Question

A resistor develops $500J$ of thermal energy in $20s$ when a current of $1.5A$ is passed through it. If the current is increased from$1.5A$ to $3A$, what will be the energy developed in the $20s$.

• A. $500J$
• B. $1000J$
• C. $2000J$
• D. $1500J$

Answer 1

The correct option is C

$2000J$

Step 1. Given Data,

Thermal energy of a resistor $H_1=500J$

Time $T=20s$

Current $I_1=1.5A$

Current $I_2=3A$,

Let $H_2$ be the energy developed in the $20s$.

Step 2. Calculate the value of $H_2$ energy developed in the $20s$.

Using Joule's law of heating, $H=I^{2}RT$

Thermal energy when a current $I_1=1.5A$passed through the rasistor,

$H_1={I_1}^{2}RT$

Thermal energy when a current $I_2=3A$passed through the rasistor,

$H_2={I_2}^{2}RT$

Keeping all other parameters same we are changing current from $1.5A$ to $3A$.

Thus, $\begin{array}{l}\frac{H_1}{H_2}={\left(\frac{I_1}{I_2}\right)}^2\end{array}$

$$\begin{gathered} \Rightarrow H_2=H_{1}x{\left(\frac{I_2}{I_1}\right)}^2 \\ \Rightarrow H_2=500x{\left(\frac{3}{1.5}\right)}^2 \\ \Rightarrow H_2=2000J \end{gathered}$$

The value of $H_2$ energy developed in the $20s$ is $2000J$

Hence, option C is correct.