Question

The resistance of a conductor is reduced to half its initial value keeping voltage applied constant .In doing so the heating effects in the conductor will be –
a) Half
b)Double
c)one-fourth
d)four times

Answer 1

Hint: Here in this question we have to use the formula for heating effect of electric current in terms of voltage because it is given in the question that voltage is constant. The heating effect is caused due to resistance so any change in resistance will also change the heating effect on a conductor. Heating effect totally depends on current; resistance and voltage.

Complete step-by-step answer:
Resistance is the property of a conductor which determines the conduction in a conductor. More is the resistance less is the conduction. Resistance of a conductor can be calculated by following relation-
$$R={\displaystyle \frac{\rho l}{A}}$$
Where $$\rho$$is the specific resistance of the conductor.
$$l$$ is the length of a conductor
A is the area of the cross section of a conductor.
Heating effect is produced in a conductor when current flows in a conductor for some interval of time. The basic cause of the heating effect is resistance; more is the resistance more is the heating effect. So it can be represented by the relation -
$$H=I^{2}Rt$$ (Equation 1)
According to ohm’s Law , $$V=IR$$
$$\Rightarrow I={\displaystyle \frac{V}{R}}$$
Put the value of I in Equation 1,we get
$$H={\displaystyle \frac{V^{2}t}{R}}$$
Heating effect of current in a conductor can be given by the above relation. It shows that heating effect is directly proportional to applied voltage and time and inversely proportional to resistance of a conductor.
According to the question, Resistance of wire gets halved then heating effect also gets affected.
As we know that
$$H={\displaystyle \frac{V^{2}t}{R}}$$
Since here applied voltage is constant. So above equation can be written as :-
$$\Rightarrow$$$$H\propto {\displaystyle \frac{1}{R}}$$(Equation1)
So according to this relation heating effect of current is inversely proportional to resistance.

When resistance gets halved so new resistance will be represented by $$R^{{}^{\prime }}$$.
$$\Rightarrow$$$$R^{{}^{\prime }}={\displaystyle \frac{R}{2}}$$
Let us assume R’ be the new resistance of the conductor.
Apply this logic to Equation 1.
Then new heating effect will calculate by -
$$\Rightarrow$$$${\displaystyle \frac{H^{{}^{\prime }}}{H}}={\displaystyle \frac{R}{R^{{}^{\prime }}}}$$
$$\Rightarrow$$$${\displaystyle \frac{H^{{}^{\prime }}}{H}}=2$$
 $$\therefore$$$$H^{{}^{\prime }}=2H$$
Heating effect of a conductor gets doubled when resistance of conductor gets halved.

So, the correct answer is “Option B”.

Note: A battery or a cell is a source of electrical energy also defined as a source of direct current. The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to flow the current through a resistor or a system of resistors connected to the battery.