Question

If the resistance of a circuit is halved and the potential difference is kept constant, then the current will become:
(A) $4$ times
(B) $8$ times
(C) double
(D) half

Answer 1

In this question, it is given that the relation between three parameters resistance, potential difference, and current. The equation which shows the relationship between the current, resistance, and the potential difference is Ohm’s law, by using this law the effect of current when resistance is halved, can be determined.

Formula used:
Ohm’s law of electricity shows the relation between the current, resistance and the potential difference is,
$V = IR$
Where $V$ is the voltage in the circuit, $I$ is the current in the circuit and $R$ is the resistance in the circuit.

Complete step by step answer:
By Ohm’s law,
$V = IR\,...................\left( 1 \right)$
Here, we need to find the current, so by keeping the current $I$ in one side and the other terms in other side, then
$\Rightarrow I = \dfrac{V}{R}$
Now, the resistance of the circuit is halved, so the resistance $R$ will become $\dfrac{R}{2}$, then
$\Rightarrow I = \dfrac{V}{{\left( {\dfrac{R}{2}} \right)}}$
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow I = \dfrac{{2V}}{R}$
The above equation is written as,
$\Rightarrow I' = \dfrac{{2V}}{R}$
Here $I'$ is the effect of current when the resistance is halved.
By substituting the term $\dfrac{V}{R}$ as the current $I$, then the above equation is written as,
$\Rightarrow I' = 2I$
Thus, the above equation shows the effect of the current when the resistance is halved.

Hence option (C) is the correct answer.

Note:
By ohm's law, when the potential difference or voltage is constant, then the current is inversely proportional to the resistance. So, if resistance decreases then the current will increase, if the resistance increases then the current will decrease.