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I'm new to electronics. So basically I already understand Ohm's law and the math behind why this is the case. What I don't get is the idea behind it.

Shouldn't the greater the load resistance be, the smaller the voltage? The resistance should 'resist' and thus lower the voltage. I hope I'm making sense.

Right here is a battery's open circuit voltage:

enter image description here

In this other scenario, I've placed a shunt of 2 ohms which should run parallel to the shunt present in the voltmeter. Which means, in my understanding, that the lower the value of your resistance, the lower the voltage drawn from the battery.

enter image description here

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  • \$\begingroup\$ If that resistance R1 is beteen V+ and some fixed load R2 to 0V, then the voltage across R1 increase with R1 or increase with R2 while the other voltage drop does the opposite both adding up to V+. So you need a reference to say which voltage is decreasing V(R2)=R2/(R1+R2) * V+ \$\endgroup\$
    – Tony Stewart EE75
    Commented May 2, 2022 at 4:46
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    \$\begingroup\$ Please show the circuit you are considering - what you are measuring where in the circuit is important. \$\endgroup\$
    – Peter Bennett
    Commented May 2, 2022 at 4:47
  • \$\begingroup\$ are you talking about greater resistance or about greater load? ... the two are inversely proportional \$\endgroup\$
    – jsotola
    Commented May 2, 2022 at 5:17
  • \$\begingroup\$ @PeterBennett Hi, so I edited the question and added a picture of the circuit. \$\endgroup\$
    – user312602
    Commented May 2, 2022 at 6:01
  • \$\begingroup\$ Consider this: "open circuit" means infinite resistance! Higher resistance actually means less "load," at least on an (approximate) voltage source like a battery. What you might be thinking about is a current source, which "works harder" (increases voltage) to overcome higher resistances to maintain current. \$\endgroup\$
    – TypeIA
    Commented May 2, 2022 at 6:12

4 Answers 4

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Your circuit is a voltage divider and behaves exactly as expected. Nothing special about one resistance being called "internal battery resistance". The circuit theory doesn't care very much about names, only about behaviors.

schematic

simulate this circuit – Schematic created using CircuitLab

R1 and R2||R3 form the voltage divider. Since R3>>R2, R3 can be ignored without introducing much error. Thus you have a voltage divider R1 and R2, whose output voltage is \$ V_1 \cdot R_2/(R_1+R_2) \$.

Shouldn't the greater the load resistance be, the smaller the voltage?

Your intuition is backwards. You should think: the greater the load conductance, the smaller the voltage, since the better the load conducts, the more it behaves like a short circuit, and a voltage across an ideal short circuit is zero.

Here, the short circuit - just like open circuit - is an abstract concept. Real short circuits never have zero resistance, unless they are superconductors that is. And even superconductors aren't ideal - they stop working when the current is too high or when they get too warm, and then "stops working" means that things go boom or that things get real expensive real quick.

Notation:

  • R2||R3 means "resistance of R2 in parallel with R3", numerically: \$ (1/R_2+1/R_3)^{-1} \$.
  • R3>>R2 means "R3 much larger than R2", typically \$ R_3 \ge 32R_2 \$, although some people are fine with a factor of 10 - it depends on how accurate your models needs to be.
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  • \$\begingroup\$ By the way, during the last electrical engineering lab, my students investigated, by the help of a simple self-made curve tracer, the IV curves of an open circuit ("nothing") and a short circuit ("a piece of wire"-:) Then, like magicians, they converted a real 1 Mohm voltmeter into an "ideal" voltmeter with "infinite" resistance ("nothing") and a real 100 ohm ammeter into an "ideal" ammeter with zero resistance ("a piece of wire"-:) \$\endgroup\$
    – Circuit fantasist
    Commented May 2, 2022 at 13:46
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    \$\begingroup\$ @Circuitfantasist That’s the nice thing about curve tracers :) \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented May 3, 2022 at 1:35
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If you had an ideal battery (ie without R1), then you would measure the same voltage whatever the load (resistors R4 and R5).

For a real battery, with an internal resistance, you can think of it as an ideal voltage source (that will always stay at 10V) and an internal resistor R1. The greater the current, the more voltage you loose in R1, therefore the voltage across the real battery decreases. And if you decrease the resistance of the load, then you increase the current (if electrons have less trouble traveling, more will do so). So the lower R4 and R5, the more current, and therefore the more voltage loss in R1, which means less apparent voltage on the battery.

Another way to see it is that you have 2 resistors in series (R1 and {R4, R5}). You split the voltage of the battery (10V) into the 2 resistors : the bigger resistor in proportion gets the biggest share. So if you decrease {R4, R5} then you have less voltage on them (and more in R1, in such way that the sum remains 10V)

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Because the resistance, should 'resist' and thus lower the voltage... Which means, in my understanding, that the lower the value of your resistance, the lower the voltage drawn from the battery.

Your assumption is true if the resistance is connected in series to the load and you are interested in the voltage across the load.

The voltage across the resistor is a loss (drop) that is subtracted from the input voltage; the rest is a useful voltage across the load. From this point of view, the loss (voltage drop) increases when the resistance increases; the gain decreases and this corresponds to your view. The network of the two elements in series forms a "voltage divider" that can be seen everywhere in this world.

In your case, the resistance is connected in parallel to the load and the relation is the opposite. Now the voltage across the resistor is useful since it is the load voltage. The network of the two elements in parallel forms a "current divider".

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Shouldn't the greater the load resistance be, the smaller the voltage? Because the resistance, should 'resist' and thus lower the voltage.

enter image description here

Josh pushing the car. Image source: Omegaman on Flickr. Creative Commons licence Attribution-NonCommercial-ShareAlike 2.0 Generic (CC BY-NC-SA 2.0).

Voltage and pressure or force are somewhat analogous. In the photo we can see that Josh is leaning at quite an angle to apply force to the car to get it to move. The more resistance the more he can lean. The higher the load, the higher the rolling resistance and the more pressure can be applied.

Similarly in an electrical circuit the higher the resistance the higher the voltage that must be applied to achieve a certain current.

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  • \$\begingroup\$ Now think about how hard you can push a ping-pong ball. You can never get to push very hard because when you push a little bit, the ball moves out of the way. \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented May 2, 2022 at 8:50
  • \$\begingroup\$ These examples illustrate the "inertia" property and are more suitable for explaining coils than resistors. \$\endgroup\$
    – Circuit fantasist
    Commented May 2, 2022 at 9:41