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(B) Since the voltmeter is ideal, it has infinite resistance, so no current flows through the branch containing the $10 \, \Omega$ resistor and the vo

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$4$

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(B) Since the voltmeter is ideal, it has infinite resistance, so no current flows through the branch containing the $10 \, \Omega$ resistor and the voltmeter.The circuit simplifies to a $20 \, \Omega$ resistor in series with a parallel combination of $(9 \, \Omega + 6 \, \Omega) = 15 \, \Omega$ and $30 \, \Omega$.The equivalent resistance of the parallel part is $R_p = \frac{15 	imes 30}{15 + 30} = \frac{450}{45} = 10 \, \Omega$.The total resistance of the circuit is $R_{eq} = 20 \, \Omega + 10 \, \Omega = 30 \, \Omega$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{30 \, V}{30 \, \Omega} = 1 \, A$.Using the current divider rule, the current through the branch with the $9 \, \Omega$ resistor is $I_1 = I 	imes \frac{30}{15 + 30} = 1 	imes \frac{30}{45} = \frac{2}{3} \, A$.The power consumed in the $9 \, \Omega$ resistor is $P = I_1^2 R = (\frac{2}{3})^2 	imes 9 = \frac{4}{9} 	imes 9 = 4 \, W$.

In the circuit shown in the figure, if the ammeter and voltmeter are ideal, then the power consumed in the $9 \, \Omega$ resistor will be .......... $W$.

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