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If a heavy current in drawn from a cell, a large number of charge carriers flow through the electrolyte sand hence more work is done. This results in more voltage drop, so terminal voltage decreases

This is what my 10th grade textbook says. But if there are more charge carriers, doesn't that also mean that the current in the external circuit would also increase, thus increasing the terminal voltage? So the voltage drop should not affect the terminal voltage?

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Although the current in the external circuit increases, it is increasing because the resistance is decreasing - so there is no unambiguous expectation that the voltage across the external circuit will actually increase.

In the simple model of a battery as a pure voltage source, whose voltage does not depend on current, in series with a pure resistance whose voltage is proportional to current, the total voltage across the terminals is the voltage of the source plus the voltage across the resistance. These voltages have opposite sign and so the total voltage across the terminals will decrease.

In effect, the internal resistance of the battery and the external resistance of the circuit it is connected to act as a voltage divider. And as the external resistance drops to zero, the voltage appears across the internal resistance.


What then becomes interesting is - why should the internal resistance be proportional to current. If one supposes that number of charge carriers is increasing, for example, then why do they not each experience the same resistance to motion, and so the voltage required to drive them be the same?

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If a heavy current is drawn from a cell, a large number of charge carriers flow through the electrolyte

You should rather say "a large number of charge carriers flow through the electrolyte per unit time" which is also known as current and its the same everywhere in the circuit.

It's good this way because simply "charge carries" doesn't mean much because that doesn't give information about the time interval.

If there are more number of charge carriers, doesn't that also mean that the current in the external circuit would also increase, thus increasing the terminal voltage?

Yes, it does but the current is the same for both external and internal circuit. But that doesn't necessarily mean the terminal voltage is also increased since other parameters like the load aren't constant.

Now, work done is given by $$P\times t=I^2Rt\tag{1}$$ so as you say if there is more current there will be more work done over time.

Now, the terminal potential difference is given by $$V(=IR)=\epsilon-Ir\tag{2}$$ $or$ $$\epsilon=I(r+R)\tag{3}$$

where $\epsilon$ is the open cell voltage and $r$ is the internal resistance, because as we know the electrolyte used in batteries offers some finite resistance to current and finally $R$ is the load resistance.

Now if current $I$ is increased while keeping the total cell voltage $\epsilon$ constant,

$1)$ The load resistance has to go down and hence $V$ will decrease

$2)$ the sum $IR+Ir$ will be constant

So its obvious that $Ir$ will increase since $I$ is increased and $r$ is the same.

Hence terminal voltage (or the voltage across the load) will decrease if the load is decreased or, if you increase the total cell voltage $\epsilon$ while keeping the resistances same, the current will increase.

Notice that the terminal voltage isn't only dependent on current so everything else has to be considered.

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The key thing which you are neglecting is that in order to increase the current through the cell you must decrease the resistance of the external circuit connected to the terminals of the cell.

If Kirchhoff's voltage law is used for your circuit one gets

$\mathcal E_{\rm cell} -I\, R_{\rm cell}-V_{\rm terminal}=0 \Rightarrow V_{\rm terminal} = \mathcal E_{\rm cell} -I R_{\rm cell}$

where $\mathcal E_{\rm cell}$ is the constant emf of the cell, $R_{\rm cell}$ is the constant (or possibly increasing) resistance of the cell, $V_{\rm terminal}$ is the terminal potential difference across the terminals of the cell and $I$ is the current in the circuit.

Note also that $V_{\rm terminal} = I R_{\rm external}$ where $R_{\rm external}$ is the resistance of the circuit connected to the terminals of the cell.

Now how is the current in the circuit going to be increased?
By decreasing $R_{\rm external}$ which means that $I \uparrow R_{\rm external}\downarrow = \mathcal E_{\rm cell} -I\uparrow R_{\rm cell}$.

If the right hand side of the equation gets smaller then so must the left hand side which is the terminal potential difference.

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The voltage at the terminals of a cell is the result of a charge separation produced by a chemical reaction within the cell. The nature of the reaction determines the open circuit voltage. As you draw current from the cell, the reaction may not be able to keep up with the demand and the terminal voltage drops. Generally, a large cell, with a lot of chemicals will show less of a drop than a smaller one. Since the drop is usually proportional to the current, this behavior is often modeled with an “internal resistor”.

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