Question

A silver, a copper and an iron wire of identical dimensions are connected in series with voltage source. ${P_{\text{s}}}$ , ${P_{\text{c}}}$ and ${P_{\text{l}}}$ are power consumed in three respectively then,
A. ${P_{\text{s}}} > {P_{\text{c}}} > {P_{\text{i}}}$
B. ${P_{\text{c}}} > {P_{\text{s}}} > {P_{\text{i}}}$
C. ${P_{\text{i}}} > {P_{\text{s}}} > {P_{\text{c}}}$
D. ${P_{\text{i}}} > {P_{\text{c}}} > {P_{\text{s}}}$

Answer 1

First of all, we will find an expression which relates resistance and resistivity. Identical dimensions mean the same length and cross-sectional area. We also apply the relation in which power is directly proportional to resistance of the wire.

Complete step by step answer:
In the given problem, we are supplied with the following data:
There are three wires given which are silver, copper and iron.
The three wires are connected in series with voltage source.
The power consumed by each wire is given as ${P_{\text{s}}}$ , ${P_{\text{c}}}$ and ${P_{\text{i}}}$ .
So, we are asked to find the order of the power consumed by each wire of different material.
To begin with, we will first analyse the arrangement of the connection. All the wires are connected in series. So, the same amount of current flows through them, as there is only a path for the current to flow. Again, we are given that the dimensions of all the three wires are exactly identical, which means that the length and the cross-sectional area of the three wires is equal to one another.
So, for now, we will find the expression of resistance which relates length and cross-sectional area of the wires, which is given below:
$R = \dfrac{{\rho l}}{A}$ …… (1)
Where,
$R$ indicates resistance.
$\rho$ indicates resistivity.
$l$ indicates length of the wire.
$A$ indicates the cross-sectional area of the wire.
But from the given condition in the question, we have the length and area of cross section is equal to one another.
So, $l$ and $A$ are taken as contents and the equation (1) becomes:
$R \propto \rho$ …… (2)
From the equation (2), we can conclude that higher the resistivity is, higher is the resistance of the wire.
Again, we have another formula which gives a relation between resistance and power:
$P = {i^2}R$ …… (3)
We know,
The resistivity of iron, copper and silver are as follows:
${\rho _{\text{i}}} = 9.7 \times {10^{ - 8}}\,\Omega{\text{ m}}$
${\rho _{\text{c}}} = 1.7 \times {10^{ - 8}}\,\Omega{\text{ m}}$
${\rho _{\text{s}}} = 1.6 \times {10^{ - 8}}\,\Omega{\text{ m}}$
So, we can conclude that resistivity of copper lies between iron and silver:
So, the order of the resistance present in the wire are given below:
${R_{\text{i}}} > {R_{\text{c}}} > {R_{\text{s}}}$
Since, power is directly proportional to resistance of the wire, so the correct order of power consumed is:
${P_{\text{i}}} > {P_{\text{c}}} > {P_{\text{s}}}$

So, the correct answer is “Option D”.

Note:
It is important to remember that in a series connection, potential differences of the elements may differ from one another, but the same amount of current will flow through them. However, in a parallel circuit, the amount of current flowing through each branch may vary depending on the resistance present in each branch, but the potential difference of all the parallel branches will be the same.