A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when t

Q: Why did we ignore the resistance of the resistor?

We didn't ignore it; we eliminated it mathematically! Since the same resistor is used in both cases, its resistance $R$ is constant. By taking the ratio of the two heat equations, $R$ cancels out, saving us an extra calculation step.

Q: Could we have calculated the resistance first and then found the final heat?

Absolutely! You could use $H_1 = I_1^2 R t_1$ to find $R = 192 / (4^2 \times 1) = 12\ \Omega$. Then, use this $R$ in the second case: $H_2 = 8^2 \times 12 \times 5 = 3840\text{ J}$. Both methods yield the exact same result, but the ratio method is faster.

Q: Why is heat proportional to the square of the current?

Electrical power is given by $P = I^2 R$. Since heat energy is power multiplied by time ($H = P \times t$), the heat dissipated depends on the square of the current. Doubling the current quadruples the power!

Q: What if the voltage was doubled instead of the current?

If voltage was given instead of current, we would use the alternate form of Joule's law: $H = \frac{V^2}{R} t$. The ratio method would still work perfectly.

Q: Does the temperature change affect the resistance in real life?

Yes, in real life, as a resistor heats up, its resistance usually increases. However, in standard textbook problems like this, we assume the resistance remains constant unless a temperature coefficient is explicitly provided.

Question

A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s is ........ J.

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are observing a simple electrical circuit with a single resistor. Initially, a current of $4	ext{ A}$ flows through this resistor for exactly $1	ext{ s}$. During this brief moment, the resistor dissipates $192	ext{ J}$ of thermal energy.

Now, the experiment is altered. The current is doubled to $8	ext{ A}$, and the duration is extended to $5	ext{ s}$. Our goal is to determine the new amount of thermal energy dissipated under these modified conditions.

### The Master Equation
To connect all these physical quantities—heat, current, resistance, and time—we rely on **Joule's Law of Heating**. The law states that the heat $H$ dissipated by a resistor is directly proportional to the square of the current $I$, the resistance $R$, and the time $t$.

Mathematically, this is expressed as:
$$H = I^2 R t$$

This equation is the cornerstone of understanding power dissipation in electrical circuits.

### Setting Up the Ratio
Since the same resistor is used in both scenarios, its resistance $R$ remains constant. This is a crucial insight! Instead of calculating the resistance explicitly, we can set up a ratio of the final heat $H_2$ to the initial heat $H_1$.

$$\frac{H_2}{H_1} = \frac{I_2^2 R t_2}{I_1^2 R t_1}$$

Notice how the resistance $R$ beautifully cancels out from the numerator and the denominator. This leaves us with a much simpler and elegant relation:

$$\frac{H_2}{H_1} = \left(\frac{I_2}{I_1}ight)^2 \left(\frac{t_2}{t_1}ight)$$

### Final Calculation
Now, we simply substitute our known values into this simplified equation. We know that $H_1 = 192	ext{ J}$, $I_1 = 4	ext{ A}$, $I_2 = 8	ext{ A}$, $t_1 = 1	ext{ s}$, and $t_2 = 5	ext{ s}$.

$$\frac{H_2}{192} = \left(\frac{8}{4}ight)^2 	imes \left(\frac{5}{1}ight)$$

Simplifying the terms inside the parentheses:

$$\frac{H_2}{192} = (2)^2 	imes 5$$

$$\frac{H_2}{192} = 4 	imes 5 = 20$$

This tells us that the new heat dissipated is exactly 20 times the initial heat. Finally, we multiply to find $H_2$:

$$H_2 = 192 	imes 20 = 3840	ext{ J}$$

**The total thermal energy dissipated in the second case is 3840 J.**

Animated Solution for Physics - Current Electricity: A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s is ........ J.

Enter Numerical Value:

Visualized Solution

Initial Conditions

Final Conditions

Joule's Law of Heating

Ratio of Heat Dissipated

Simplified Ratio

Substituting Values

Calculation

Final Answer

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

Analyzing the Setup Imagine you are observing a simple electrical circuit with a single resistor

The Master Equation To connect all these physical quantities—heat, current, resistance, and time—we rely on Joule's Law of Heating

Setting Up the Ratio Since the same resistor is used in both scenarios, its resistance $R$ remains constant

Final Calculation Now, we simply substitute our known values into this simplified equation

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