Question

A heater coil is rated 100W, 200V. It is cut into two identical parts. Both parts are connected together in parallel, to the same source of 200V. The energy liberated per second in the new combination is?
${\text{A}}{\text{. 100J}} \\\ {\text{B}}{\text{. 200J}} \\\ {\text{C}}{\text{. 300J}} \\\ {\text{D}}{\text{. 400J}} \\\$

Answer 1

Hint: To find the energy liberated, we start off by calculating the resistance of the initial coil using the given values of power and voltage. We then calculate the individual resistances of the identical parts and the effective resistance of the circuit of two identical parts connected in parallel to compute the new power of this circuit.

Complete step-by-step solution -

Given Data,
Power of the coil = 100W
Voltage = 200V
The coil is cut into two

Formula Used:
${\text{P = }}\dfrac{{{{\text{V}}^2}}}{{\text{R}}}$, where P is the power, V is the voltage and R is the resistance of the coil respectively.

Power of a coil is defined as the ability to do something. It is the amount of energy changed per unit time. To find the power or work done by a coil, we use the formula ${\text{P = }}\dfrac{{{{\text{V}}^2}}}{{\text{R}}}$.
So using this formula we find the resistance of the given coil –
${\text{P = }}\dfrac{{{{\text{V}}^2}}}{{\text{R}}} \Rightarrow 100 = \dfrac{{{{200}^2}}}{{\text{R}}} \Rightarrow {\text{R = }}\dfrac{{40000}}{{100}} = 400\Omega$

Then it is said the coil is divided into two equal halves that means the resistance of the coil is changed as it is dependent on the dimensions of the coil and the power of the coil is also changed as it is dependent on the resistance.
The voltage across the coil still remains the same because it is connected in parallel connection.
Now the original coil is cut into two equal halves, i.e. the individual resistances of the coils become half of the original resistance ‘R = 400Ω’.
The new individual resistances are $\dfrac{{\text{R}}}{2},\dfrac{{\text{R}}}{2}$.
$\Rightarrow \dfrac{{\text{R}}}{2} = \dfrac{{400}}{2} = 200\Omega$

These two resistances are connected in a parallel connection. Therefore the effective resistance of the coil should be found to determine the word done by it per second.
We know the effective resistance in a parallel connection is given by –
$\dfrac{1}{{{{\text{R}}_{{\text{effective}}}}}} = \dfrac{1}{{{{\text{R}}_1}}} + \dfrac{1}{{{{\text{R}}_2}}}$
Our individual resistances are 200Ω each, therefore the effective resistance of the new circuit is given by:
$\Rightarrow \dfrac{1}{{{{\text{R}}_{{\text{effective}}}}}} = \dfrac{1}{{200}} + \dfrac{1}{{200}} \\\ \Rightarrow \dfrac{1}{{{{\text{R}}_{{\text{effective}}}}}} = \dfrac{2}{{200}} \\\ \Rightarrow {{\text{R}}_{{\text{effective}}}} = 100\Omega \\\$
Hence the effective resistance of the two coils when connected in parallel is 100Ω.

Now the work done by the new coil per second is given by –
$\Rightarrow {\text{P = }}\dfrac{{{{\text{V}}^2}}}{{\text{R}}} \\\ \Rightarrow {\text{P = }}\dfrac{{{{200}^2}}}{{100}} \\\ \Rightarrow {\text{P = }}\dfrac{{40000}}{{100}} \\\ \Rightarrow {\text{P = 400W}} \\\$
Hence the work done per second by the two coils in parallel connection is 400J.
Option D is the correct answer.

Note – In order to answer this type of question the key is to know the formula which gives us the relation between power, voltage and resistance. It is a vital step to observe that the resistance is halved and voltage remains the same when the coils are cut and connected in parallel.
Resistance of a material is given by the formula ${\text{R = }}\dfrac{{\rho {\text{l}}}}{{\text{A}}}$, where l is the length and A is the area of cross section of the material.
This means resistance of a material is directly proportional to its length. This is why the resistance of each coil is also halved when they are cut into two equal parts.
Power is generally expressed in the unit ‘Watt’ or ‘Joules per unit second’ in the international system of units.