JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 17)
Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is :
1 : 1
1 : 27
1 : 3
3 : 1
Explanation
For parallel connection, potential difference is same $(v)$
$$ \begin{aligned} & P_{1}=\left(\frac{v^{2}}{R_{1}}\right) \\\\ & P_{2}=\left(\frac{v^{2}}{R_{2}}\right) \\\\ & \frac{P_{1}}{P_{2}}=\frac{H_{1}}{H_{2}}=\left(\frac{R_{2}}{R_{1}}\right)=\frac{3 R}{R}=(3: 1) \end{aligned} $$
$$ \begin{aligned} & P_{1}=\left(\frac{v^{2}}{R_{1}}\right) \\\\ & P_{2}=\left(\frac{v^{2}}{R_{2}}\right) \\\\ & \frac{P_{1}}{P_{2}}=\frac{H_{1}}{H_{2}}=\left(\frac{R_{2}}{R_{1}}\right)=\frac{3 R}{R}=(3: 1) \end{aligned} $$
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