Question

The three resistors configuration $R_1,R_2$ and $R_3$ are connected to $3V$ battery as shown in figure. The power dissipated by the configuration $R_1,R_2$ and $R_3$ is $P_1,P_2$ and $P_3$ respectively. Then –

A) $P_1>P_2>P_3$
B) $P_1>P_3>P_2$
C) $P_2>P_1>P_3$
D) $P_3>P_2>P_1$

Answer 1

Hint: -Find the equivalent resistances for all the configurations in $R_1,R_2$ and $R_3$. Now, by using the relation between power and resistance we will find the order of power in different configurations.

Complete Step by Step Solution: -
In the first configuration of $R_1$,

From this configuration, we can conclude that there is an equivalent wheatstone bridge circuit.
$\therefore {\displaystyle \frac{1\mathrm{\Omega }}{1\mathrm{\Omega }}}={\displaystyle \frac{1\mathrm{\Omega }}{1\mathrm{\Omega }}}=1\mathrm{\Omega }$
Therefore, the equivalent resistance in the configuration $R_1$ is $1\mathrm{\Omega }$.
Now, we have to calculate the equivalent resistance in the configuration $R_2$ -

From the above figure, the equivalent resistance can be calculated by –
$$\begin{gathered} \Rightarrow R_2={\left({\displaystyle \frac{1}{1+1}}+1+{\displaystyle \frac{1}{1+1}}\right)}^{-1} \\ \therefore R_2=2^{-1}=0.5\mathrm{\Omega } \end{gathered}$$
Therefore, the equivalent resistance in the configuration $R_2$ is $0.5\mathrm{\Omega }$.
Now, we have to calculate the equivalent resistance in the configuration $R_3$ -

From the above figure, the equivalent resistance for the configuration $R_3$ is –
$$\begin{gathered} \Rightarrow R_3=1+{\displaystyle \frac{\left(1+1\right)\left(1+1\right)}{\left(1+1\right)+\left(1+1\right)}} \\ \therefore R_3=1+1=2\mathrm{\Omega } \end{gathered}$$
Therefore, the equivalent resistance in the configuration $R_3$ is $2\mathrm{\Omega }$.
We know that, relationship between power, voltage and resistance is –
$\Rightarrow P={\displaystyle \frac{V^2}{R}}$
So, the power $P_1$ for configuration $R_1$ is –
$\Rightarrow P_1={\displaystyle \frac{3^2}{1}}=9W$
Power $P_2$ for configuration $R_2$ is –
$$\begin{gathered} P_2={\displaystyle \frac{3^2}{0.5}}={\displaystyle \frac{9}{0.5}} \\ \therefore P_2=18W \end{gathered}$$
Power $P_3$ for configuration $R_3$ is –
$$\begin{gathered} P_3={\displaystyle \frac{3^2}{2}}={\displaystyle \frac{9}{2}} \\ \therefore P_3=4.5W \end{gathered}$$
Now, we have calculated the power for each configuration for $R_1,R_2$ and $R_3$ which are –
$$\begin{gathered} \Rightarrow P_1=9W \\ \Rightarrow P_2=18W \\ \Rightarrow P_3=4.5W \end{gathered}$$
Therefore, we can conclude that –
$P_2>P_1>P_3$

Hence, the correct option is (C).

Note: -Wheatstone bridge is the circuit which is connected in form of the bridge and is composed of two known resistors, one unknown resistor and one variable resistor connected in the form of a bridge. This bridge gives accurate measurements, so it is most reliable. This bridge is used for the precise measurement of low resistance and is also used to measure inductance, capacitance and impedance by using the variations.