Unable to understand how higher current results in lower p.d

[vinci]

To start with this is not a numerical but merely a basic concept I fail to understand. I will quote what the CIE book says
"Consider a battery of emf 3 V and of internal resistacne 1 ohm. The maximum current that can be drawn from this battery is I=e/r = 3/1=3 A
The terminal p.d of the battery dpeends on the resistance of the external resistor. For an external resistance of 1 ohm the terminal pd is 1.5v (1/2 of emf). The terminal p.d approaches the value of emf when the external resistance R is very much greater than internal resistance. E,g a resistor of resistance 1k oh connected to battery will give a terminal p.d of 2.997V. this is almost equal to emf of battery. The more current a battery supplies, the more its terminal pd will decrease."

The last line (more current = less p.d ) doesn't go well with the formula V=IR which says more voltage will result in more current. If more voltage was to yield less current wouldn't Voltage and Current be inversely proportional?
Another confusion is "A resistance of 1k will give a terminal p.d of 2.997V", that doesn't make much sense either. It is a known fact that resistors decrease the amount of current flowing and a higher resistance will result in a lower current and since voltage and current are directly proportional wouldn't the voltage be low too?

 

[stevendaryl]

Okay, this might help explain what is going on: Let $\mathcal{E}$ be the emf of the battery, and let $r$ be its internal resistance. Let it be connected to an external resistance of $R$. In that case, the total resistance is $r+R$, and so the current flowing through the wire will be given by:

$\mathcal{E} = I (r+R)$, or $I = \frac{\mathcal{E}}{r + R}$

The voltage across just the external resistance, $R$ will be given by: $V = I R = \frac{\mathcal{E} R}{r+R} = \frac{\mathcal{E}}{1 + \frac{r}{R}}$

So $V < \mathcal{E}$. It only becomes equal to $\mathcal{E}$ if $\frac{r}{R} \rightarrow 0$, which means that $r$ is very small compared to $R$.

 

[haruspex]

Steven has supplied the answer in terms of equations, but let me offer a more descriptive answer.
In order to get more current from the same battery, you would need to reduce the external resistance. But you would find that the current does not increase as quickly as you would expect if it were simply V=IR with a constant V. The explanation is that the measured V is also reducing somewhat (or to put it another way, that there is internal resistance that needs to be taken into account).

 

[vinci]

What still confuses me is this
"n order to get more current from the same battery, you would need to reduce the external resistance"
Less resistance is equal to a smaller potential drop across the resistor, no? How would a smaller potential drop yield a larger current when the two are directly proportional?

 

[cnh1995]

Less resistance is equal to a smaller potential drop across the resistor, no?

Yes. But as the resistance decreases, V/R ratio increases. Decrease in voltage is less compared to decrease in resistance. This is because of the internal resistance of the source. Voltage across the internal resistance increases, this means current increases.

 

[stevendaryl]

What still confuses me is this
"n order to get more current from the same battery, you would need to reduce the external resistance"
Less resistance is equal to a smaller potential drop across the resistor, no? How would a smaller potential drop yield a larger current when the two are directly proportional?

In the expression $V = I R$, you can see that $I$ is proportional to $V$ if $R$ is held constant. But we're not holding $R$ constant, we're changing it. We're changing both $V$ and $R$.

I just went through the math: $V$, the voltage drop across the external resistance, varies with $R$ according to:

$V = \frac{\mathcal{E}}{1 + \frac{r}{R}}$

where $\mathcal{E}$ is the emf, and $r$ is the internal resistance, and $R$ is the external resistance. So $V$ depends on $R$. So it's not possible to vary $V$ and hold $R$ constant.