Circuits practical test · how to find the resistor under the tape

Name: Circuits practical test · how to find the resistor under the tape
Duration: 5 min 13 s
Channel: Thomas Schuurmans
Description: Build the circuit as in the diagram, smallest loop first. Measure a series of voltages and currents, and take the total resistance from the gradient of your (U,I) graph, here 132 ohms. Work back with one divided by R total equals one divided by R1 plus one divided by R2, and flip the result: 220 ohms.

Build the circuit as in the diagram, smallest loop first. Measure a series of voltages and currents, and take the total resistance from the gradient of your (U,I) graph, here 132 ohms. Work back with one divided by R total equals one divided by R1 plus one divided by R2, and flip the result: 220 ohms.

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At the practical test there is a source on your table, two multimeters, three resistors and a bunch of leads. One resistor is under tape, and you have to determine its value. In this video I go through the whole route once, from the loose leads to the resistor whose value you are not allowed to see.

Don't start with leads, start with the circuit diagram. Lay out your equipment the way it sits in the drawing, build the smallest loop first (source, ammeter and R1) and then add the second branch. The ammeter goes in the main current, before the current splits. The voltmeter comes last, alongside and not in between.

Before you switch the source on, choose the terminal and the range. R1 is 330 ohms, so at 6.0 volts you expect 18 milliamperes: that fits easily in the mA terminal, which goes up to 200 milliamperes.

Then the series of measurements. Read your own meter and write down what it says, with the zeros: 1.02 volts at 7.7 milliamperes, and so on up to six volts. In the (U,I) graph you draw the best straight line, and its gradient gives the total resistance: 5.94 volts divided by 45.0 milliamperes is 132 ohms.

That is where the step comes that almost everyone forgets. With one divided by R total equals one divided by R1 plus one divided by R2 you find 0.0045. That is not R2, that is one divided by R2. Flip it and you get 220 ohms.

Is that right? Rebuild into a series circuit with the 470 ohm resistor: 6.01 volts and 8.71 milliamperes give 690 ohms together, minus 470 is 220 ohms. Only then may the tape come off. Red, red, brown, gold: 220 ohms, plus or minus 5 percent.

For upper secondary physics, in electricity and the circuits practical.

In this video

The full explanation, in writing

What you can do after this video

0:00After this video you can use an ammeter and a voltmeter to determine an unknown resistance. It takes three steps. You build the circuit as in the circuit diagram. You measure a series of voltages and currents. And from your graph you work back to that one resistance.

What is on your table

0:19This is on your table at the practical test. A source, two meters, three resistors and a bunch of leads. And you have thirty minutes to take a series of readings. I'll do it once all the way through: from the loose leads to the resistor whose value you are not allowed to see.

Draw first, then lay out as in the drawing

0:38Don't start with leads. Start with the diagram, because that is your blueprint. Lay out your equipment the way it sits in the drawing. The source on the left. The ammeter at the top, in the main current, before the current splits. The two resistors one above the other, in parallel: both between the same two points.

0:58And the voltmeter above them, across the resistors.

The smallest loop first, then the second path

1:02Now build, and first the smallest loop: the source, the ammeter and R one. A red lead from the positive to the mA terminal of the yellow meter. From the COM terminal to the start of R one. And from the end of R one a black lead, along the bottom back to the negative.

1:21Follow the loop with your finger: you pass through the ammeter. Then it's right. Then add the second branch: R two. Start to start, end to end. Now the current splits after the ammeter, and comes back together before it returns.

Which terminal, which range

1:38Two things on the yellow meter: the terminal and the range. How much current can you expect? R one is three hundred and thirty ohms. At six point zero volts, six point zero divided by three hundred and thirty goes through it: zero point zero one eight amperes, eighteen milliamperes.

1:56That fits easily in the mA terminal, which goes up to two hundred milliamperes. Rotary switch to direct current, range two hundred milliamperes.

Last of all, and alongside

2:06Last of all, the voltmeter. It goes alongside, not in between. The V terminal on the positive side, COM on the negative side, both across the resistors. A good check: if you take the voltmeter away, nothing changes in the circuit. The current simply keeps flowing.

2:26Raise your hand to get it checked. Only then does the source go on.

Read it the way the meter says it

2:30Turn the knob up until the blue meter shows about one volt. Don't look at the power supply: your own meter is the measurement. One point zero two volts, seven point seven milliamperes. Write it down as it appears, with the zeros. Then two volts, three, and so on up to six.

U up, I sideways

2:51Now the data processing, which you do on your own. A U,I graph: U goes vertical, I horizontal. Choose a scale that fills your paper, and plot your measurements as crosses. With your set square, draw the best straight line, through the origin.

3:05Don't join the dots. For the gradient, pick two points on the line, far apart, not two measured points. Five point nine four volts divided by forty-five milliamperes. Convert those mA first: zero point zero four five zero amperes. That gives one hundred and thirty-two ohms.

3:26That is R total.

The step everyone forgets

3:27You know R one and you know R total. In parallel: one divided by R total equals one divided by R one plus one divided by R two. So one divided by R two is one divided by one hundred and thirty-two, minus one divided by three hundred and thirty.

3:44Zero point zero zero four five. And this is where many people stop: R two is zero point zero zero four five ohms. Right? A resistance of less than five thousandths of an ohm, on your table? You don't have R two yet, you have one divided by R two.

4:01Flip it: one divided by that number. Two hundred and twenty ohms.

Is it right? Rebuild and see

4:07Is that right? You can check it with a second circuit. Unplug every lead and lay out the equipment as in the new drawing: R two in series with R three, which is four hundred and seventy ohms. At six point zero one volts, eight point seven one milliamperes flow.

4:25Together that is six hundred and ninety ohms. Minus four hundred and seventy: two hundred and twenty. Only now, tape off. Red, red, brown, gold: two hundred and twenty ohms, plus or minus five percent. Somewhere between two hundred and nine and two hundred and thirty-one.

4:43Your two hundred and twenty lies in between.

Three things

4:46Three things. The ammeter goes in the circuit, the voltmeter alongside. Read your own meter, with the zeros. And after one divided by R two you're only halfway: flip it.

Draw first, then build

4:59Want to know more about this? Then watch the videos on series and parallel on this channel. And at the practical test: first draw, then build.

Frequently asked questions

How do you determine an unknown resistance at a practical test?

Measure the current at a few voltages, plot the points in a (U,I) graph and draw the best straight line. The gradient is the resistance. If the unknown resistor is in parallel with a known one, that gradient is the total resistance, and you then work back to the unknown one.

Where do you connect the ammeter and the voltmeter?

The ammeter in the circuit, in the main current before the split, in the mA terminal and the COM terminal. The voltmeter last, alongside across the resistors: the V terminal on the positive side, COM on the negative side. If you take the voltmeter away, the current simply keeps flowing.

Which range do you choose on the ammeter?

Estimate the current first. At 330 ohms and 6.0 volts that is 6.0 divided by 330, about 18 milliamperes. That fits easily in the mA terminal with the 200 milliampere range.

Why do I get a resistance of 0.0045 ohms?

Because you are not finished yet. With one divided by R total minus one divided by R1 you calculate one divided by R2, not R2 itself. Flip that result: one divided by 0.004545 is 220 ohms.

How do you check whether your answer is right?

Build a second circuit: the unknown resistor in series with a known one of 470 ohms. At 6.01 volts and 8.71 milliamperes the total resistance is 690 ohms, minus 470 is 220 ohms. Then the tape may come off: red, red, brown, gold is 220 ohms plus or minus 5 percent, so between 209 and 231 ohms.

Want to see the theory behind it again: the videos on the series circuit, the parallel circuit and combined circuits are on this channel.

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Physics with Thomas: physics explained with pictures where they add something, and without them where they distract.

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About me

My name is Thomas Schuurmans, and I have a passion for physics and for understanding why things work the way they do. What I want is to get young people asking “why is that?” and “how does that work?” more often, and to hand them the tools of physics in a way that is simple and that you can see.

I graduated in applied physics at Delft University of Technology in 2003. After that I spent more than twenty years outside education: first at TNO, the Dutch applied research institute, then at a design agency, and eventually founding Proportion Global, through which I work on innovation questions in Africa, Latin America and South Asia. That work resembles physics more than you would expect: don't start from a solution, first understand what is going on, try something, be wrong, and look again.

Since 2026 I have been teaching physics to both lower and upper secondary classes, and I started a master's at the University of Amsterdam for my full teaching qualification. That is where I learned that a secondary school pupil's real attention for new material lasts about seven minutes. And my own weakness happens to be telling too many side stories.

That is where the idea came from: videos that explain one topic sharply and visually, inside those seven minutes. I make them for my own pupils. Then I publish them, because good explanation should be within reach of anyone who needs it, wherever you live and whatever language you think and speak in.