In the circuit shown in the figure,the power | vedclass

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(B) Let the total current entering the parallel combination be $i$. The current $i$ splits into $I_1$ and $I_2$ through the $1 \Omega$ and $2 \Omega$

Vedclass · Accepted Answer

$4 : 2 : 27$

Vedclass · Answer

(B) Let the total current entering the parallel combination be $i$. The current $i$ splits into $I_1$ and $I_2$ through the $1 \Omega$ and $2 \Omega$ resistors respectively.Using the current divider rule:$I_1 = i 	imes \frac{2}{1+2} = \frac{2}{3} i$$I_2 = i 	imes \frac{1}{1+2} = \frac{1}{3} i$The current through the $3 \Omega$ resistor is the total current $i$.The power developed across each resistor is given by $P = I^2 R$.For $1 \Omega$ resistor: $P_1 = I_1^2 	imes 1 = (\frac{2}{3} i)^2 	imes 1 = \frac{4}{9} i^2$For $2 \Omega$ resistor: $P_2 = I_2^2 	imes 2 = (\frac{1}{3} i)^2 	imes 2 = \frac{2}{9} i^2$For $3 \Omega$ resistor: $P_3 = i^2 	imes 3 = 3 i^2 = \frac{27}{9} i^2$The ratio of powers is $P_1 : P_2 : P_3 = \frac{4}{9} i^2 : \frac{2}{9} i^2 : \frac{27}{9} i^2 = 4 : 2 : 27$.

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

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