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 \left( \frac{2}{1+2} ight) = \frac{2I}{3}$$I_2 = I \left( \frac{1}{1+2} ight) = \frac{I}{3}$The current flowing through the $3 \Omega$ resistor is the total current $I$.The power developed across a resistor is given by $P = I^2 R$.$P_1 = I_1^2 	imes 1 = \left( \frac{2I}{3} ight)^2 	imes 1 = \frac{4I^2}{9}$$P_2 = I_2^2 	imes 2 = \left( \frac{I}{3} ight)^2 	imes 2 = \frac{2I^2}{9}$$P_3 = I^2 	imes 3 = 3I^2$Now,the ratio $P_1 : P_2 : P_3 = \frac{4I^2}{9} : \frac{2I^2}{9} : 3I^2 = \frac{4}{9} : \frac{2}{9} : 3$.Multiplying by $9$,we get $4 : 2 : 27$.

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

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