Question

Compare the power used in $2$ ohm resistor in each of the following circuits
(a) A $6\;V$ battery in series with $1$ ohm and $2$ ohm resistors.
(b) A $4\;V$ battery in parallel with $12$ ohm and $2$ ohm resistor.

Answer 1

In a circuit when there is a voltage supply connected to the circuit elements in that circuit then due to the voltage drop across the connected elements and the current flowing through the circuit there is a certain amount of power that is dissipated. The concept of resistances in series and parallel must be used in order to find the equivalent resistance of the circuit. The formula for the power in an electrical circuit in terms of its voltage current and resistance needs to be applied.

Formula used:
The formula for the power generated is given by the equation:
$P = VI$
The ohm’s law equation is as follows:
$V = IR$
Where, $V$ is the voltage, $I$ is the current and $R$ is the resistance.

Complete answer:
First, we must know the concept of power in order to solve the above problem. Power is defined as the rate at which the work is done by a source of emf or voltage supply in maintaining an electric current throughout the circuit. The power is a measure of the heat that is dissipated due to the production of current when there is a voltage supply connected across its circuit elements. Hence the equation for the power generated is said to be:
$P = VI$ ---------($1$)
In the above question we are asked to find out the power in two different circuits and then compare their values.

(a) Since there is a voltage supply connected to the circuit there will be a potential difference produced across the circuit elements and hence there will also be a current flowing through the circuit. Hence we apply the ohm’s law equation in order to give the relationship between voltage and current in the circuit. The ohm’s law equation is as follows:
$V = IR$

The current that is flowing in the circuit has not been given and hence we must first find the value of current out. Since the circuit is said to be consisting of two resistors connected in series to each other, there will be a need to find out its equivalent, that is, the total resistance of the circuit.Hence the above equation will become as follows:
$V = I{R_{eq}}$
Since we are required to find the current, by rearranging the terms we get:
$I = \dfrac{V}{{{R_{eq}}}}$ --------($2$)
The voltage value is already given to be $6\;V$ and hence $V = 6\;V$. We now need to calculate the equivalent resistance in order to reduce the circuit, that is, replace the series combination of resistors by a single resistance value. The resistance values are given to be $1$ ohm and $2$ ohm respectively and hence ${R_1} = 1\Omega$ and ${R_2} = 2\Omega$.

By the concept of resistances in series, the equivalent resistance is given by the sum of the resistors connected in series. Hence the resistances in series equation is given as follows:
${R_s} = {R_1} + {R_2} + ...{R_n}$
Since there are two resistors connected in series we get:
${R_{eq}} = {R_1} + {R_2}$
$\Rightarrow {R_{eq}} = 1 + 2$
$\Rightarrow {R_{eq}} = 3\Omega$
By substituting these values in equation ($2$) we get:
$I = \dfrac{6}{3}$
$\Rightarrow I = 2{\rm A}$
The current across a series combination of resistors is constant the series resistances are connected by a single wire and there won’t be any splitting up of current across any junctions.

When putting the ohm’s law equation in the power equation given by equation ($1$) we get:
$P = \left( {IR} \right)I$
$\Rightarrow P = {I^2}R$ -------($3$)
This is done because we are required to find out the power in one of the resistors and hence we need an equation for power in terms of its current and resistance. The power generated in the $2\Omega$ is required to be found out and the value of current has been obtained above.
Hence by substituting the values in equation ($3$) we get:
$\Rightarrow P = {\left( 2 \right)^2} \times 2$
$\Rightarrow P = 4 \times 2$
$\therefore P = 8\;W$
Hence, the power in the $2\Omega$ resistor is $8\;W$.

(b) We need to find out the power consumed in the $2\Omega$ resistor by using the ohm’s law equation as well as the power equation. Here, there is no need of finding out the equivalent resistance of the resistances in parallel because we can instead construct an equation for power in terms of voltage and resistance.
From equation ($2$) we get:
$I = \dfrac{V}{R}$
We substitute the above equation in equation ($1$) which is the power equation and we get:
$P = V\left( {\dfrac{V}{R}} \right)$
$\Rightarrow P = \dfrac{{{V^2}}}{R}$ ---------($4$)

Since, the power across the $2\Omega$ resistor needs to be found, $R = 2\Omega$
By the concept of resistances in parallel the voltage across them will remain the same. In a parallel combination, the potential difference across the resistors will be equal since the terminals or the junctions connecting them are at the same potential. This voltage is given to be $4\;V$.
Hence we substitute these values in equation ($4$) we get:
$P = \dfrac{{{{\left( 4 \right)}^2}}}{2}$
$\Rightarrow P = \dfrac{{16}}{2}$
$\therefore P = 8\;W$

Hence from part a) and part b) by comparing the power generated in the $2\Omega$ resistor in both of the given circuits we can clearly see that the power is the same in both which is $8\;W$.

Note: In a series combination of resistances in a circuit, the potential difference and the power consumed, that is the heat generated will be larger in the higher resistance. In the parallel combination of resistances the current and the power consumed will be larger in the resistor having the smallest resistance.