Question

A battery of emf 4 volts and internal resistance $1 \Omega$ is connected in parallel with another battery of emf 1-volt Combination is used to send current through an external resistance $2 \Omega .$ Calculate current through external resistance and potential difference across external resistance.

Answer 1

We know that the electrical resistance of a circuit component or device is defined as the ratio of the voltage applied to the electric current which flows through it: If the resistance is constant over a considerable range of voltage, then Ohm's law, $I=V/R$, can be used to predict the behaviour of the material. An electric current flow when electrons move through a conductor, such as a metal wire. The moving electrons can collide with the ions in the metal. This makes it more difficult for the current to flow, and causes resistance. Based on this concept we have to solve this question.

Complete step by step answer
We know that ohm's Law states that the current flowing in a circuit is directly proportional to the applied potential difference and inversely proportional to the resistance in the circuit. In other words, by doubling the voltage across a circuit the current will also double.
It should be known to us if the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations. Because electric potential difference is expressed in units of volts, it is sometimes referred to as the voltage.
The diagram for this question is given below:

Let us first write the given values:
$\mathrm{V}_{1}=4 \mathrm{v}$
$\mathrm{R}_{1}=1 \Omega$
$\mathrm{V}_{2}=1 \mathrm{v}$
$\mathrm{R}_{2}=0 \Omega$
So, the total current from the combination = current flowing $2 \Omega=\mathrm{I}$
Hence, we can say that the current is given as:
$\therefore \mathrm{I}=\dfrac{\mathrm{v}_{1}-\mathrm{v}_{2}}{\mathrm{R}_{1}-\mathrm{R}_{2}}=\dfrac{4-1}{1+0} \mathrm{A}$
$=3 \mathrm{A}$
The potential difference is given as:
Potential difference across $2 \Omega=(3 \times 2) \mathrm{v}$

After the evaluation is given as 6 v.

Note: It should be known to us that when a voltage is connected across a wire, an electric field is produced in the wire. Metal wire is a conductor. Some electrons around the metal atoms are free to move from atom to atom. This causes a difference in energy across the component, which is known as an electrical potential difference. Electrical potential difference is the difference in the amount of potential energy a particle has due to its position between two locations in an electric field. The unit of potential difference generated between two points is called the Volt and is generally defined as being the potential difference dropped across a fixed resistance of one ohm with a current of one ampere flowing through it.