Exam

JEE Main

Year

2024

Date

5 April

Shift

Evening Shift

Subject

Physics

Chapter

Current Electricity

JEE Main 2024 · 5 April · Evening Shift · Physics · Current Electricity

The ratio of heat dissipated per second through the resistance 5...

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

Step-by-step solution

$$\begin{aligned} & \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{10}{5}=\frac{2}{1} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{i}_1^2 \mathrm{R}_1}{\mathrm{i}_2^2 \mathrm{R}_2}=\left(\frac{2}{1}\right)^2 \times \frac{5}{10}=\frac{2}{1}\end{aligned}$$