# Resistance in internal battery, voltage source

I was reading the post Which dissipates more power, a small or big resistor? and DanielSank's reply said

Now, any circuit you would reasonably call a "voltage source" must have a low internal resistance compared to typical load resistances. If it didn't then the voltage across the load would depend on the load resistance, which would mean your source isn't doing a good job of being a fixed voltage source.

If we're connecting a load to the battery, this would be in series.

Ri = 1 Ohms Fixed voltage source at 9V I = V/Ri = 9A

If we add R load in the mix, it is series so the current is the same in the circuit. How are we going to get the voltage to be fixed other than R load to be identical or similar values to Ri? Wouldn't the voltage drop between two resistance add up to the total voltage?

• @DanielSank, this is for you. – Alfred Centauri May 25 '17 at 3:05
• I fixed up this post and removed the second question which was directed at Daniel (BTW @AlfredCentauri you can only ping people who have already commented on the post, for the most part). Ace8888, you're welcome to post that as a separate question, but it's best if you don't ask a specific person to answer it - just post the question, and whoever knows how to answer can do so. – David Z May 25 '17 at 3:25
• @davidz , I've to reword my question, please update the title. – Ace8888 May 25 '17 at 3:38
• I'd encourage you to change the title yourself as part of your edit. Unless you can't think of a good title. – David Z May 25 '17 at 3:39
• @davidz, i just did. Sorry I'm still learning how to navigate through the site :) – Ace8888 May 25 '17 at 3:41

If we model the real battery as an ideal voltage source together with an internal resistance $R_i$, and we connect a load resistor $R$ across the battery, the current is $I=\frac{V}{R+R_i}$. This is a voltage divider: the voltage across the load is $V_{load} = IR = V\frac{R}{R+R_i}$.

A good voltage source has low internal resistance (low compared to the load resistor): $R \gg R_i$, in which case $V_{load} \approx V$; the voltage across the load is (approximately) constant and does not depend strongly on the load resistance.

But if the load resistor is comparable to the internal resistance, you have $V_{load} \approx \frac{V}{2}$; the voltage source is not a very good voltage source any more: the voltage across the load will depend strongly on the load resistance.

Here's a graph of what $V_{load}$ looks like for a battery voltage of $9V$ with an internal resistance of $1\Omega$. As you can see for $R \lt 10\Omega$ or so, the voltage across the load varies substantially, but as $R$ increases, $V_{load}$ approaches the voltage of the battery asymptotically:

• So when R(i) is low resistance compared to the R(L), the V(L) is roughly the same as V(i). I'm assumed that by the definition of voltage source meaning that whatever R(L) that is being used, we want the V(L) to be close to the voltage supply as much as possible to be consider a good "voltage source" – Ace8888 May 26 '17 at 3:03
• That is correct. – NickD May 31 '17 at 22:29

$R_{\rm i} = 1\,\Omega$
Fixed voltage source at $9 \rm V$
$I = \frac {V}{R_{\rm i}} = 9\,\rm A$

With a load resistor of $8\,\Omega$
The new current is $\frac {9} {8+1} = 1\,\rm A$ and the voltage across the load resistor is $8\times 1= 8\,\rm V$

With a load resistor of $89\,\Omega$
The new current is $\frac{9}{89+1}=0.1\,\rm A$ and the voltage across the load resistor is $89\times 0.1= 8.9\,\rm V$

With a load resistor of $899\,\Omega$
The new current is $\frac{9}{899+1}=0.01\,\rm A$ and the voltage across the load resistor is $899\times 0.01= 8.99\,\rm V$.

You will notice that as the load resistor gets larger and larger compared to the internal resistance of the power supply, the voltage across the load resistor gets closer and closer to the voltage of the power supply.