In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are in the ratio.

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Aicharya · Accepted Answer

Current distribution is as shown in figure

$\therefore$ Powers are $ \begin{aligned} & P_1=I_1^2 R_1=\frac{4}{9} i^2, P_2=I_2^2 R_2=\frac{2}{9} i^2, \text { and } P_3=I_3^2 R_3=3 i^2 \\ & =\frac{27}{9} i^2 . \\ & \Rightarrow \quad P_1: P_2: P_3:: 4: 2: 27 \end{aligned} $

In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are…

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