000×175=0.175.ΩmΩmΩΩ

So, now we can put in our values into the equation for multiplier resistance: 𝑅=150.020.175.MVAΩ

Volts over amperes become ohms, making the equation 7500.175=749.825.ΩΩΩ

Rounding to the nearest ohm, we see that the galvanometer resistance is so small that it does not change the final answer at all!

The final answer is 750 ohms.

Let’s summarize what we have learned in this explainer.

Key Points

  • A voltmeter can be made by connecting a galvanometer in series with a multiplier resistor.
  • The multiplier resistor increases the range of voltages that the galvanometer can measure.
  • The maximum voltage, 𝑉max, a voltmeter made from a galvanometer can measure is 𝑉=𝐼(𝑅+𝑅),maxGGM where 𝐼G is the maximum current the galvanometer can measure, 𝑅G is the resistance of the galvanometer, and 𝑅M is the resistance of the multiplier resistor.
  • The equation to find the resistance of the multiplier resistor, 𝑅M, is 𝑅=𝑉𝐼𝑅,MmaxGG where 𝑉max is the maximum measurable voltage of the galvanometer, 𝐼G is the maximum measurable current of the galvanometer, and 𝑅G is the resistance of the galvanometer.

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