# Simple multimeter test

I have a very simple issue. Maybe it is because I am very new to electronics.

Say I have 9 V DC source and a 1 kΩ resistor. Negative is connected to one end of resistor and positive to another end. Using a multimeter between two ends of the resistor shows me the 9 V which to my understanding is correct. Actually we have about 7 V drop after the resistor but the multimeter is showing a voltage difference between positive and negative and not the voltage drop of the resistor.

1. How can I use a multimeter in this very basic case to see the voltage after the resistor?
2. Don't we have about 2 V after this resistor actually? Shouldn't the multimeter show the voltage difference of about 7 V? Added a simple circuit

Added yet another picture to clarify my question.

• What are you trying to measure? Why are you expecting to see 7 volts somewhere if it's a 9V battery? Commented Apr 15, 2023 at 8:58
• Draw a diagram of what you have connected. Either your description has left something out, or you have badly misunderstood something.
– JRE
Commented Apr 15, 2023 at 9:13
– Amir
Commented Apr 15, 2023 at 9:37
• You keep on saying 'after the resistor', which is meaningless. Your diagram shows the voltage across the resistor, and across the battery. This is how voltage should be measured. If you mean to measure the current flowing through the resistor, then break the circuit, and place the meter, on a mA range, in series with the resistor and battery. Commented Apr 15, 2023 at 9:43

simulate this circuit – Schematic created using CircuitLab

Figure 1. The circuit schematic.

Redrawing your cartoon as a circuit schematic makes it more obvious that the voltmeter is measuring the voltage across the battery / resistor parallel combination. Ignoring the very small wiring resistance, all points on the top rail will be at the same +9 V with respect to the battery negative. All points on the bottom rail will be 0 V. The difference between the rails will be 9 - 0 = 9 V.

The CircuitLab simulator defaults to a 2 Ω internal resistance for a 9 V battery so when you hook up a 1 kΩ resistor and draw 9 mA from the battery there will be a voltage drop at the battery terminals. You can calculate this using Ohm's law, $$\ V_{drop}=IR_{int} = 9m \times 2 = 18 \text {mV} \$$.

How can I use a multimeter in this very basic case to see the voltage after the resistor?

Everything is in parallel in your circuit. There is no "after the resistor". Current will flow from the battery and split between R1 and the meter in inverse proportion to their resistances. (i.e., most of the current will flow through R1.)

Don't we have about 2 V after this resistor actually? Shouldn't the multimeter show the voltage difference of about 7 V?

You give no clue about how you came to the number of 7 V. Hopefully you can now see that this is not the case.

Update after question edited.

simulate this circuit

Figure 2. (a) shows the cartoon redrawn as a schematic.

I encourage you to learn to draw proper shcematics that read from left to right with higher voltages at the top and ground or negative at the bottom. You can then trace current flow and voltage drop from top to bottom much more easily.

With Figure 1a it's easy to see that the voltmeter is measuring the voltage drop across the LED. If you want to measure the voltage drop across the resistor then wire it as shown in 1b.

simulate this circuit

Figure 3. How to measure the voltage "before and after the resistor".

• In Figure 3a VM1 is measuring the voltage at the top of R1 with respect to ground (the battery negative).
• VM2 is measuring the voltage "after" R1 with respect to ground. Since this point is actually connected to ground the voltages on both meter terminals are the same and so it reads 0 V.
• Because you were measuring the battery voltage. Read my answer again and have a look at the update. Commented Apr 15, 2023 at 11:43
• No, it's because you now have two components in series and there is a voltage drop across each. In Figure 1 there is 9 mA flowing through R1. In Figure 2 ther will be about 7 mA running through both the resistor and the LED. Commented Apr 15, 2023 at 12:03
• No! Figure 2b shows you how to measure the LED voltage drop. Voltage drop is the difference in voltage between two points on a circuit. The voltage drop across the LED is the voltage between its two terminals, the anode (+) and the cathode (-). Commented Apr 15, 2023 at 12:43
• Figure 1 is measuring the voltage drop across the resistor and it's (almost) 9 V. Figure 2b is measuring the voltage drop across the resistor and it's (almost) > 7 V. The voltmeter measures the difference in voltage between two points. I'll post another edit. Commented Apr 15, 2023 at 13:00
• @Amir The voltage drop across the resistor is a function of the entire circuit, not just the resistor. With the LED there, it changes the current through the resistor such that the voltage drop is different from when the LED isn't there. Commented Apr 15, 2023 at 15:10

Here's the test schematic which also includes the internal resistance of the battery.

Considering 9 V nominal open-circuit battery voltage and 1.7 Ω nominal internal impedance,

Load current 'I' = 9 / 1001.7 = 0.00898 A and

Load voltage 'V' = 0.00898 x 1000 = 8.98 V.

If you had a BA-Separator and 2 voltmeters you would get the wanted readings:

A single voltmeter alone cannot make the separation, but shows only the total voltage. Of course, if you have only one meter you can connect it in turns to the separator outputs.

• What is a "BA-Separator"? I've never heard of it in 50 years of electronics. I'm guessing that this is a spoof answer and that BA stands for "before-after". Commented Apr 15, 2023 at 10:45
• before/after what? Lol. There's only one single resistor in here. There's no before and after Commented Apr 15, 2023 at 21:24