0
\$\begingroup\$

I'm very new to electronics and have been trying to learn the basics of electricity as a starting point. The mental model I've developed so far is the standard water/pipes analogy. I.e. voltage is like the pressure, amperage is flow of water, and ohms is the constricting or pinching of the pipe.

What I struggle with is how, if the circuit is open, the voltage would be unaffected by the resistors. As an example:

enter image description here

In this circuit if you measure the voltage between A-D, B-D, and C-D when the circuit is open like this, the voltage is always 5V. If connected however, the voltage becomes 5V, 2.5V, 0V, respectively.

Why are the open circuit voltages different? I understand that voltage represents a potential difference, but when we measure the voltage between B-D and C-D, shouldn't the resistance restrict some of that potential? It's hard for me to fit this into my mental model of water and pipes.

I know that Ohm's law requires current, and maybe that has something to do with it? But how can 5V be measured with the multimeter between C-D without the resistors affecting the potential during the measurement?

\$\endgroup\$
1
  • \$\begingroup\$ The water analogy isn't all that great, but even so, your understanding of it is wrong. No matter the size of the restriction in the pipe, if there's no water flowing hen the pressure will be the same on both sides. For as long as there's a pressure difference, water will flow to equalize the pressure. Remember, the open end of a wire is not like the open end of a pipe - electrons don't run out into the air (much). An open wire is like a closed-off pipe. \$\endgroup\$
    – brhans
    Mar 31, 2022 at 20:56

3 Answers 3

5
\$\begingroup\$

Why are the open circuit voltages different? I understand that voltage represents a potential difference, but when we measure the voltage between B-D and C-D, shouldn't the resistance restrict some of that potential? It's hard for me to fit this into my mental model of water and pipes.

The voltage drop through a resistor is given by Ohm's Law: \$V = IR\$. If \$I\$ is zero, then there is no voltage drop. Because no current can flow in an open circuit, \$I\$ is zero.

I know that Ohm's law requires current, and maybe that has something to do with it? But how can 5V be measured with the multimeter between C-D without the resistors affecting the potential during the measurement?

A voltmeter (or multimeter set to voltage mode) is specifically designed to measure a voltage while drawing as little current as possible. We can say that its input impedance is very large, on the order of megaohms.

You may solve the following circuit and convince yourself that the voltage across C-D is approximately 5 V:

enter image description here

(the 1 megaohm resistor represents your voltmeter's leakage, and is an under-estimate; in reality a good multimeter will be much closer to an open circuit)

\$\endgroup\$
1
\$\begingroup\$

Why are the open circuit voltages different?

They are not different. They are just on the extreme end of the spectrum. So let's work your way to that extreme end from the end you are familiar with using limits:

Suppose the voltmeter was constructed so that the internal resistance across its terminals was 10Ohms. If you connected it to the circuit to try and measure the voltage source then \$I=\frac{V}{R}=\frac{5}{10\Omega+10\Omega+10\Omega}=167mA\$ would flow. That would make a voltage drop of \$V=IR=(167mA)(10\Omega+10\Omega)=3.34V\$ lost. So \$5V-3.34V = 1.66V\$ would appear at the voltmeter terminals and that is what would be measured. So your error in attempting to measure the source voltage is rather large.

Now increase the volmeter internal resistance (which is formally called the input resistance) to 1000 Ohms, then 1MOhm, then 10OHms, all the way to infinite.

Every time you increase the voltmeter input resistance, the current decreases which means the voltage drop across the resistors decreases which means more of the source voltage appears at the voltmeter terminals to be measured.

The conclusion is that at a theoretical infinite Ohms, no current will flows so no voltage is dropped across the resistors which means all the source voltage appears at the voltmeter inputs to be measured. Real voltmeters strive to achieve this by having an input resistance as high as reasonably possible.

\$\endgroup\$
0
\$\begingroup\$

A digital multimeter (DVM) does have a finite impedance, and when you measure voltage between C and D, Current does flow through the meter. However, consider the impedance of a modern electronic multimeter, and compare it to that of two 10Ω resistors.

Input impedance for a DVM may be as low as 10MΩ to over 100MΩ. Use Ohm's law to calculate the voltage drop across 20Ω in series with 10MΩ with a 5VDC supply. That amount is finite, but so small that it is beyond the precision of all but the best laboratory meters to measure.

Analog multimeters using the D’Arsonval movement are devices that require higher current, and it might be possible to note the very slight drop across the circuit with resistors vs. with them shorted, if the meter required a few mA full scale... or perhaps not.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.