A resistor develops 500 J of thermal energy in 20 s, when a current of 1.5 A is passed through it. I

Q: Why don't we need to calculate the value of the resistance $R$?

Because we are comparing two states of the exact same resistor, the resistance $R$ is constant. When we set up a ratio between the two states, the $R$ terms cancel each other out completely.

Q: What if the time was also doubled along with the current?

If the time was doubled, the heat would increase even more. Since $H = I^2Rt$, doubling the current gives a factor of $2^2 = 4$, and doubling the time gives another factor of $2$. The total heat would be $4 \times 2 = 8$ times the original heat.

Q: Can we use the formula $H = \frac{V^2}{R} t$ instead?

You could, but it would be much harder here. We are given the current, not the voltage. If you change the current in a fixed resistor, the voltage across it also changes (since $V = IR$). Using $H = I^2Rt$ is the most direct path since $I$ is explicitly given.

Q: Why does doubling the current quadruple the heat?

This is the mathematical nature of a quadratic relationship. Because the current term $I$ is squared in Joule's Law ($I^2$), any change to the current is squared in its effect on the heat. $(2I)^2 = 4I^2$.

Q: Is this thermal energy lost forever?

In the context of the electrical circuit, yes. The electrical potential energy is converted into thermal energy (heat), which dissipates into the surrounding environment. This is the principle behind electric heaters and toasters!

Question

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

Options:

(A) 1500 J

(B) 1000 J

(C) 500 J

(D) 2000 J

Sigma Percentile · Accepted Answer

### The Setup: A Resistor in Action

Imagine a simple electrical circuit with a resistor. When current flows through this resistor, it encounters opposition, and this electrical friction generates heat.

In our specific problem, a current of $1.5 	ext{ A}$ flows through a resistor for $20 	ext{ s}$, and it dissipates $500 	ext{ J}$ of thermal energy. The question asks: what happens to the heat generated if we double the current to $3 	ext{ A}$ for the exact same duration?

### The Master Equation: Joule's Law

To understand the relationship between current and heat, we turn to **Joule's Law of Heating**. This fundamental principle states that the heat energy ($H$) produced by a resistor is given by the equation:

$$H = I^2 R t$$

Here, $I$ is the current, $R$ is the resistance, and $t$ is the time the current flows.

### The Power of Proportionality

Look closely at the variables in our scenario. We are using the exact same resistor, which means the resistance $R$ is constant. The time $t$ is also kept constant at $20 	ext{ s}$.

Because $R$ and $t$ do not change, the heat generated is directly proportional to the square of the current:

$$H \propto I^2$$

This is a crucial insight! It means we don't even need to know the actual value of the resistance to solve the problem. We can simply set up a ratio comparing the initial state (State 1) to the final state (State 2):

$$\frac{H_1}{H_2} = \left(\frac{I_1}{I_2}ight)^2$$

### The Final Calculation

Now, let's plug in the numbers we were given. The initial heat $H_1$ is $500 	ext{ J}$, the initial current $I_1$ is $1.5 	ext{ A}$, and the final current $I_2$ is $3 	ext{ A}$:

$$\frac{500}{H_2} = \left(\frac{1.5}{3}ight)^2$$

Notice how beautifully the numbers simplify. $1.5$ is exactly half of $3$, so the fraction inside the parentheses reduces to $\frac{1}{2}$. Squaring this fraction gives us $\frac{1}{4}$:

$$\frac{500}{H_2} = \frac{1}{4}$$

Finally, a quick cross-multiplication reveals our answer:

$$H_2 = 500 	imes 4 = 2000 	ext{ J}$$

**The core takeaway:** Because heat is proportional to the *square* of the current, doubling the current doesn't just double the heat—it quadruples it!

Animated Solution for Physics - Current Electricity: A resistor develops 500 J of thermal energy in 20 s, when a current of 1.5 A is passed through it. If the current is increased from 1.5 A to 3 A, what will be the energy developed in 20 s?

Select Answer:

Visualized Solution

Visualizing the Setup

Joule's Law of Heating

Identifying Constants

Setting Up the Ratio

Substituting Values

Simplifying the Ratio

Final Calculation

The Way Forward

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Setup

The Master Equation

The Power of Proportionality

The Final Calculation

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