Question

A wire when connected to 220 V mains supply has power dissipation ${P_1}$. Now, the wire is cut into two equal pieces, which are connected in parallel to the same supply. Power dissipation in this case is ${P_2}$. Then ${P_2}:{P_1}$ is
A. 1
B. 4
C. 2
D. 3

Answer 1

Any resistor in a circuit that has voltage drop across it dissipates electrical power.
Dissipates, means loss of energy in the form of heat.
The electrical power is converted into heat energy.
Formula for power dissipation or loss-

Using the formula of electrical power
$P = V \times I$, where, V & I are the voltage difference across the resistor and current flowing through the resistor respectively.
From Ohm's law, we know the following relation between the voltage drop, current and resistance.
$V = I \times R$.

Substituting the value of $V$in the equation of power dissipation-
$\Rightarrow P = \left( {I \times R} \right) \times I \\\ {\text{ = }}{I^2} \times R \\\$
$\therefore P = {I^2}R$
Writing in terms of $V\& R$-
$\Rightarrow P = {\left( {\dfrac{V}{R}} \right)^2} \times R \\\ \Rightarrow P = \dfrac{{{V^2}}}{R} \\\$

The unit of power is Joule/sec or watt.
$\Rightarrow 1W = 1J/s$

The approach to the problem given-
1. The resistance of a conductor of uniform cross section is directly proportional to its length and inversely proportional to its area of cross section-
Mathematically,
$R = \dfrac{{\rho L}}{A}$, where $\rho ,{\text{ is the resistivity of the material and is a constant for the given material}}{\text{.}} \\\ {\text{L, is the length of conductor}}{\text{.}} \\\ {\text{A, is the area of cross section}} \\\$

$\because$the cross-sectional area and resistivity of material is constant, so we can safely say that the resistance of this wire is directly proportional to its length.
If the length is halved, then the resistance will also be halved.
As the question demands, the wire is cut into two equal lengths.

2. If resistors (say, ${R_1},{R_2},{R_3}$) are connected in parallel, then the equivalent resistance ${R_{eqv}}$will be-
$\Rightarrow \dfrac{1}{{{R_{eqv}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{R{}_2}} + \dfrac{1}{{{R_3}}}$

Complete step by step solution:
We have the following information at hand-
1. The potential difference across the wire $= 220V$
2. The power dissipation across the wire is ${P_1}$
The power dissipation after the wire is cut into two equal parts and connected across the power source parallelly is ${P_2}$ and that the ratio of ${P_1}\& {P_2}$ ought to be calculated.

Let us consider that the original length of wire be $L$and its original resistance be $R$.
$\Rightarrow R = \dfrac{{\rho L}}{A}$
After the wire is cut into halves, the resistance of each halves is ${R_1}\& {R_2}$.
$\Rightarrow {R_1} = \dfrac{{\rho \left( {\dfrac{L}{2}} \right)}}{A}{\text{ and }}{R_2} = \dfrac{{\rho \left( {\dfrac{L}{2}} \right)}}{A} \\\ \Rightarrow {R_1} = {R_2} = \dfrac{{\rho L}}{A} \times \dfrac{1}{2} = \dfrac{R}{2} \\\$
The resistors ${R_1}{\text{ and }}{R_2}{\text{ are connected in parallel, }}$
Let the equivalent resistance be ${R_{eqv}}$
$\Rightarrow \dfrac{1}{{{R_{eqv}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
$\Rightarrow \dfrac{1}{{R{}_{eqv}}} = \dfrac{1}{{\left( {\dfrac{R}{2}} \right)}} + \dfrac{1}{{\left( {\dfrac{R}{2}} \right)}} = \dfrac{2}{R} + \dfrac{2}{R}$
$\Rightarrow \dfrac{1}{{{R_{eqv}}}} = \dfrac{4}{R} \\\ \Rightarrow {R_{eav}} = \dfrac{R}{4} \\\$

Now it’s time to compare the power dissipation between the two cases-
Case1- when the length of resistor was L
Power dissipation $= {P_1}$
$\Rightarrow {P_1} = \dfrac{{{V^2}}}{R}$

Case2- when the length was halved-
Power dissipation $= {P_2}$
$\Rightarrow {P_2} = \dfrac{{{V^2}}}{{\left( {\dfrac{R}{4}} \right)}} = \dfrac{{4{V^2}}}{R}$
$\Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{\left( {\dfrac{{4{V^2}}}{R}} \right)}}{{\left( {\dfrac{{{V^2}}}{R}} \right)}} = 4$

Thus ${P_2}:{P_1} = 4$

$\therefore$Option B is the correct option.

Note: These types of questions can be quickly solved by just considering the factors on which a quantity depends and creating a mathematical equation for it and then comparing the cases-
For ex- in our question, it is evident that only the length is changed, so
$R = CL$, where C stands for constant,
Since the length has been halved, it makes sense to infer that the resistance will be halved too.