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what is it about the design of a good ammeter that allows you to connect

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In this explainer, we will larn how to draw the combination of a galvanometer with a shunt resistor to design a DC ammeter.

An ammeter is a device that tin be used to measure the current in a circuit. As we shall see, we can brand such a device using a galvanometer along with a resistor.

Since the ammeter's design makes use of a galvanometer, allow's begin by reminding ourselves how a galvanometer behaves.

A galvanometer is a device that responds to the management and magnitude of electric current. The following figure shows a galvanometer.

Equally shown in the sketch, the galvanometer has a needle that can deflect to either side of zero. This needle deflects whenever there is a current through the galvanometer.

Then, if we apply a potential difference beyond the galvanometer, resulting in a electric current through it, the needle will deflect to one side of the nothing. This is shown in the sketch beneath.

As shown in the sketch, if nosotros reverse the polarity of the potential departure such that the electric current goes the opposite way through the galvanometer, the needle will deflect to the opposite side of the zero.

So, nosotros tin can see that a galvanometer provides a means of measuring the amount of current through it. In fact, it turns out that the deflection of the galvanometer'south needle abroad from the central null is proportional to the magnitude of the current, up until the signal at which the needle reaches the end of the calibration.

This limit, at which the needle is pointed fully to one stop of the scale, is illustrated below.

In this situation, we say that the needle has a maximum deflection. The electric current that causes the needle to just reach this maximum deflection is the maximum value of electric current that can be measured using this galvanometer. The exact maximum current that can be measured using a galvanometer varies depending on the device simply is typically on the order of microamperes or milliamperes.

For example, if we have a galvanometer with a full-scale deflection of 500 μA, and so this particular device would exist able to mensurate the intensity and direction of a electric current as long as this current is smaller than 500 μA.

If we, however, desire to use a galvanometer every bit an ammeter, nosotros will run into two issues.

The first problem is that there is clearly a limit on the intensity of the currents we can measure out with a galvanometer. Specifically, we are limited to currents below the full-scale deflection current of the galvanometer. This represents an upshot if we want to measure currents exterior this range.

We might think that if nosotros could somehow extend the range of a galvanometer, so this would provide an accurate way of measuring larger currents also. However, there is a second problem: the galvanometer has its own internal resistance. Permit's see why this represents a problem by considering a simple circuit.

In this excursion, the prison cell provides a potential deviation 𝑉 across a resistor of resistance 𝑅 , and and then there is a current 𝐼 .

We can call up that Ohm's police force tells us that, for such a circuit, nosotros have 𝑉 = 𝐼 𝑅 .

We tin can rearrange this by dividing both sides by the resistance 𝑅 to go an expression for the electric current in the circuit in terms of the potential divergence and resistance: 𝐼 = 𝑉 𝑅 .

Allow's now see what happens when nosotros attempt to utilise a galvanometer in order to measure out the value of 𝐼 . Imagine that the range of the galvanometer and the values of 𝑉 and 𝑅 are such that 𝐼 is less than the full-scale deflection current of the galvanometer.

Calculation in a galvanometer in series with the other components, we go the following circuit.

However, we said earlier that the galvanometer has some resistance of its own. Permit's make that clear by explicitly cartoon in this resistance in our circuit.

We have labeled the resistance of the galvanometer 𝑅 . We can run into that nosotros at present have two resistors connected in serial.

Nosotros can remember that when we have two resistors continued in series, the total resistance is given by the sum of the individual resistances. So, in our case, if we label the total resistance 𝑅 , then we take that 𝑅 = 𝑅 + 𝑅 .

If we at present apply Ohm'south law to the circuit equally a whole, nosotros tin encounter exactly why this is a problem. Permit'south use the grade of the equation in which the current is the subject. We volition label this current 𝐼 . The total resistance of our circuit is now 𝑅 , not merely 𝑅 . Substituting this into Ohm'due south law, nosotros accept 𝐼 = 𝑉 𝑅 .

Then, we can substitute in 𝑅 = 𝑅 + 𝑅 to get 𝐼 = 𝑉 𝑅 + 𝑅 .

Let's now compare this equation to our original equation without the galvanometer, in which 𝐼 = 𝑉 𝑅 . The fact that the total resistance of the circuit has changed means that the intensity of the current has also changed.

This means that the galvanometer, the device we were trying to use in order to mensurate the current, has actually changed the value of the current nosotros wanted to measure. This is a bit like having a weighing scale that changes the mass of the object yous place on information technology, or a ruler that changes the length of the thing you were trying to measure out with it.

Fortunately, it turns out that there is a mode nosotros tin can deal with this problem of the galvanometer changing the electric current in the circuit. Nosotros can practice this by adding a resistor in parallel with the galvanometer.

This resistor is referred to as a shunt resistor, and we accept labeled its resistance 𝑅 .

We can recall that when we add a parallel branch to a circuit, the potential difference across each branch will be the same. Meanwhile, the current gets carve up such that there is a electric current in each of the ii parallel branches. And then, instead of all of the electric current flowing through the galvanometer, some of the current now follows the other path through the shunt resistor instead.

Let'due south look at a quick example.

Case 1: Finding the Potential Difference beyond Two Parallel Branches in a Circuit with a Galvanometer

The circuit diagram represents a galvanometer combined with a shunt resistor. The emf of the source continued to the galvanometer and the shunt is 3.0 V. The circuit does not represent a circuit in which the galvanometer and shunt correctly part equally an ammeter.

  1. What is the potential departure across the shunt? Answer to ane decimal identify.
  2. What is the potential difference across the galvanometer? Answer to one decimal place.

Reply

Part 1

In the diagram for this question, we have a circuit containing a galvanometer in parallel with a shunt resistor. In this case, the resistance of the galvanometer has not been explicitly included equally a resistor in the circuit diagram, but nosotros know that it does take some resistance.

This get-go part of the question is request u.s. to notice the potential difference across the shunt resistor in this diagram.

Nosotros tin recall that the emf of the voltage source in a circuit is equal to the full potential difference beyond the components that the current travels through as it goes around the excursion.

In this case, nosotros have two parallel branches, meaning that there are two possible routes for the current to take. The full potential difference across each of these loops must exist equal to the emf of the voltage source.

The loop passing through the shunt resistor is marked in pink in the diagram beneath.

We know that the total potential difference across this loop must be equal to the emf of the source, which is iii.0 V.

Nosotros will presume that the wires have no resistance then that all of the resistance along the loop marked is in the shunt resistor.

This means that all of the 3.0 V potential departure over the pink loop is across the shunt resistor. Then, we know that the potential difference across the shunt resistor is equal to 3.0 V.

Part two

This 2nd function of the question is asking us for the potential difference across the galvanometer.

The second of the two consummate loops in the diagram, which is the loop passing through the galvanometer, is marked in orange in the diagram below.

As with the first function of the question, we know that the full potential difference across the loop must be equal to the emf of the source, which is iii.0 Five.

Over again assuming that the wires have no resistance, this means that all of the resistance along the orangish loop is in the galvanometer.

Therefore, all of the 3.0 V potential difference that is across the orange loop is across the galvanometer. So, nosotros know that the potential departure beyond the galvanometer is equal to 3.0 V.

As illustrated in this case, when we connect a shunt resistor in parallel with the galvanometer, the potential deviation across the shunt resistor will exist equal to the potential divergence across the galvanometer.

Nosotros too know that the current splits into the ii parallel branches. To detect out how much current is in each branch, we can apply Ohm's police force to each branch separately.

We will label the resistance of the galvanometer, the current through the galvanometer, and the potential difference across information technology with a subscript 𝐺 , and we volition label these same quantities for the shunt resistor with a subscript 𝑆 .

And then, Ohm's law for each co-operative gives usa the following two equations: 𝐼 = 𝑉 𝑅 , 𝐼 = 𝑉 𝑅 .

We know that both branches have the same potential departure. That is, in these equations, nosotros know that 𝑉 = 𝑉 .

Let's consider what happens we choose a shunt resistor with a resistance much smaller than that of the galvanometer, that is, when we pick a shunt resistor such that 𝑅 ≪ 𝑅 .

For 2 resistors continued in parallel, the overall resistance is lower than the lowest of the ii individual resistances. The full resistance of the galvanometer and shunt resistor in parallel is therefore smaller than 𝑅 and, hence, much smaller than 𝑅 .

This means that the overall outcome of the combination of galvanometer and shunt resistor on the electric current in the excursion is very small. In other words, past adding a shunt resistor in parallel with the galvanometer, nosotros accept overcome the problem that the resistance of the galvanometer would affect the current in the circuit that it was trying to measure. Any consequence on the current in the circuit will now exist much smaller.

We tin too find from the ii Ohm's law equations that the expressions for 𝐼 and 𝐼 take the aforementioned values in the numerator, but the denominator in the 𝐼 expression is much larger than in the 𝐼 expression. What this means is that if 𝑅 ≪ 𝑅 , then 𝐼 ≫ 𝐼 .

In other words, most of the electric current goes through the path containing the shunt resistor. Meanwhile, in that location is a pocket-size, constant proportion of the current through the galvanometer. This means that the deflection of the galvanometer'southward needle volition be proportional to the current in the circuit. Hence, the combination of the galvanometer and shunt resistor may exist used to measure out the current in the excursion.

So, everything within the orangish box in the diagram below together functions as an ammeter.

When building an ammeter in this way, it is important to carefully cull the resistance 𝑅 of the shunt resistor to give the best results. Remember that the resistance 𝑅 of the galvanometer has a fixed value. Changing the value of 𝑅 changes the fraction of the current that passes through the galvanometer. We want a value of 𝑅 such that the current through the galvanometer is high enough that the needle displays a articulate reading, just besides low enough that the needle does not quite hit full-scale deflection.

To piece of work out the best value for 𝑅 , we can once over again make apply of Ohm's police force, which states that, for a potential difference 𝑉 , resistance 𝑅 , and current 𝐼 , 𝑉 = 𝐼 𝑅 .

Since nosotros are trying to find a resistance, we want to make 𝑅 the subject. Dividing both sides by 𝐼 gives united states 𝑅 = 𝑉 𝐼 .

We are trying to work out what value we should utilize for the shunt resistance 𝑅 . So, let's supercede 𝑅 in Ohm'due south law with this shunt resistance 𝑅 . Nosotros should besides utilise the potential divergence across the shunt resistor, 𝑉 , in place of 𝑉 and the current through the shunt resistor, 𝐼 , in place of 𝐼 .

Making these substitutions gives us 𝑅 = 𝑉 𝐼 .

Nosotros can make some substitutions in this equation in gild to brand it a flake more useful. First off, nosotros have already said that the potential difference beyond the shunt resistor, 𝑉 , is equal to the potential divergence across the galvanometer, 𝑉 . And so, nosotros tin supplant 𝑉 by 𝑉 in our equation: 𝑅 = 𝑉 𝐼 .

We besides know that the total electric current in the circuit, 𝐼 , splits into the two currents 𝐼 and 𝐼 such that 𝐼 = 𝐼 + 𝐼 . Alternatively, subtracting 𝐼 from both sides of this, nosotros have that 𝐼 = 𝐼 − 𝐼 .

Substituting this expression for 𝐼 into our Ohm's law equation, we get 𝑅 = 𝑉 𝐼 − 𝐼 .

Now, we can apply Ohm'due south law once more in social club to replace 𝑉 , the potential deviation beyond the galvanometer. We know that the current through the galvanometer is 𝐼 and the resistance of the galvanometer is 𝑅 . Ohm's law tells us that 𝑉 = 𝐼 𝑅 .

Substituting this in for 𝑉 in our expression for 𝑅 gives the states 𝑅 = 𝐼 𝑅 𝐼 − 𝐼 .

One way to empathise what this equation means is to consider a specific value of electric current in the galvanometer. In detail, there will be a value of this current 𝐼 that gives a full deflection of the galvanometer's needle. This value is the full-scale deflection current.

We also know that 𝐼 is a pocket-size but constant proportion of the total electric current 𝐼 . So, this maximum value of 𝐼 that can be recorded corresponds to a maximum value of the current 𝐼 that tin can be measured by the ammeter.

With this in mind, nosotros arrive at the post-obit estimation of our equation for 𝑅 .

Equation: The Resistance of a Shunt Resistor in an Ammeter

Suppose nosotros take an ammeter consisting of a galvanometer with a total-scale deflection electric current 𝐼 and a resistance 𝑅 connected in parallel with a shunt resistor.

In social club to exist able to measure out a maximum current of 𝐼 using this ammeter, we must use a shunt resistor with a resistance 𝑅 given by 𝑅 = 𝐼 𝑅 𝐼 − 𝐼 .

Now allow's take a look at an instance problem.

Example 2: Finding the Required Resistance of a Shunt Resistor in an Ammeter

A galvanometer has a resistance of 15 mΩ. A current of 125 mA produces a full-scale deflection of the galvanometer. Detect the resistance of a shunt that when connected in parallel with the galvanometer, allows it to be used every bit an ammeter that can mensurate a maximum electric current of 12 A. Answer to the nearest microhm.

Reply

Permit'due south brainstorm by drawing a circuit diagram and labeling the values that nosotros have been given.

We have explicitly fatigued in a resistor representing the resistance of the galvanometer. This resistance is 𝑅 = ane 5 yard Ω . We know that the total-scale deflection electric current of the galvanometer is 𝐼 = one 2 5 k A , and we want to use this setup as an ammeter to measure a maximum electric current of 12 A.

We are asked to work out what value of the shunt resistance, which we have labeled 𝑅 , will permit united states to measure out this maximum current.

We can call up that we know an equation for the required shunt resistance 𝑅 in order to exist able to measure a maximum current 𝐼 , given an ammeter using a galvanometer with full-scale deflection electric current 𝐼 and resistance 𝑅 : 𝑅 = 𝐼 𝑅 𝐼 − 𝐼 .

Before substituting in our values for the quantities on the right-manus side, nosotros need to convert them and so that they all have consistent units. If nosotros measure the currents in amperes and the galvanometer's resistance in ohms, then nosotros volition get a shunt resistance with units of ohms.

The value of 𝐼 = i 2 A is already given in amperes, and so we merely need to convert 𝐼 and 𝑅 .

Nosotros have that 𝐼 = one 2 5 = 0 . ane 2 v yard A A and 𝑅 = 1 v = 1 . 5 × 1 0 m Ω Ω .

Substituting in our values for 𝐼 , 𝑅 , and 𝐼 , we get that 𝑅 = ( 0 . 1 2 5 ) × 1 . 5 × 1 0 1 2 − 0 . 1 2 v . A Ω A A

Evaluating the right-hand side of this expression, we notice that 𝑅 = 1 . 5 7 8 ix … × i 0 . Ω

Hither, an ellipsis is used to indicate that there are farther decimal places.

Finally, we note that the question asks us for our respond in units of microhms, to the nearest microhm. Recalling that ane = 1 0 µ Ω Ω , we tin can give our answer for the required shunt resistance as 𝑅 = ane 5 8 . µ Ω

We can also take our equation for the shunt resistance 𝑅 and look at it in another way.

If we rearrange the equation to make the electric current 𝐼 the subject, then nosotros become an equation that tells usa the maximum current we tin can mensurate with an ammeter, given the properties of its components.

Let's come across how nosotros tin make 𝐼 the subject. Call back that nosotros outset with the post-obit equation: 𝑅 = 𝐼 𝑅 𝐼 − 𝐼 .

We want to get the current 𝐼 out of the denominator of the fraction, so nosotros will begin past multiplying both sides of the equation past 𝐼 − 𝐼 : 𝑅 ( 𝐼 − 𝐼 ) = 𝐼 𝑅 ( 𝐼 − 𝐼 ) 𝐼 − 𝐼 𝑅 ( 𝐼 − 𝐼 ) = 𝐼 𝑅 .

In the second line, nosotros have canceled the 𝐼 − 𝐼 term that appears in the numerator and denominator on the right-mitt side.

Adjacent, we divide both sides of the equation by 𝑅 : 𝐼 − 𝐼 = 𝐼 𝑅 𝑅 .

Finally, we add 𝐼 to both sides: 𝐼 = 𝐼 𝑅 𝑅 + 𝐼 .

Equation: The Measurement Range of an Ammeter

Suppose we have an ammeter consisting of a galvanometer with a full-scale deflection electric current 𝐼 and a resistance 𝑅 connected in parallel with a shunt resistor of resistance 𝑅 .

The maximum current that can be measured using this ammeter, also known as the measurement range of the ammeter, is and so given by 𝐼 = 𝐼 𝑅 𝑅 + 𝐼 .

Permit's finish up by taking a look at a couple more instance questions.

Example three: Calculating the Measurement Range of an Ammeter

A galvanometer has a resistance of 12 mΩ. A current of 150 mA produces full-scale deflection of the galvanometer. A shunt is connected in parallel with the galvanometer to catechumen information technology into an ammeter. The resistance of the shunt is 70 µΩ. What is the greatest current that the ammeter tin can measure? Reply to 1 decimal place.

Reply

Let'south brainstorm by drawing a circuit diagram and labeling the values that we have been given

We have explicitly drawn in a resistor representing the resistance of the galvanometer. This resistance is 𝑅 = 1 2 1000 Ω . Nosotros know that the full-scale deflection current of the galvanometer is 𝐼 = i 5 0 1000 A and that the resistance of the shunt resistor is 𝑅 = 7 0 µ Ω .

We are asked to piece of work out what the greatest current is that this ammeter can measure.

Nosotros can think that we have an equation for the maximum current that an ammeter tin can measure out given the properties of its components: 𝐼 = 𝐼 𝑅 𝑅 + 𝐼 .

We know the values of all of the quantities on the correct-mitt side of this equation.

However, earlier substituting in our values, we need to make the units compatible. If we use units of amperes for 𝐼 and units of ohms for both resistances 𝑅 and 𝑅 , and so we will get a current 𝐼 in units of amperes.

Converting our values into these units, we have that 𝐼 = 1 v 0 = 0 . 1 5 m A A , 𝑅 = 1 two = 1 . two × 1 0 g Ω Ω , and 𝑅 = seven 0 = 7 × ane 0 µ Ω Ω .

Substituting in these values into our expression for 𝐼 , we get that 𝐼 = ( 0 . ane v ) × 1 . 2 × 1 0 7 × i 0 + 0 . ane 5 . A Ω Ω A

Evaluating the correct-mitt side gives us that 𝐼 = 2 five . eight 6 iv … , A where the ellipsis indicates that the value has further decimal places.

Finally, we note that we are asked to give our answer to one decimal place. Rounding to i decimal place, we have that the greatest electric current that the ammeter tin mensurate is given by 𝐼 = 2 5 . ix . A

Case 4: Calculating the Currents through the Galvanometer and Shunt Resistor in an Ammeter

The current 𝐼 in the circuit shown is 2.v mA, which is the greatest current that can be measured using the ammeter continued to the circuit. The resistance of the galvanometer is ten times the resistance of the shunt.

  1. Find 𝐼 , the current through the galvanometer. Answer to the nearest microampere.
  2. Observe 𝐼 , the current through the shunt. Reply to two decimal places.

Answer

Office one

Nosotros are given a circuit that acts equally an ammeter. We are asked to work out the electric current 𝐼 through the galvanometer, given that the maximum electric current that this ammeter can measure is 𝐼 = ii . 5 m A .

We are not told the actual values of the resistances of the galvanometer and shunt resistor. However, we are told that the galvanometer'south resistance, which we volition call 𝑅 , is 10 times the shunt resistance, which we will telephone call 𝑅 . In other words, we have that 𝑅 = 1 0 𝑅 .

We can recall that nosotros know an equation linking the current 𝐼 in a circuit, such every bit the one in this question, and the current 𝐼 through the galvanometer: 𝐼 = 𝐼 𝑅 𝑅 + 𝐼 .

Nosotros too know that, for this ammeter, 𝑅 = 1 0 𝑅 . And then, we can supercede 𝑅 by 1 0 𝑅 in our expression for 𝐼 to requite us that 𝐼 = 𝐼 × ( 1 0 𝑅 ) 𝑅 + 𝐼 .

We tin can now cancel 𝑅 in the numerator of the fraction on the right-hand side with 𝑅 in the denominator of this fraction: 𝐼 = i 0 𝐼 + 𝐼 = 1 ane 𝐼 .

Since in this example we know the electric current 𝐼 and we want to find the value of 𝐼 , we can make 𝐼 the subject by dividing both sides of this equation past 11: 𝐼 = 𝐼 i one .

Finally, substituting in 𝐼 = ii . 5 m A gives us our outcome for 𝐼 , the electric current through the galvanometer: 𝐼 = ii . 5 1 one = 0 . ii two vii … 𝐼 = 2 2 7 . k A m A μ A

Hither, we accept given our answer in microamperes to the nearest microampere, as requested by the question.

It is worth noticing that, since we were told that 𝐼 is the maximum current that tin exist measured by this ammeter, then we know that 𝐼 is the full-scale deflection current of the galvanometer.

Function 2

Nosotros were told in the question that the current in the circuit is 𝐼 = two . five m A . Nosotros accept also at present worked out that the current in the galvanometer is 𝐼 = 2 2 7 = 0 . two two 7 μ A m A .

This second part of the question is asking the states to observe the electric current 𝐼 through the shunt resistor.

We can retrieve that the full current into a junction must be equal to the current out of it.

Looking back at the diagram in the question, nosotros see that we have a current 𝐼 into the junction, besides as currents 𝐼 and 𝐼 each coming out of the junction forth dissimilar branches. So, the current 𝐼 is getting split at the junction into the currents 𝐼 and 𝐼 .

In other words, we know that nosotros must accept 𝐼 = 𝐼 + 𝐼 .

We are trying to find the value of 𝐼 , so we should rearrange this equation by subtracting 𝐼 from both sides to make 𝐼 the subject area: 𝐼 = 𝐼 − 𝐼 .

When substituting in values for 𝐼 and 𝐼 , we need to brand certain we utilize the same units for both. Using units of milliamperes, we have that 𝐼 = 2 . 5 one thousand A and 𝐼 = 0 . ii 2 7 thousand A . Substituting these values in gives u.s. our issue for the current 𝐼 through the shunt resistor: 𝐼 = 2 . 5 − 0 . 2 2 vii 𝐼 = 2 . 2 7 . m A m A m A

Nosotros take given our answer to ii decimal places, equally requested by the question.

Finally, let'south summarize what nosotros have learned in this explainer.

Fundamental Points

  • An ammeter tin can exist made by connecting a galvanometer and a resistor, known every bit a shunt resistor, in parallel.
  • To brand an ammeter in this way, the resistance 𝑅 of the shunt resistor must be much less than the resistance 𝑅 of the galvanometer: 𝑅 ≪ 𝑅 .
  • If we take an ammeter consisting of a galvanometer with a total-scale deflection current of 𝐼 and a resistance of 𝑅 connected in parallel with a shunt resistor, and so in order to be able to measure out a current 𝐼 , the required value of the shunt resistance 𝑅 is given by 𝑅 = 𝐼 𝑅 𝐼 − 𝐼 .
  • We can rearrange this equation to give us an expression for the maximum current 𝐼 that we tin measure with a particular ammeter, given the properties of its components: 𝐼 = 𝐼 𝑅 𝑅 + 𝐼 .

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