A potential difference of $20{\text{V}}$ is needed to make a current of $0.05{\text{A}}$ flow through a resistor. What potential difference is needed to make a current of $300{\text{mA}}$ flow through the same resistor?
(A) \[60{\text{V}}\]
(B) $120{\text{V}}$
(C) \[40{\text{V}}\]
(D) $150{\text{V}}$

Hint: To solve this question, we need to use the Ohm’s law to find out the resistance of the given resistor from the given values of the potential difference and the current. Then, using the same law, we can determine the potential difference for the given value of the current through the resistor.

Formula used: The formula used to solve this question is given by
$V = IR$, here $V$ is the potential difference across a resistor of resistance $R$ through which a current of $I$ is flowing.

Complete step-by-step solution:
From the Ohm’s law we know that the potential drop across a resistor is directly proportional to the current flowing through it. The constant of this proportionality is known as the resistance of the resistor. So the relation between the current and the potential difference can be given by
$V = IR$.................(1)
According to the question, a potential difference of $20{\text{V}}$ is needed to make a current of $0.05{\text{A}}$ flow through the given resistance. So we can write
$20 = 0.05R$
$R = \dfrac{{20}}{{0.05}}\Omega $
On solving we get
$R = 400\Omega $.............(2)
Now, the current through the resistor is equal to $300{\text{mA}}$. So we have
$I = 300{\text{mA}}$
We know that $1{\text{mA}} = {10^{ - 3}}{\text{A}}$. So we can write
$I = 300 \times {10^{ - 3}}{\text{A}}$
On solving we get
$I = 0.3{\text{A}}$.............(3)
Substituting (2) and (3) in (1) we get
$V = 0.3 \times 400$
$ \Rightarrow V = 120{\text{V}}$
Thus, the required value of the potential difference is equal to $120{\text{V}}$.

Hence, the correct answer is option B.

Note: Do not forget to write the value of the current given in the question into the SI unit. The value given is in milli amperes. So it has to be converted into amperes.