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(A) First,simplify the circuit. The resistors $R_3 = 60 \, \Omega$ and $R_4 = 30 \, \Omega$ are in parallel. Their equivalent resistance $R_p$ is give

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$R_1$

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(A) First,simplify the circuit. The resistors $R_3 = 60 \, \Omega$ and $R_4 = 30 \, \Omega$ are in parallel. Their equivalent resistance $R_p$ is given by:$R_p = \frac{60 	imes 30}{60 + 30} = \frac{1800}{90} = 20 \, \Omega$.This combination is in series with $R_5 = 20 \, \Omega$,so the equivalent resistance of this branch is $R_{branch} = 20 + 20 = 40 \, \Omega$.Now,this branch is in parallel with $R_2 = 40 \, \Omega$. The equivalent resistance of this parallel part is $R_{eq'} = \frac{40 	imes 40}{40 + 40} = 20 \, \Omega$.Finally,this is in series with $R_1 = 40 \, \Omega$. The total resistance of the circuit is $R_{total} = 40 + 20 = 60 \, \Omega$.The total current from the battery is $I = \frac{V}{R_{total}} = \frac{3.0}{60} = 0.05 \, A$.Power dissipated in $R_1$ is $P_1 = I^2 R_1 = (0.05)^2 	imes 40 = 0.0025 	imes 40 = 0.1 \, W$.The current $I$ splits equally into the two parallel branches ($R_2$ and the branch containing $R_3, R_4, R_5$) because both have a resistance of $40 \, \Omega$. Thus,$I_2 = I_{branch} = 0.025 \, A$.Power in $R_2$ is $P_2 = (0.025)^2 	imes 40 = 0.000625 	imes 40 = 0.025 \, W$.In the branch with $R_3, R_4, R_5$,the current is $0.025 \, A$. Power in $R_5$ is $P_5 = (0.025)^2 	imes 20 = 0.0125 \, W$.The voltage across the parallel combination of $R_3$ and $R_4$ is $V_p = I_{branch} 	imes R_p = 0.025 	imes 20 = 0.5 \, V$.Power in $R_3$ is $P_3 = \frac{V_p^2}{R_3} = \frac{0.5^2}{60} = \frac{0.25}{60} \approx 0.0042 \, W$.Power in $R_4$ is $P_4 = \frac{V_p^2}{R_4} = \frac{0.5^2}{30} = \frac{0.25}{30} \approx 0.0083 \, W$.Comparing all powers,$P_1 = 0.1 \, W$ is the maximum. Therefore,$R_1$ dissipates the most power.

In the circuit shown below,the resistances are given (in $\Omega$) and the battery is assumed ideal with an electromotive force (emf) equal to $3.0 \, V$. The resistor that dissipates the most power is

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