Potential difference of an emf with internal resistance

In summary, a battery of emf 2V and internal resistance 0.1ohm is being charged with a current of 5 ampere. The current flows inside the battery from the negative terminal to the positive terminal, which discharges the battery. When recharging, the direction of current is reversed and flows from the positive terminal to the negative terminal. The potential difference between the two terminals is calculated using the formula V = E - ir, which yields a value of 1.5V. However, the answer given is 2.5V, which is due to the difference in current direction. The terms "anode" and "cathode" may be used differently depending on the operating mode of the battery
  • #1
Tanishq Nandan
122
5

Homework Statement


A battery of emf 2V and internal resistance 0.1ohm is being charged with a current of 5 ampere.
In what direction will the current flow inside the battery?What is the potential difference between the two terminals of the battery?

Homework Equations


If a battery of emf E and an internal resistance r has a current i flowing through it,potential difference across it's terminals,say V is defined as:
V=E-ir

The Attempt at a Solution


When we connect a battery in a circuit,INSIDE THE BATTERY,current flows from the negative terminal to the positive terminal,which discharges the battery(correct me if I'm wrong here please)
So,while recharging,the direction should be reversed,shouldn't it?
Also,the formula stated above,yields the answer to be 1.5V
But,the answer given is:
Current flows from the anode to the cathode
V.d=2.5 Volts
I am guessing if I just get the reasoning for the first part,the formula stated could be modifed to yield the correct answer,but why does the current inside the battery flow from the negative terminal to the positive terminal,wouldn't that discharge the battery?
 
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  • #2
Tanishq Nandan said:
When we connect a battery in a circuit,INSIDE THE BATTERY,current flows from the negative terminal to the positive terminal,which discharges the battery(correct me if I'm wrong here please)
Conventional current (movement of positive charges) flows from the positive terminal, though the external circuit, and back to the negative terminal of the battery. Internal to the battery, conventional current is "pumped" from its negative terminal to its positive terminal. When analyzing circuits it's in your best interests to express things in terms of conventional current as that is the accepted standard and less likely to cause confusion amongst those reading your equations and explanations.
So,while recharging,the direction should be reversed,shouldn't it?
Correct.
Also,the formula stated above,yields the answer to be 1.5V
But,the answer given is:
Current flows from the anode to the cathode
V.d=2.5 Volts
I am guessing if I just get the reasoning for the first part,the formula stated could be modifed to yield the correct answer,but why does the current inside the battery flow from the negative terminal to the positive terminal,wouldn't that discharge the battery?
You don't need to modify the formula, just get the current direction correct. That means assigning the appropriate sign to the current value to reflect its direction with respect to the direction assumed for the current definition in place when the formula in question was written.
 
  • #3
gneill said:
Internal to the battery, conventional current is "pumped" from its negative terminals to its positive terminal
Ok,I got that.
Now,if I am correct and the direction does get reversed,doesn't that mean that conventional current is moving from the positive to the negative terminal now?
But,as I stated in the post,the answer given is:current flows from anode to cathode.
 
  • #4
Tanishq Nandan said:
Ok,I got that.
Now,if I am correct and the direction does get reversed,doesn't that mean that conventional current is moving from the positive to the negative terminal now?
But,as I stated in the post,the answer given is:current flows from anode to cathode.
The terms "anode" and "cathode" can sometimes be tricky to interpret when a device is not being used in its normal operating mode. For a battery that is driving current into a circuit the positive terminal is designated as the cathode. If the battery is being charged and current is thus flowing into its positive terminal, that terminal is then referred to as the anode.

Take a look at the Wikipedia article:
https://en.wikipedia.org/wiki/Anode
 
  • #5
gneill said:
The terms "anode" and "cathode" can sometimes be tricky to interpret when a device is not being used in its normal operating mode. For a battery that is driving current into a circuit the positive terminal is designated as the cathode. If the battery is being charged and current is thus flowing into its positive terminal, that terminal is then referred to as the anode.

Take a look at the Wikipedia article:
https://en.wikipedia.org/wiki/Anode

Ok,got it.Crystal clear now,thanks!
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electrical potential between two points in a circuit. It is measured in volts (V) and represents the amount of energy required to move a unit of electric charge from one point to another.

2. What is emf?

Emf stands for electromotive force and is a measure of the energy per unit charge that is supplied by a source, such as a battery or generator, to establish an electric current. It is also measured in volts (V).

3. What is internal resistance?

Internal resistance is the inherent resistance within a source of emf, such as a battery. It is caused by the resistance of the materials and components within the source, and it works against the flow of current. Internal resistance is also measured in ohms (Ω).

4. How does internal resistance affect potential difference?

Internal resistance can cause a decrease in the potential difference of a source, meaning that the voltage measured across the terminals of the source will be less than the emf. This is due to the voltage drop across the internal resistance, which decreases the overall potential difference of the circuit.

5. How is the potential difference of an emf with internal resistance calculated?

The potential difference of an emf with internal resistance can be calculated using Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R). In this case, the resistance is the sum of the internal resistance (r) and the external resistance (R) in the circuit. So the equation is V = I(R+r).

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