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For a test on circuits that I just completed, a question asked about a voltage divider circuit (see below.)

However, we were told to assume that the load had no resistance and was parallel with the second resistor.

Wouldn't this mean that all of the current would be diverted through the load, so that no current would run through the second resistor? And wouldn't that mean that there is no output voltage (V_R) since there isn't any resistance after R1?

The question seems to imply that there would be a nonzero voltage VR, however, and it seems nonsensical if it is equal to zero, but shouldn't we be assuming that the load has infinite resistance so that we are investigating an ideal voltage divider?

Voltage Divider Circuit

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    \$\begingroup\$ I can only assume poor wording and that they actually meant "no conductance" or "infinite resistance". The alternative doesn't make sense. \$\endgroup\$
    – Reinderien
    Jan 21 '21 at 0:34
  • \$\begingroup\$ This tutorial about voltage divider power source loading might help clarify things: learningaboutelectronics.com/Articles/Voltage-follower. Cheers. \$\endgroup\$
    – tlfong01
    Jan 21 '21 at 1:13
  • \$\begingroup\$ Sounds like very bad english. What does the rest of the text say beyond the paragraph you have provided? If the question for such a simple concept is that muddied then it's going to be problematic going deeper into more complex subject matter. \$\endgroup\$
    – mhaselup
    Jan 21 '21 at 4:21
  • \$\begingroup\$ Crummy question from the instructor. \$\endgroup\$
    – JRE
    Jan 21 '21 at 6:56
  • \$\begingroup\$ If in doubt and in a test, solve both scenarios. \$\endgroup\$
    – Andy aka
    Jan 21 '21 at 10:18
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Infinite resistance means no current, which means it can be ignored. Its a really good trick for analyzing circuits because it can make estimation much faster (and also a trick for opamp circuits).

If the load hand no resistance (0Ω) then all of the current would be diverted through the load and none through R2

A good exercise is to find the voltage for the no load (ideal) condition, then do it with a small load.

What one will find is a load pull down the divider, so a resistor divider doesn't do well with a load if voltage maintenance is the goal

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  • \$\begingroup\$ Yes, that’s true, but doesn’t saying the load has no current imply that the 2nd resistor is completely pointless? And then there is no division of voltage, which usually is the point of this type of circuit? \$\endgroup\$
    – COsborne
    Jan 21 '21 at 0:36
  • \$\begingroup\$ @COsborne the "only" point of this type of circuit \$\endgroup\$
    – mhaselup
    Jan 21 '21 at 4:22
  • \$\begingroup\$ @COsborne: The "LOAD" might be a very high input impedance multimeter, for example. \$\endgroup\$
    – Transistor
    Jan 21 '21 at 7:45
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If not already, one of the things you should become very familiar with is idea of the Thevenin equivalent for the resistor divider case. It really helps in better understanding a case like this one. (By this, I mean understanding it despite poor wording, should that ever happen to you.)

Your schematic can be converted to this one:

schematic

simulate this circuit – Schematic created using CircuitLab

Where \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$ and \$V_\text{TH}=V_{\text{B}_1}\cdot\frac{R_2}{R_1+R_2}\$.

Now you can easily imagine what happens throughout the range of \$0\:\Omega\le R_\text{LOAD}\le \infty\: \Omega\$. It will be easy to work out the current, etc.

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