JEE MAIN - Physics (2024 - 5th April Evening Shift - No. 14)

The ratio of heat dissipated per second through the resistance $$5 \Omega$$ and $$10 \Omega$$ in the circuit given below is:

JEE Main 2024 (Online) 5th April Evening Shift Physics - Current Electricity Question 17 English

4 : 1
2 : 1
1 : 2
1 : 1

Explanation

To determine the ratio of heat dissipated per second through resistances $ 5 \Omega $ and $ 10 \Omega $ in the given parallel circuit, let's break it down step by step.

Understanding the Problem

When resistors are connected in parallel:

  1. The voltage across each resistor is the same.

  2. The power dissipated in each resistor can be found using the formula:

$ P = \frac{V^2}{R} $

Applying the Power Formula in Parallel Resistors

Given:

  • $ R_1 = 5 \Omega $

  • $ R_2 = 10 \Omega $

Let $ V $ be the voltage across the resistors.

Power Dissipated in Each Resistor

For the $ 5 \Omega $ resistor:

$ P_{5} = \frac{V^2}{5} $

For the $ 10 \Omega $ resistor:

$ P_{10} = \frac{V^2}{10} $

Finding the Ratio of Powers

Now, let's find the ratio of the power dissipated through the $ 5 \Omega $ resistor to the power dissipated through the $ 10 \Omega $ resistor:

$ \frac{P_{5}}{P_{10}} = \frac{\frac{V^2}{5}}{\frac{V^2}{10}} = \frac{10}{5} = 2 $

So, the ratio of power dissipated through the $ 5 \Omega $ resistor to the $ 10 \Omega $ resistor is $ 2:1 $.

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