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Question: A resistor \(R\) is connected to a cell of emf \(e\) and internal resistance \(r\). Potential differ...

A resistor RR is connected to a cell of emf ee and internal resistance rr. Potential difference across the resistor RR is found to be VV. State the relation between ee, VV, RR and rr.

Explanation

Solution

Here first we have to see what internal resistance, emf and potential difference is –
Internal resistance refers to the opposition to the current flow generated by the cells and batteries themselves, which contribute to heat generation.
Potential difference is often known as work conducted from one point to another to transfer a unit charge.
EMF is an electromotive force, calculated in charge coulombs. It is pressure generated from an electrical energy or source.

Complete step by step solution:
Given,
A resistor RR is connected to a cell of emf ee and internal resistance rr
Potential difference across the resistor RR is found to be VV .
Now let us see the relationship between emf ee, current II, resistor RR and internal resistance rr .
Emf, e=I(R+r)e = I\left( {R + r} \right)
Also,
I=eR+rI = \dfrac{e}{{R + r}} …… (1)
We know that,
According to ohm’s law-
V=IRV = IR, where VV is the voltage of the circuit, II is the current and RR is the resistance.
Therefore, I=VRI = \dfrac{V}{R} …… (2)
Putting equation (2) in equation (1), we get-
VR=eR+r V=eRR+r  \dfrac{V}{R} = \dfrac{e}{{R + r}} \\\ V = \dfrac{{eR}}{{R + r}} \\\
Hence, the relation between ee, VV, RR and rr is:
VR=eR+r\dfrac{V}{R} = \dfrac{e}{{R + r}}

Additional information:
Internal resistance occurs when there is current present in the system or the electrical circuit and there is a voltage reduction in the source voltage or source battery. It is caused by electrolytic content in the battery or other sources of voltage.
When no current flows, the EMF is the potential difference of the source. A voltage source’s internal resistance rr influences the output voltage when current flows.

Note: Here we have to remember the Ohm’s law to get the final output. Also, we have to derive II from the relation e=I(R+r)e = I\left( {R + r} \right). So, we can see that the current is directly proportional to the emf which is very important to remember the equation of emf and current.