Question

Compare the power used in the $2\mathrm{\Omega }$ resistor in each of the following circuits:

(i) A $6\mathrm{V}$ battery in series with $1\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors.

(ii) A $4\mathrm{V}$ battery in parallel with $12\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors.

Answer 1

Step 1: Given data

(i) Potential difference ${\mathrm{V}}_1=6\mathrm{V}$, Resistance values ${\mathrm{R}}_1=1\mathrm{\Omega }$ and ${\mathrm{R}}_2=2\mathrm{\Omega }$ resistors

(ii) Potential difference ${\mathrm{V}}_2=4\mathrm{V}$, Resistance values ${\mathrm{R}}_3=12\mathrm{\Omega }$ and ${\mathrm{R}}_4=2\mathrm{\Omega }$ resistors.

Step 2: Formula used

According to Ohm's law,

$\mathrm{V}=\mathrm{IR}$

$\mathrm{P}={\mathrm{I}}^2\mathrm{R}$

Where,

$\mathrm{V}$ is the voltage.

$\mathrm{I}$ is the current.

$\mathrm{R}$ is the resistance.

$\mathrm{P}$ is power

Step 3: Calculating power when a $6\mathrm{V}$ battery in series with $1\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors

Potential differences $\left({\mathrm{V}}_1\right)=6\mathrm{V}$

$1\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors are connected in series.

The equivalent resistance of the circuit,

$$\begin{gathered} \mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2 \\ \mathrm{R}=1+2 \\ \mathrm{R}=3\mathrm{\Omega } \end{gathered}$$

According to Ohm’s law,

$\mathrm{V}=\mathrm{IR}$

$\mathrm{V}$ is the voltage.

$\mathrm{I}$ is the current.

$\mathrm{R}$ is the resistance.

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{V}}_1}{\mathrm{R}} \\ \mathrm{I}=\frac{6}{3} \\ \mathrm{I}=2\mathrm{A} \end{gathered}$$

1. Because there is no current division in series circuits, this current will pass through each component. As a result, the $2\mathrm{\Omega }$ resistor has a current of $2\mathrm{A}$ flowing through it.
2. Whenever the current across the circuit elements remains constant, the below formula is implemented.

By the given expression, power is

$$\begin{gathered} {\mathrm{P}}_1={\mathrm{I}}^2\mathrm{R} \\ {\mathrm{P}}_1=2^2\times 2 \\ {\mathrm{P}}_1=8\mathrm{W} \end{gathered}$$

Hence, the power used in the $2\mathrm{\Omega }$ resistor a $6\mathrm{V}$ battery in series with $1\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors is $8\mathrm{W}$.

Step 4: Calculating power when a $4\mathrm{V}$ battery in parallel with $12\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors

Potential differences $\left({\mathrm{V}}_2\right)=4\mathrm{V}$

$12\mathrm{\Omega }$ and $2\mathrm{\Omega }$ resistors are connected in parallel.

1. In a parallel circuit, the voltage across each component remains constant. As a result, the voltage across a $2\mathrm{\Omega }$ resistor will be $4\mathrm{V}$.
2. Whenever the voltage across the circuit components is constant, the below formula is utilized.

Power consumed by $2\mathrm{\Omega }$ the resistor is given by

$$\begin{gathered} {\mathrm{P}}_2=\frac{{\left({\mathrm{V}}_2\right)}^2}{\mathrm{R}} \\ {\mathrm{P}}_2=\frac{4^2}{2} \\ {\mathrm{P}}_2=8\mathrm{W} \end{gathered}$$

Therefore, the power used by $2\mathrm{\Omega }$ resistor in both the circuits is $8\mathrm{W}$.