2
\$\begingroup\$

This is kind of a silly question I think. I'm about 95% sure I know the answer, but I'm going to ask it as if I don't have a clue. I just want confirmation that what I think is true is indeed true.

Consider the following voltage divider:

schematic

simulate this circuit – Schematic created using CircuitLab

When both switches are closed, the load resistor, R3 will see about 24v.

What voltage will R3 see when SW1 is open and SW2 is closed?

What voltage will R3 see when SW2 is open and SW1 is closed?

Again, I believe I know the answers to these two questions, but I just want to confirm.

What I think happens:

If switch 1 is open, R3 will see 0V.

If switch 2 is open though, I'm not entirely sure whether R3 will see 0 or 240V.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ What is the reference point? \$\endgroup\$ Jul 17, 2022 at 20:22
  • 1
    \$\begingroup\$ With either switch open there can be no current flow and therefore with either switch open there can be no voltage drop across any of the resistors. \$\endgroup\$
    – user173271
    Jul 17, 2022 at 20:31
  • \$\begingroup\$ It will indeed be 24v. This is a simple voltage divider, and 1800 ohms is 9x 200 ohms, which yields a 10% output. Or rather, it will be 23.9.... volts, but that's a rounding error. \$\endgroup\$ Jul 17, 2022 at 20:47
  • 3
    \$\begingroup\$ @All - I've deleted comments which seem to either be misunderstandings, meta, or are now otherwise obsolete. Some other comments may have been deleted due to referring to another now-deleted comment. If your comment was deleted, but you believe you still have a valid, current point which is allowed by the commenting policy and fully complies with the Code of Conduct, then you can consider posting it again. It is OK to ask for more details, but you must be nice when asking or responding. Thanks. \$\endgroup\$
    – SamGibson
    Jul 17, 2022 at 21:19
  • 1
    \$\begingroup\$ Interesting anyway. Just a problem of "safety". If the negative wire of the 240 V is grounded ... I would not touch any resistors when SW1 is connected to the positive wire of the 240V power supply ... \$\endgroup\$
    – Antonio51
    Jul 17, 2022 at 21:21

2 Answers 2

3
\$\begingroup\$

If switch 2 is open though, I'm not entirely sure whether R3 will see 0 or 240V.

There will be 0V across R3 terminals but there will be 240V between either resistor terminal and the 0V battery terminal.

\$\endgroup\$
8
  • \$\begingroup\$ So, in other words, if a voltmeter were placed across R3, it would read 0v when either switch is open, yes? \$\endgroup\$ Jul 17, 2022 at 21:40
  • 1
    \$\begingroup\$ Exactly. All the nodes in the circuit (the resistors) would be at 240V (in relation to the battery negative terminal) when SW2 is open and SW1 is closed. Since all resistor terminals are at the same potential, you would measure 0V across any of them. \$\endgroup\$
    – devnull
    Jul 17, 2022 at 21:41
  • 1
    \$\begingroup\$ Very good. The voltage divider will be used to feed voltage monitoring relays that will trigger stuff elsewhere in the system at various battery voltage levels. Sw2 will actually be a n-channel mosfet. I just needed to make sure that opening the circuit on the low side wouldn’t cause my voltage monitors to read 240v all of the sudden. If they read 0v we’re golden. \$\endgroup\$ Jul 17, 2022 at 21:55
  • \$\begingroup\$ I see. But make sure the voltage monitor can withstand the 240V common mode (if referenced to the same 0V as the input). \$\endgroup\$
    – devnull
    Jul 17, 2022 at 21:57
  • 1
    \$\begingroup\$ Yes, the 3 voltmeters shown in Transistor's answer show exactly what I mean. The problem is I don't know what the "voltage monitor" you mentioned is. If it is a two terminal (battery powered?) device like a normal voltmeter, which floats to the voltage being measured (like VM1), no problem. If it is some kind of circuit, like the one which controls the NMOS you mentioned, and is powered by (referenced to) another non-isolated power supply, the common mode voltage could, and probably will, destroy it. \$\endgroup\$
    – devnull
    Jul 18, 2022 at 10:16
3
\$\begingroup\$

You can check this quite easily in the simulator.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The simulator requires the circuit to be ground referenced so it's added to BAT1 negative and this is now your 0 V reference.

With SW1 closed and SW2 open VM1 shows no potential difference across R31 while VM2 and VM3 show that both ends of R3 are at 240 V above ground.

1 It actually shows about 48 μV due to the small measurement current drawn by VM3. From this we can estimate that the simulator is using a value of 240 / 48μ = 5 MΩ for VM3's internal resistance. This is reasonable.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks for that walkthrough, you’re absolutely right that I need to get myself better familiarized with the simulator. First time I tried it I ran into some likely user-error difficulties and haven’t dove back in since. \$\endgroup\$ Jul 17, 2022 at 21:57
  • \$\begingroup\$ Derek, the best simulator I know for learning is the Falstad Circuit Simulator. \$\endgroup\$
    – ee_student
    Jul 17, 2022 at 22:00
  • 1
    \$\begingroup\$ I suspect it's more likely to be the current through the open switch, than through the voltmeter. I've never seen a simulator that simulates the burden impedance of a voltmeter, but they usually do model leakage through open switches. \$\endgroup\$
    – Hearth
    Jul 18, 2022 at 14:51
  • 1
    \$\begingroup\$ @Hearth It is indeed the case with LTSpice, at least. But for this simulator, connecting another switch in parallel with SW2 did not affect the voltage indicated by VM1. Conversely, connecting another voltmeter in parallel with VM3, doubled the voltage indicated by VM1. \$\endgroup\$
    – devnull
    Jul 18, 2022 at 15:00

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.