Ammeter vs Voltmeter: What Is the Difference?

An ammeter measures current at one point in a circuit, so it is wired in series and an ideal one has zero resistance. A voltmeter measures potential difference between two points, so it is wired in parallel and an ideal one has infinite resistance. Swap them and the circuit itself changes.

AP Physics: Unit 11 (topics 11.2 Simple Circuits, 11.5 Compound Direct Current (DC) Circuits). The meters are learning objective 11.5.C of AP Physics 2 Topic 11.5, describe the measurement of current and potential difference in a circuit. Its complete essential knowledge is: ammeters measure current at a specific point (11.5.C.1), must be connected in series with the element whose current is measured (11.5.C.1.i), and ideal ones have zero resistance so they do not affect that current (11.5.C.1.ii); voltmeters measure electric potential difference between two points (11.5.C.2), must be connected in parallel with the element across which the potential difference is measured (11.5.C.2.i), and ideal ones have infinite resistance so no charge flows through them (11.5.C.2.ii); and nonideal ammeters and voltmeters will change the properties of the circuit being measured (11.5.C.3). The Topic 11.5 boundary statement has three clauses: AP Physics 2 only expects students to qualitatively discuss how a nonideal ammeter or voltmeter will affect the results of measurements, unless otherwise stated all batteries, wires, and meters are assumed to be ideal, and circuits with batteries of different potential differences connected in parallel will not be assessed. Meters also appear in Topic 11.2, where 11.2.A.1 lists them among circuit elements and 11.2.A.4.ii gives their schematic symbols, an A in a circle and a V in a circle, with the topic's boundary statement adding that unless otherwise specified all schematics use conventional current. Unit 11, Electric Circuits, carries 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods. AP Physics C: Electricity and Magnetism repeats all seven statements of 11.5.C word for word but its boundary statement omits the qualitative-only restriction on nonideal meters.

The distinction, stated once

An ammeter answers a question about one place. Essential knowledge 11.5.C.1 in AP Physics 2 says ammeters are used to measure current at a specific point in a circuit. To count the charge passing a point each second, the meter has to be in the path, so 11.5.C.1.i says ammeters must be connected in series with the element in which current is being measured. Being in the path means the meter's own resistance would change what it is measuring, so 11.5.C.1.ii says ideal ammeters have zero resistance.

A voltmeter answers a question about two places. Essential knowledge 11.5.C.2 says voltmeters are used to measure electric potential difference between two points in a circuit. A difference between two points needs both points, so 11.5.C.2.i says voltmeters must be connected in parallel with the element across which potential difference is being measured. Being connected across an element gives charge a second route, so 11.5.C.2.ii says ideal voltmeters have infinite resistance, and no charge flows through them.

Every other row on this page is a consequence of those two paragraphs, and the whole thing collapses into one line: an ammeter is spliced in, a voltmeter is touched on. You have to cut the circuit open to insert an ammeter. You never have to cut anything to use a voltmeter.

The ideal resistances are the part worth memorising, because they point in opposite directions for the same reason. Zero for the ammeter and infinite for the voltmeter are both ways of saying "do not disturb what you are measuring". A meter in the path must not impede the flow; a meter across a gap must not offer a new route.

Side by side

	Ammeter	Voltmeter
Measures	Current at a specific point (11.5.C.1)	Potential difference between two points (11.5.C.2)
Reads in	Amperes, coulombs per second	Volts, joules per coulomb
Wired	In series with the element (11.5.C.1.i)	In parallel with the element (11.5.C.2.i)
Ideal resistance	Zero (11.5.C.1.ii)	Infinite (11.5.C.2.ii)
Current through the meter	The same current as the element	None, for an ideal meter
Potential difference across the meter	None, for an ideal meter	The same as across the element
Circuit must be broken to fit it	Yes	No
Schematic symbol (11.2.A.4.ii)	A circle containing A	A circle containing V
One reading or many, in a series loop	One, the same everywhere	A different one for every pair of points
Effect of connecting it wrongly	In parallel it short-circuits the element	In series it stops the current
What a nonideal one does	Reduces the current it is reading	Draws some current, reducing the potential difference it is reading

The two rows in the middle are the definitional pair, and it is worth seeing that they are exact mirrors. An ideal ammeter carries the full current and drops no voltage. An ideal voltmeter drops the full voltage and carries no current. Neither meter, in its ideal form, is a place where anything is used up.

The row about many readings is the one that catches people in a lab. Move an ideal ammeter to any point in a single series loop and the reading does not change, because a series loop has one current. Move a voltmeter's probes and the reading changes every time, because every pair of points has its own potential difference. If a series-circuit question gives you three ammeter readings that differ, something in the circuit is not what you think it is.

What the CED requires, boundary statement included

The meters live in learning objective 11.5.C of Topic 11.5, Compound Direct Current (DC) Circuits: describe the measurement of current and potential difference in a circuit. Its essential knowledge is the six statements quoted above plus one more.

11.5.C.3: nonideal ammeters and voltmeters will change the properties of the circuit being measured.

That is the complete essential knowledge of 11.5.C, checked item by item against the page: 11.5.C.1 with sub-points i and ii, 11.5.C.2 with sub-points i and ii, and 11.5.C.3 with none.

The boundary statement on Topic 11.5 has three clauses, and all three matter:

> AP Physics 2 only expects students to qualitatively discuss how a nonideal ammeter or voltmeter will affect the results of measurements. Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed.

The first clause is the one that shapes your exam answers: nonideal meters are a describe it topic, not a compute it topic. If you find yourself calculating the loading error of a voltmeter with a finite resistance, you have gone past what AP Physics 2 asks. The second clause is the default assumption you work under unless a problem says otherwise, and it covers batteries and wires as well as meters. The third clause removes a class of circuit from the exam altogether.

Meters also appear earlier, in Topic 11.2. Essential knowledge 11.2.A.1 lists the elements a circuit may include as wires, batteries, resistors, lightbulbs, capacitors, switches, ammeters, and voltmeters, and 11.2.A.4.ii gives the schematic symbols, an A in a circle for the ammeter and a V in a circle for the voltmeter, with variable elements indicated by a diagonal strikethrough arrow across the standard symbol. The Topic 11.2 boundary statement adds that unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.

AP Physics C: Electricity and Magnetism carries objective 11.5.C with all seven of those essential knowledge statements identical word for word. Its boundary statement is shorter: it keeps the ideal-by-default assumption and the parallel-batteries exclusion, and it drops the clause restricting nonideal meter analysis to qualitative discussion. So the calculus course can ask you to go further on nonideal meters than AP Physics 2 can.

The case that separates them: the two ways to get it wrong

Take the simplest circuit that can hold both meters, a battery and one resistor, and misconnect one meter at a time. Do not misconnect both at once: the two errors do different things, and putting them together hides the more interesting one, because an infinite resistance anywhere in the loop stops everything.

Use a battery of emf $9.0 \ \mathrm{V}$ with an internal resistance of $r = 0.50 \ \Omega$ , stated explicitly because the Topic 11.5 boundary statement otherwise assumes the battery is ideal, and a $30 \ \Omega$ resistor.

	Both meters correct	Ammeter moved into parallel with the resistor	Voltmeter moved into series in the loop
Ammeter	In series with the resistor	Across the resistor	Still in series
Voltmeter	Across the resistor	Still across the resistor	In series in the loop
Resistance the battery drives	$30 \ \Omega$ plus $0.50 \ \Omega$ internal	$0.50 \ \Omega$ internal only, the resistor is bypassed	Infinite, the voltmeter blocks the loop
Current in the loop	$0.30 \ \mathrm{A}$	$18 \ \mathrm{A}$	Zero
Ammeter reads	$0.30 \ \mathrm{A}$	$18 \ \mathrm{A}$ , a short-circuit current	$0 \ \mathrm{A}$
Voltmeter reads	$8.9 \ \mathrm{V}$	$0 \ \mathrm{V}$	$9.0 \ \mathrm{V}$ , the full emf
What the resistor does	Dissipates energy normally	Almost nothing; no current takes that route	Nothing; there is no current anywhere

The two failure modes are different in kind, which is why doing them one at a time is worth the trouble.

An ammeter in parallel short-circuits the element. Its zero resistance is a route that bypasses the resistor entirely, and by 11.2.A.2.iii a short circuit is one in which charges would be able to flow with no change in potential difference. Almost all the current takes the meter’s path, the element it was supposed to be measuring gets essentially none, and the current is limited only by whatever resistance is left in the loop. Here that is the battery’s internal $0.50 \ \Omega$ , giving $18 \ \mathrm{A}$ , sixty times the correct reading. With an ideal battery, which is the Topic 11.5 default, there would be nothing left at all and the model predicts an unbounded current. That is why this mistake damages equipment rather than just producing a wrong number.

A voltmeter in series stops the circuit. Its infinite resistance is now in the only path, so the current falls to zero. Nothing lights, nothing warms, the correctly placed ammeter reads zero, and the voltmeter itself reads the full $9.0 \ \mathrm{V}$ : with no current, the terminal relation $\Delta V_{\mathrm{terminal}} = \mathcal{E} - Ir$ gives the whole emf, the resistor drops none of it, and all of it appears across the gap the meter occupies. A reading of essentially the full source voltage in a circuit where nothing is working is the signature of this mistake.

Notice the diagnostic value of the pair. One error gives a wildly large ammeter reading and zero volts across the element; the other gives zero amps and the full source voltage. Neither looks like a small measurement error, which is the one merciful thing about both of them.

Where you put it decides what it reads

For an ammeter, position matters only in the sense of which branch. For a voltmeter, position matters in the sense of which two points. That asymmetry is the whole of circuit meter reading, and it falls straight out of the series and parallel rules in 11.5.A.1.i and 11.5.A.1.ii.

In a series loop, 11.5.A.1.i says any charge passing through one circuit element must proceed through all elements in that connection and has no other path available, and the current in each element in series must be the same. So one ammeter reading serves the whole loop. Meanwhile the potential differences divide up, so a voltmeter has a different reading for every element and yet another for the pair of them together.

In a parallel section, 11.5.A.1.ii says charges may flow through one of two or more paths, and across each path the potential difference is the same. So one voltmeter reading serves every branch. Meanwhile the current splits, so an ammeter reads something different in each branch and something different again in the main line feeding them.

The practical rule that comes out of this:

To place an ammeter, find the branch. Ask which current the question wants, then break that branch and insert the meter there. An ammeter in the main line does not tell you a branch current.
To place a voltmeter, find the two nodes. Ask which pair of points the question wants, then touch the probes there. A voltmeter across two elements reads the sum of their potential differences, not either one.
An ideal voltmeter across an ideal wire reads zero. By 11.5.B.1 ideal wires have negligible resistance, so $\Delta V = IR$ gives essentially nothing across them. This is worth knowing because it tells you that two points joined by ideal wire are electrically the same point.

The series vs parallel circuits guide carries the reduction procedure for working out the currents and potential differences in the first place, and voltage vs current covers the across-and-through distinction the meters are built around.

Ideal and nonideal, and how far the CED asks you to go

The two ideal resistances are stated as design goals with reasons attached, and the reasons are the useful part.

An ideal ammeter has zero resistance so that it does not affect the current in the element that it is in series with (11.5.C.1.ii). Adding resistance in series would reduce the current, so the meter would read less than the current that flowed before it arrived.
An ideal voltmeter has infinite resistance so that no charge flows through them (11.5.C.2.ii). Offering a finite parallel path would draw some current, which reduces the potential difference across the element, so the meter would read less than the potential difference that existed before it arrived.

In both cases the nonideal meter reads low, and in both cases the reason is that the meter has become part of the circuit rather than an observer of it. That single sentence is the qualitative answer 11.5.C.3 and the boundary statement are asking for.

What the boundary statement then does is stop you there. AP Physics 2 only expects a qualitative discussion of how a nonideal meter affects the results, so the expected answer names the direction of the error and the mechanism, not a corrected value. A well-formed exam answer looks like: the voltmeter's finite resistance provides a parallel path, so it draws current, which reduces the equivalent resistance of that section and therefore the potential difference across it, so the reading is lower than the value with no meter attached.

One useful consequence of the ideal assumptions is that ideal meters are invisible in the energy accounting. An ideal ammeter has no potential difference across it, so by the sheet's power relation $P = I \Delta V$ it dissipates nothing. An ideal voltmeter has no current through it, so it also dissipates nothing. Neither meter takes energy out of the circuit, which is what makes them legitimate instruments rather than extra components.

When it costs a mark

Putting the ammeter in parallel or the voltmeter in series. The two consequences are described above, and both are asked about directly. On a diagram question, an A in a circle drawn across a resistor is worth no credit even if the arithmetic that follows is correct.

Giving the ammeter a resistance in a problem that assumes ideal meters. The Topic 11.5 boundary statement says that unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Adding an ammeter resistance you invented inflates the equivalent resistance and lowers every current in the circuit.

Computing a numerical correction for a nonideal meter. AP Physics 2 asks only for a qualitative discussion. A number there is out of scope, and the time spent producing it is time not spent on the explanation the rubric wants.

Reading an ammeter in the main line as a branch current. In a parallel section the main line carries the sum of the branch currents. If a question asks for the current in the $120 \ \Omega$ branch, the meter has to be in that branch.

Reading a voltmeter across two elements as one element's potential difference. Probes spanning a series pair read the sum. Move one probe to the node between them to get either element on its own.

Saying an ideal voltmeter reads zero because no current flows through it. No current flows through it and its reading is the full potential difference across the element. Confusing the current through the meter with the quantity the meter reports is a common wrong justification attached to a right answer, and justification questions mark the reasoning.

Assuming the ammeter reading changes along a series loop. It does not. 11.5.A.1.i makes the current in each series element the same, so an ideal ammeter reads the same wherever you break that loop to insert it.

When the two readings coincide, and why that lulls you

The two meters are never measuring the same physical quantity, so their readings only coincide numerically, and there is exactly one circumstance in which that happens by design.

When the resistance is $1 \ \Omega$ . Then $I = \Delta V / R$ returns the same digits for both, and an ammeter reading $9.0 \ \mathrm{A}$ sits next to a voltmeter reading $9.0 \ \mathrm{V}$ . It is a coincidence of units and nothing more, but it makes a worked example look as though the two instruments are interchangeable. Change the resistor and the coincidence evaporates.

Two more situations blur the distinction without making the readings equal.

A single resistor on a single battery. There is one current and one potential difference, and either measurement plus Ohm's law gives you the other. Nothing in that circuit forces you to think about where a meter goes, because there is only one place to put each of them. This is the circuit most people learn on, which is why the placement rules feel arbitrary until a second element appears.

A circuit with the meters already drawn in. When a schematic hands you both symbols in position, the placement decision has been made for you, and the question reduces to arithmetic. That is a comfortable exercise and a poor test of whether you understand the pair, which is why exam questions so often ask you to add a meter to a diagram instead.

The distinction turns on the moment there are two elements and you have to choose where a meter goes. Once a circuit has a series pair or a parallel branch, every question about it forces you to say which quantity you want and therefore which meter and which position. Everything before that is the special case.

Where this sits on the AP exam

The meters are objective 11.5.C in Topic 11.5, inside Unit 11, Electric Circuits, which the CED weights at 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods. The suggested skills the CED lists for Topic 11.5 are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Skill 1.A is the reason meter questions so often arrive as a schematic to complete rather than a number to compute: the CED wants you to draw the meter in the right place. Skill 2.C is the reason the three-column misconnection comparison in this page is worth practising, since comparing scenarios is precisely what a misplaced-meter question asks.

The unit's exam guidance is also unusually direct about vocabulary. The CED notes that students can inadvertently miscommunicate their answers by using words incorrectly, and that they should know the differences in meaning between current, potential difference, resistance, resistivity, and capacitance. Writing "the voltage through the resistor" or "the current across the resistor" is the version of that error this pair produces, and it is the version a grader notices first.

For the quantities themselves, see voltage vs current. For the difference between the object property the meters are usually probing and the material property behind it, see resistance vs resistivity. The Ohm's law guide covers the arithmetic that turns one meter reading into the other, and the Ohm's law calculator does it numerically.

Correct wiring, then each misconnection on its own

A battery with emf $\mathcal{E} = 9.0 \ \mathrm{V}$ and internal resistance $r = 0.50 \ \Omega$ is connected by ideal wires to a $30 \ \Omega$ resistor, with an ideal ammeter in series and an ideal voltmeter across the resistor. (a) Find both readings. (b) The ammeter alone is moved so that it sits in parallel across the resistor, the voltmeter staying where it was. Find the loop current and both readings. (c) Starting again from the correct circuit, the voltmeter alone is moved into the loop in series, the ammeter staying in series too. Find the loop current and both readings. (d) Say which error is more damaging and why.

(a) The ideal ammeter has zero resistance and the ideal wires have negligible resistance (11.5.B.1), so the total resistance the battery drives is $30 + 0.50 = 30.5 \ \Omega$ .
Current: $I = \mathcal{E} / (R + r) = 9.0 / 30.5 = 0.295 \ \mathrm{A}$ , which is $0.30 \ \mathrm{A}$ to two significant figures. The ammeter reads this, and it would read the same wherever in the loop it were placed, by 11.5.A.1.i.
The voltmeter is across the resistor, so it reads the resistor's own potential difference: $\Delta V = IR = (0.295)(30) = 8.85 \ \mathrm{V}$ , or $8.9 \ \mathrm{V}$ to two significant figures.
Check with the CED's terminal relation $\Delta V_{\mathrm{terminal}} = \mathcal{E} - Ir$ from 11.5.B.3: $9.0 - (0.295)(0.50) = 9.0 - 0.148 = 8.85 \ \mathrm{V}$ . The resistor is the only element outside the battery, so its potential difference must equal the terminal potential difference, and the two routes agree.
(b) The ammeter's zero resistance now sits directly across the $30 \ \Omega$ resistor, which is a short circuit by 11.2.A.2.iii: charges can flow from one side to the other with no change in potential difference. Two resistances in parallel, one of them zero, give an equivalent resistance of zero, so the resistor is bypassed.
The only resistance left in the loop is the battery's internal $0.50 \ \Omega$ . Current: $I = \mathcal{E} / r = 9.0 / 0.50 = 18 \ \mathrm{A}$ .
Readings: the ammeter carries essentially all of that, so it reads $18 \ \mathrm{A}$ , sixty times the $0.30 \ \mathrm{A}$ of part (a). The voltmeter is still across the resistor, which is now bridged by a zero-resistance path, so it reads $0 \ \mathrm{V}$ . Terminal check: $\Delta V_{\mathrm{terminal}} = 9.0 - (18)(0.50) = 9.0 - 9.0 = 0 \ \mathrm{V}$ , consistent.
(c) Now the voltmeter is in series in the loop, and an ideal voltmeter has infinite resistance (11.5.C.2.ii). An infinite resistance in the only path means no charge can complete the loop, so $I = 0$ .
Readings: the ammeter, still correctly in series, reads $0 \ \mathrm{A}$ . With $I = 0$ the terminal relation gives $\Delta V_{\mathrm{terminal}} = 9.0 - (0)(0.50) = 9.0 \ \mathrm{V}$ , and the resistor drops $IR = 0$ , so the loop rule puts the whole $9.0 \ \mathrm{V}$ across the voltmeter. It reads $9.0 \ \mathrm{V}$ , which looks like a perfectly sensible measurement of the battery and is not a measurement of the resistor at all.
(d) The ammeter-in-parallel error is more damaging. It raised the current from $0.30$ to $18 \ \mathrm{A}$ , and with the battery treated as ideal, which the Topic 11.5 boundary statement assumes by default, $r$ would be zero and the model would predict an unbounded current. The voltmeter-in-series error merely stops the circuit: nothing is harmed, and its signature is a near-full-voltage reading in a circuit where nothing works.

Correct: ammeter $0.30 \ \mathrm{A}$ , voltmeter $8.9 \ \mathrm{V}$ . Ammeter moved into parallel: the loop current jumps to $18 \ \mathrm{A}$ , the ammeter reads that short-circuit current and the voltmeter reads $0 \ \mathrm{V}$ . Voltmeter moved into series: the current falls to zero, the ammeter reads $0 \ \mathrm{A}$ and the voltmeter reads the full $9.0 \ \mathrm{V}$ . Two errors, two opposite symptoms.

One series loop, one ammeter reading, three voltmeter readings

A $40 \ \Omega$ resistor and an $80 \ \Omega$ resistor are connected in series across an ideal $12 \ \mathrm{V}$ battery with ideal wires and ideal meters. (a) Find the reading of an ideal ammeter placed between the battery and the $40 \ \Omega$ resistor, and of a second ideal ammeter placed between the two resistors. (b) Find the reading of an ideal voltmeter placed across the $40 \ \Omega$ , across the $80 \ \Omega$ , across both together, and across one of the ideal wires. (c) State the general rule each part illustrates.

(a) In series the resistances add, so $R_{\mathrm{eq,s}} = 40 + 80 = 120 \ \Omega$ , using the sheet relation $R_{\mathrm{eq,s}} = \sum_i R_i$ .
$I = \Delta V / R_{\mathrm{eq}} = 12 / 120 = 0.10 \ \mathrm{A}$ . Both ammeters read $0.10 \ \mathrm{A}$ , because 11.5.A.1.i requires the current in each element in series to be the same and an ideal ammeter adds no resistance to change it.
(b) Across the $40 \ \Omega$ : $\Delta V = IR = (0.10)(40) = 4.0 \ \mathrm{V}$ .
Across the $80 \ \Omega$ : $\Delta V = (0.10)(80) = 8.0 \ \mathrm{V}$ .
Across both together: the probes now span a $120 \ \Omega$ combination carrying $0.10 \ \mathrm{A}$ , so $\Delta V = (0.10)(120) = 12 \ \mathrm{V}$ . Equivalently, $4.0 + 8.0 = 12 \ \mathrm{V}$ , which matches the battery and satisfies Kirchhoff's loop rule.
Across an ideal wire: ideal wires have negligible resistance (11.5.B.1), so $\Delta V = IR$ with $R \approx 0$ gives essentially $0 \ \mathrm{V}$ .
(c) The ammeter rule is that a single series loop has one current, so one reading serves the whole loop and moving the meter changes nothing. The voltmeter rule is that every pair of points has its own potential difference, so the reading depends entirely on where the probes sit, and probes spanning two elements report the sum rather than either part.

Both ammeters read $0.10 \ \mathrm{A}$ . The voltmeter reads $4.0 \ \mathrm{V}$ across the $40 \ \Omega$ , $8.0 \ \mathrm{V}$ across the $80 \ \Omega$ , $12 \ \mathrm{V}$ across both, and $0 \ \mathrm{V}$ across an ideal wire. One current, four different potential differences, in the same circuit.

Parallel branches: one voltmeter reading, three ammeter readings

A $40 \ \Omega$ resistor and a $120 \ \Omega$ resistor are connected in parallel across an ideal $12 \ \mathrm{V}$ battery. All wires and meters are ideal. (a) Find the reading of an ideal voltmeter across each resistor. (b) Find the reading of an ideal ammeter placed in the $40 \ \Omega$ branch, in the $120 \ \Omega$ branch, and in the main line next to the battery. (c) Check the branch currents against the equivalent resistance. (d) Say what an ammeter reading in the main line does not tell you.

(a) Both resistors span the same two nodes, so by 11.5.A.1.ii the potential difference across each path is the same, and both are connected straight across the ideal battery. The voltmeter reads $12 \ \mathrm{V}$ across each.
(b) Branch currents from $I = \Delta V / R$ : in the $40 \ \Omega$ branch, $I = 12/40 = 0.30 \ \mathrm{A}$ ; in the $120 \ \Omega$ branch, $I = 12/120 = 0.10 \ \mathrm{A}$ .
Main line: by Kirchhoff's junction rule the battery must supply the sum, $0.30 + 0.10 = 0.40 \ \mathrm{A}$ .
(c) Confirm with the sheet relation $1/R_{\mathrm{eq,p}} = \sum_i 1/R_i$ : $1/R_{\mathrm{eq}} = 1/40 + 1/120 = 3/120 + 1/120 = 4/120$ , so $R_{\mathrm{eq}} = 120/4 = 30 \ \Omega$ .
Then $I = \Delta V / R_{\mathrm{eq}} = 12/30 = 0.40 \ \mathrm{A}$ , matching the sum of the branch currents exactly. Note also that $30 \ \Omega$ is less than either individual resistance, as 11.5.A.2.iii requires: adding parallel paths increases the number of routes available to charges and decreases the equivalent resistance.
(d) The main-line reading of $0.40 \ \mathrm{A}$ tells you the total but neither branch. The smaller resistance takes the larger share, three times as much here, and no measurement in the main line can separate the two. To get a branch current the meter must be broken into that branch.

The voltmeter reads $12 \ \mathrm{V}$ across both resistors. The ammeter reads $0.30 \ \mathrm{A}$ in the $40 \ \Omega$ branch, $0.10 \ \mathrm{A}$ in the $120 \ \Omega$ branch and $0.40 \ \mathrm{A}$ in the main line, confirmed by $R_{\mathrm{eq}} = 30 \ \Omega$ . One potential difference, three currents, which is the exact reverse of the series case.

Frequently asked questions

What is the difference between an ammeter and a voltmeter?

An ammeter measures current at a specific point in a circuit and must be connected in series with the element whose current is being measured; an ideal one has zero resistance. A voltmeter measures electric potential difference between two points and must be connected in parallel with the element across which the potential difference is being measured; an ideal one has infinite resistance. Those are essential knowledge statements 11.5.C.1 and 11.5.C.2 in AP Physics 2. In short, an ammeter is spliced into the path and a voltmeter is touched across a gap.

Why is an ammeter connected in series and a voltmeter in parallel?

Because of what each quantity is. Current is the rate charge passes one point, so the meter has to sit in the path for the same charge to pass through it, which means in series. Potential difference is a comparison between two points, so the meter has to touch both points, which means in parallel with whatever lies between them. The ideal resistances follow from the same logic: an ammeter must not impede the flow it sits in, so zero resistance, and a voltmeter must not provide a new route for charge, so infinite resistance.

What happens if you connect an ammeter in parallel?

You short-circuit the element you were trying to measure. An ideal ammeter has zero resistance, so placing it across a resistor gives charge a route with no potential difference along it, which is exactly the CED's definition of a short circuit in 11.2.A.2.iii. Nearly all the current takes the meter's path, the element carries almost none, and the total current is limited only by whatever resistance remains in the loop. With an ideal battery the model predicts an unbounded current, which is why this mistake damages equipment rather than just giving a wrong number.

What happens if you connect a voltmeter in series?

The current stops. An ideal voltmeter has infinite resistance, and placing it in the only path means no charge can complete the loop. Nothing in the circuit operates, and the voltmeter itself reads close to the full source potential difference, because with no current everywhere else the entire potential difference appears across the gap the meter occupies. A meter reading nearly the full battery voltage in a circuit where nothing works is the signature of this error.

Why does an ideal ammeter have zero resistance and an ideal voltmeter infinite resistance?

So that neither one changes what it is measuring. Essential knowledge 11.5.C.1.ii says ideal ammeters have zero resistance so that they do not affect the current in the element they are in series with, and 11.5.C.2.ii says ideal voltmeters have infinite resistance so that no charge flows through them. A real ammeter adds a little series resistance, which lowers the current, and a real voltmeter offers a finite parallel path, which lowers the potential difference. Both nonideal meters therefore read low, and essential knowledge 11.5.C.3 states that nonideal meters change the properties of the circuit being measured.

Does an ammeter read the same everywhere in a series circuit?

Yes. Essential knowledge 11.5.A.1.i states that in a series connection any charge passing through one element must proceed through all elements in that connection, and that the current in each element in series must be the same. An ideal ammeter adds no resistance, so inserting it anywhere in that loop gives the same reading. A voltmeter is the opposite: its reading changes with every pair of points, because each pair has its own potential difference.

How much does AP Physics 2 expect about nonideal meters?

A qualitative discussion only. The boundary statement on Topic 11.5 says AP Physics 2 only expects students to qualitatively discuss how a nonideal ammeter or voltmeter will affect the results of measurements, that unless otherwise stated all batteries, wires, and meters are assumed to be ideal, and that circuits with batteries of different potential differences connected in parallel will not be assessed. So name the direction of the error and the mechanism, and do not compute a correction. The AP Physics C: Electricity and Magnetism boundary statement drops the qualitative-only clause, so the calculus course can ask for more.