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(C) Let $i$ be the total current flowing through $R_1$. The current $i$ splits into $i_1$ and $i_2$ through $R_2$ and $R_3$ respectively.Since $R_2$ a

Vedclass · Accepted Answer

$R_2 = R_3$ and $R_1 = \frac{1}{4}R_2$

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(C) Let $i$ be the total current flowing through $R_1$. The current $i$ splits into $i_1$ and $i_2$ through $R_2$ and $R_3$ respectively.Since $R_2$ and $R_3$ are in parallel,the potential difference across them is the same. For equal heat energy $H = \frac{V^2}{R}t$,we must have $R_2 = R_3$.If $R_2 = R_3$,the current $i$ splits equally,so $i_1 = i_2 = \frac{i}{2}$.The heat energy produced in $R_1$ is $H_1 = i^2 R_1 t$.The heat energy produced in $R_2$ is $H_2 = (\frac{i}{2})^2 R_2 t = \frac{i^2 R_2 t}{4}$.For equal heat energy,$H_1 = H_2$,so $i^2 R_1 t = \frac{i^2 R_2 t}{4}$.This gives $R_1 = \frac{R_2}{4}$.

What should be the relationship between the values of the three resistors shown in the figure so that they produce equal heat energy?

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