Question

A wire when connected to 220V 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

Hint: After cutting the wire into equal pieces, the resistance of each wire will be halved and then connecting them in parallel will give us a new equivalent resistance. Since they are connected in parallel the voltage across their ends shall remain the same as before. We will use this approach to calculate the power dissipated initially and then after we have cut the wire into two pieces.

Complete answer:
Let us first assign some terms that we are going to use in our solution.
Let the mains power supply which is a constant value of 220 Volts be denoted by ‘V’. And, let the resistance in the wire initially be denoted by R.
Now, as we know that resistance in a uniform wire having a length ‘l’, area of cross-section ‘A’ and resistivity ‘$\rho $ ’ can be written using the following formula:
$\Rightarrow R=\rho \dfrac{l}{A}$
From the above equation, we can see that:
$\Rightarrow R\propto l$
Thus, the resistances of the two equal pieces of wire is equal to $\dfrac{R}{2}$ .
Thus, the equivalent resistance of the new system is equal to:
$\Rightarrow {{R}_{net}}=\dfrac{\dfrac{R}{2}\times \dfrac{R}{2}}{\dfrac{R}{2}+\dfrac{R}{2}}$

Now, let the power dissipated initially be ${{P}_{1}}$ and after the wire has been cut be equal to ${{P}_{2}}$.
Then,
$\Rightarrow {{P}_{1}}=\dfrac{{{V}^{2}}}{R}$ [Let this expression be equation number (1)]
$\Rightarrow {{P}_{2}}=\dfrac{{{V}^{2}}}{\dfrac{R}{4}}$
 $\therefore {{P}_{2}}=\dfrac{4{{V}^{2}}}{R}$ [Let this expression be equation number (2)]
On dividing the equation number (2) by equation number (1), we get:
$\begin{align}
 & \Rightarrow \dfrac{{{P}_{2}}}{{{P}_{1}}}=\dfrac{\dfrac{4{{V}^{2}}}{R}}{\dfrac{{{V}^{2}}}{R}} \\
 & \therefore \dfrac{{{P}_{2}}}{{{P}_{1}}}=4 \\
\end{align}$
Hence, the ratio of power dissipated after cutting the wire and initially, that is, ${{P}_{2}}:{{P}_{1}}$ comes out to be 4.

So, the correct answer is “Option B”.

Note: We should always know the dependence of parameters on factors as questions are often framed by changing one of these factors and asking us the new value for our parameter. Also, in problems involving ratio, we should be very careful to give the correct ratio as an answer and not the inverse of it.