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

In summary, the maximum current that can be drawn from a battery is determined by its emf and internal resistance. The terminal voltage of the battery depends on the external resistance, with a smaller external resistance resulting in a higher terminal voltage. This is because there is also an internal resistance in the circuit that needs to be taken into account. To get more current from the same battery, the external resistance needs to be decreased, which will also cause a decrease in the terminal voltage. This is due to the fact that the voltage across the internal resistance increases as the external resistance decreases.
  • #1
vinci
12
4
Member advised to use the homework template for posts in the homework sections of PF.
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?
 
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  • #2
Okay, this might help explain what is going on: Let [itex]\mathcal{E}[/itex] be the emf of the battery, and let [itex]r[/itex] be its internal resistance. Let it be connected to an external resistance of [itex]R[/itex]. In that case, the total resistance is [itex]r+R[/itex], and so the current flowing through the wire will be given by:

[itex]\mathcal{E} = I (r+R)[/itex], or [itex]I = \frac{\mathcal{E}}{r + R}[/itex]

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

So [itex]V < \mathcal{E}[/itex]. It only becomes equal to [itex]\mathcal{E}[/itex] if [itex]\frac{r}{R} \rightarrow 0[/itex], which means that [itex]r[/itex] is very small compared to [itex]R[/itex].
 
  • #3
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).
 
  • #4
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?
 
  • #5
vinci said:
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.
 
  • #6
vinci said:
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 [itex]V = I R[/itex], you can see that [itex]I[/itex] is proportional to [itex]V[/itex] if [itex]R[/itex] is held constant. But we're not holding [itex]R[/itex] constant, we're changing it. We're changing both [itex]V[/itex] and [itex]R[/itex].

I just went through the math: [itex]V[/itex], the voltage drop across the external resistance, varies with [itex]R[/itex] according to:

[itex]V = \frac{\mathcal{E}}{1 + \frac{r}{R}}[/itex]

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

What is the relationship between current and potential difference?

The relationship between current (I) and potential difference (V) is known as Ohm's Law. It states that the current flowing through a conductor is directly proportional to the potential difference applied across it, as long as the temperature and other physical conditions remain constant. In other words, as current increases, potential difference increases proportionally.

Why does a higher current result in a lower potential difference?

This phenomenon can be explained by the concept of electrical resistance. In a circuit, the flow of current encounters resistance from the material it is passing through. As the current increases, the resistance also increases, leading to a decrease in potential difference. This is similar to how water flowing through a narrow pipe experiences more resistance compared to a wider pipe.

What factors affect the relationship between current and potential difference?

The main factors that affect this relationship are the material of the conductor, its length, and its cross-sectional area. Different materials have different levels of resistance, and a longer conductor will have more resistance compared to a shorter one. A larger cross-sectional area allows for more current to flow through, resulting in a lower resistance and a higher potential difference.

Can a circuit have a higher current and a higher potential difference at the same time?

Yes, it is possible for a circuit to have both a high current and a high potential difference. This can happen when there is a lower resistance in the circuit, allowing for more current to flow through, while at the same time, a higher potential difference is being applied. However, the relationship between current and potential difference will still follow Ohm's Law.

How does this relationship impact everyday electronic devices?

The relationship between current and potential difference is crucial in understanding how electronic devices function. By controlling the current and potential difference, we can regulate the amount of power being used by a device. This allows for devices to operate safely and efficiently, as well as providing a means for us to measure and troubleshoot any issues that may arise in the circuit.

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