Question

The maximum power rating of a$20\Omega $resistor is $1kW.$ The resistor will melt if it is connected across a DC source of voltage-
(A) $160{\text{V}}$
(B) $140{\text{V}}$
(C) $100{\text{V}}$
(D) $120{\text{V}}$

Answer 1

Hint
To solve this question, we need to find the expression of power dissipated across a resistor in terms of the voltage and the resistance. Then, using the value of the maximum power rating, compute the maximum DC voltage, which the resistance can withstand.
$\Rightarrow P = {I^2}R$
$\Rightarrow V = IR$
$P$ is the power dissipated, $V$ is the voltage, $I$ is the current, and $R$ is the resistance.

Complete step by step answer
We know that the power dissipated across a resistance is given by
$\Rightarrow P = {I^2}R$ (1)
Also, we know from Ohm’s law that
$\Rightarrow V = IR$
Dividing by$R$ both sides, we have
$\Rightarrow I = \dfrac{V}{R}$ (2)
Substituting (2) in (1), we get
$\Rightarrow P = {\left( {\dfrac{V}{R}} \right)^2}R$
On simplifying, we get
$\Rightarrow P = \dfrac{{{V^2}}}{R}$
So, for a given value of the resistance, the power is maximum when the applied DC voltage is maximum.
That is, ${P_{\max }} = \dfrac{{{V_{\max }}^2}}{R}$ (3)
According to the question, the maximum power is
$\Rightarrow {P_{\max }} = 1kW = 1000W$
Also, the resistance of the resistor is given as $R = 20\Omega $
Substituting these in (3), we get
$\Rightarrow 1000 = \dfrac{{{V_{\max }}^2}}{{20}}$
Multiplying by$20$both sides, we get
$\Rightarrow {V_{{{\max }^2}}} = 20000$
Taking square root, we get
$\Rightarrow {V_{\max }} = \sqrt {20000} $
which finally gives
$\Rightarrow {V_{\max }} = 1.4 \times {10^2}{\text{V}}$
Or
$\Rightarrow {V_{\max }} = 140{\text{V}}$
So when a DC source voltage of more than $140{\text{V}}$ is applied, the power dissipated across the given resistance will exceed the maximum power limit of $1kW.$
Therefore, the resistance will melt if it is connected across a DC voltage of more than $140{\text{V}}$
Hence, the correct answer is option (A), $160{\text{V}}$.

Note
We may argue that the answer to this question should be $140{\text{V}}$. As this is the beginning voltage from which the resistor will start melting. But we should not forget the value $160{\text{V}}$, which is more than $140{\text{V}}$. So the resistor will definitely melt at this voltage. If a value more than $140{\text{V}}$ had not been given in the options, then we would have chosen $140{\text{V}}$ as our correct answer.