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(D) Let the current through resistor $R$ be $I_1$ and the current through resistor $2R$ be $I_2$. Since $R$ and $2R$ are connected in parallel,the pot

Vedclass · Accepted Answer

$4/27$

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(D) Let the current through resistor $R$ be $I_1$ and the current through resistor $2R$ be $I_2$. Since $R$ and $2R$ are connected in parallel,the potential difference across them is the same,so $V_{AB} = V_{CD}$.Therefore,$I_1 R = I_2 (2R)$,which gives $I_2 = I_1 / 2$.The total current flowing through the $3R$ resistor is $I_3 = I_1 + I_2 = I_1 + I_1 / 2 = (3/2) I_1$.The power dissipated in resistor $R$ is $P_1 = I_1^2 R$.The power dissipated in resistor $3R$ is $P_3 = I_3^2 (3R) = ((3/2) I_1)^2 (3R) = (9/4) I_1^2 (3R) = (27/4) I_1^2 R$.The ratio of power dissipated in $R$ to that in $3R$ is $\frac{P_1}{P_3} = \frac{I_1^2 R}{(27/4) I_1^2 R} = \frac{4}{27}$.

The ratio of powers dissipated respectively in $R$ and $3R$,as shown in the figure,is:

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