NEET 2022 Physics Thermal Energy in Resistors MCQ Question

To solve the problem of finding the ratio of thermal energy developed in two resistors of resistances $R_1 = 100 \, \Omega$ and $R_2 = 200 \, \Omega$ connected in parallel, we need to understand the relationship between resistance, current, and thermal energy.

Step 1: Understanding the Circuit Configuration

When resistors are connected in parallel, the voltage across each resistor is the same. Let’s denote the voltage across the parallel combination as $V$.

Step 2: Calculate the Current through Each Resistor

Using Ohm's Law $I = \frac{V}{R}$, we can calculate the current through each resistor:

1. For the 100 Ω resistor: $I_1 = \frac{V}{R_1} = \frac{V}{100}$

2. For the 200 Ω resistor: $I_2 = \frac{V}{R_2} = \frac{V}{200}$

Step 3: Calculate the Thermal Energy Developed in Each Resistor

The thermal energy ($Q$) developed in a resistor is given by the formula: $Q = I^2 R t$ where $I$ is the current through the resistor, $R$ is its resistance, and $t$ is the time duration.

Now, let's calculate the thermal energy for each resistor:

1. For the 100 Ω resistor:

   $$Q_1 = I_1^2 R_1 t = \left(\frac{V}{100}\right)^2 \cdot 100 \cdot t = \frac{V^2}{10000} \cdot 100 \cdot t = \frac{V^2 t}{100}$$

2. For the 200 Ω resistor:

   $$Q_2 = I_2^2 R_2 t = \left(\frac{V}{200}\right)^2 \cdot 200 \cdot t = \frac{V^2}{40000} \cdot 200 \cdot t = \frac{V^2 t}{200}$$

Step 4: Calculate the Ratio of Thermal Energies

Now we can find the ratio of the thermal energies developed in the 100 Ω resistor to the 200 Ω resistor:

$$\text{Ratio} = \frac{Q_1}{Q_2} = \frac{\frac{V^2 t}{100}}{\frac{V^2 t}{200}} = \frac{200}{100} = 2$$

Therefore, the ratio of thermal energy developed in the 100 Ω resistor to that in the 200 Ω resistor is:

$Q_1 : Q_2 = 2 : 1$

Conclusion

The correct answer is indeed B) 2 : 1.

Clarifying Incorrect Options

• Option A (1 : 4): This would imply that the 100 Ω resistor produces significantly less thermal energy compared to the 200 Ω resistor, which is incorrect based on our calculations.
• Option C (1 : 2): This would suggest that the 100 Ω resistor produces half the thermal energy of the 200 Ω resistor, which also contradicts our findings.
• Option D (2 : 1): This is misleading in phrasing since it suggests that the 200 Ω resistor produces twice the thermal energy, which is the opposite of what we have derived.

Thus, the ratio of thermal energy developed in the 100 Ω resistor to that in the 200 Ω resistor is correctly given by option B: 2 : 1.