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Electromotive force is equal to internal resistance times current plus resistance of other components in the circuit times current.

$$\varepsilon = rI + RI$$ Therefore: $$V= \varepsilon-rI$$

What I'm really confused about is this graph: enter image description here

This graph relates to battery being used up. This means decrease of voltage (why?). According to the graph as voltage decreases, current increases. The only way I can explain it using the equation V=e-rI is that for some reason internal resistance r increases and as eloctromotive force stays the same, this means decrease in voltage V so both sides equal each other again. But wait! According to the graph current was supposed to increase and the above reasoning doesn't account for that. Please explain in your own words step by step how you can explain this graph using the equation V=e-rI.

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    \$\begingroup\$ You have exchanged cause and effect. The voltage decreases because of the increasing current through the internal resistance. \$\endgroup\$ – Peter Bennett Sep 16 '15 at 20:06
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    \$\begingroup\$ I was always taught that it's the VOLTAGE that CAUSES current to flow, not the other way around. So why current increases in the first place? \$\endgroup\$ – Richard Smith Sep 16 '15 at 20:14
  • \$\begingroup\$ the battery emf causes the current, not the terminal voltage. If you short-circuit the battery, the emf drives a large current through the internal resistance and the short-circuit, but the terminal voltage is zero. \$\endgroup\$ – Chu Sep 16 '15 at 20:39
  • \$\begingroup\$ If you increase the load on a battery (decrease load resistance, add more light bulbs in parallel...) the current delivered by the battery will increase, causing an increased voltage drop across the battery's internal resistance and reducing the voltage measured between the battery terminals. \$\endgroup\$ – Peter Bennett Sep 16 '15 at 20:49
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    \$\begingroup\$ This graph does not relate to the battery being used up. \$\endgroup\$ – immibis Sep 17 '15 at 2:06
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If you load up a motor with a physical force the motor starts to decrease speed as the torque from the physical load ramps up. Eventually the motor slows down so much it stalls and you are left with zero speed and maximum torque.

It's the same for a battery. Without a load it runs at full speed (open circuit voltage) and as you load it up the terminal voltage lowers as the current taken increases. Eventually, with a shorted out battery the current taken is at maximum but the terminal voltage is zero.

The internal resistance of the cell causes this to happen. If a cell didn't have internal resistance it could supply any amount of current without the terminal voltage falling (an impossibility of course).

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  • \$\begingroup\$ Oh, now I think that I misunderstood the whole thing when asking the question- I read this in a context of discharging batteries and I thought that what the graph shows is what's happening if you leave the battery until it discharges (in the paragraph above on the site was written that voltage decreases as battery discharges and I thought this graph illustrates that - but it turned out to describe what happens if you do affect the circuit e.g. what happens if you decrease resistance and thus increase current). \$\endgroup\$ – Richard Smith Sep 16 '15 at 21:18
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The graph that you have there it shows the LOAD line, the voltage at current equal zero is the open voltage current of the cell and the current at voltage equal zero is the short circuit current. So it shows all the possible values of the voltage seen at the output of the cell (V=EMF-rI) as a function of the current. So in other words the graphs you show the bounds of the battery. Now when you apply a load R, you lock that load line to an operating point. And you can even go ahead and plot V=IR for a particular R on top of the load line and the point where they crossed will be the operating point. Now remember, that a model for a battery is an ideal voltage source, internal resistance. when you start pulling current from the battery and complete the load there will be a voltage drop rI corresponding to the voltage drop due to the internal resistance this will cause the voltage of the cell to be lower than the voltage of the voltage source. Now lets look at the load line when you look at the open circuit volatage (current = 0) that means that there is current being drawn the the voltage source equals the voltage source. Now at the other end at the short circuit current the Voltage is zero, it means you have drawn out so much current that the voltage created by rI or (voltage drop due to the internal) resistance is equal to the electromotive force and a equilibrium is reached.

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The graph shows the increase in voltage drop as the current drawn from the battery increases.

V = e - IR

When current drawn I increases in the above equation voltage V will decrease if e is constant.

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  • \$\begingroup\$ Same question as to Peter Bennett - why would current increase? We can't control the amount of current because it is dependent on voltage and resistance, right? \$\endgroup\$ – Richard Smith Sep 16 '15 at 20:17
  • \$\begingroup\$ Yes, current depends upon resistance and voltage. You can increase the current by decreasing the resistance in the external circuit. Thus, the circuit will draw more current while the internal resistance is constant. \$\endgroup\$ – Murali Sep 16 '15 at 20:20

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