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A battery of emf 15 V and internal resistance 3 ohms is connected to two resistors of resistances 3 ohm and 6 ohms in series. The potential difference between the terminals of the battery is: (A) 12.5 V(B) 9 V(C) 11.25 V(D) 10 V

Last updated date: 20th Jun 2024
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Hint: 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,
$I\text{ }=\text{ }V/R$, 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.

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 total resistance of a circuit connected to the given cell of emf $15 \mathrm{V}$ is the sum of all the resistances from various sources like, resistors, ammeter, voltmeter, etc., connected in series along with the internal resistance of the cell.
In this case, the total resistance is given as $3\Omega +3\Omega +6\Omega =12\Omega$. $3\Omega +3\Omega +6\Omega =12\Omega$
From the ohm's law the total current can be calculated from the
formula $\mathrm{I}=\mathrm{V} / \mathrm{R}$
That is, $\mathrm{I}=\dfrac{\mathrm{V}}{\mathrm{R}}=\dfrac{15}{12 \mathrm{A}}=1.25 \quad \mathrm{A}$
Hence, the current through the battery is 1.25 amperes. The voltage output of a device is measured across its terminals and is called its terminal voltage V. Terminal voltage is given by the equation $\mathrm{V}=\mathrm{emf}-\mathrm{Ir}$ where, $\mathrm{r}$ is the internal resistance and I is the current flowing at the time of the measurement.
Therefore, $\mathrm{V}=15 \quad \mathrm{V}-(1.25 \times 3)=11.25 \quad \mathrm{V}$.
Hence, the potential difference across the terminals of the cell is 11.25 V.

Hence, option C is the correct answer.

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.