Simple multimeter test

Question

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.

Answer 1

^{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.

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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.

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^{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.

Answer 2

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.

Answer 3

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.