In the circuit shown,the resistances are give | vedclass

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(B) The power dissipated in a resistor is given by $P = I^2 R$ or $P = V^2 / R$.First,calculate the equivalent resistance of the parallel combination

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$R_2$

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(B) The power dissipated in a resistor is given by $P = I^2 R$ or $P = V^2 / R$.First,calculate the equivalent resistance of the parallel combination of $R_3$ $(60 \ \Omega)$ and $R_4$ $(30 \ \Omega)$: $R_{34} = (60 	imes 30) / (60 + 30) = 1800 / 90 = 20 \ \Omega$.Next,this $R_{34}$ is in series with $R_2$ $(50 \ \Omega)$,so the resistance of this branch is $R_{234} = 50 + 20 = 70 \ \Omega$.Now,$R_{234}$ is in parallel with nothing else,but the total circuit resistance is $R_{total} = R_1 + R_{234} = 50 + 70 = 120 \ \Omega$.The total current from the battery is $I_{total} = V / R_{total} = 3.0 / 120 = 0.025 \ A$.Power in $R_1$: $P_1 = I_{total}^2 R_1 = (0.025)^2 	imes 50 = 0.000625 	imes 50 = 0.03125 \ W$.The voltage across the parallel combination ($R_2$ and $R_{34}$) is $V_p = I_{total} 	imes R_{234} = 0.025 	imes 70 = 1.75 \ V$.Power in $R_2$: $P_2 = V_p^2 / R_2 = (1.75)^2 / 50 = 3.0625 / 50 = 0.06125 \ W$.The current through the $R_3-R_4$ branch is $I_{branch} = V_p / R_{234} = 1.75 / 70 = 0.025 \ A$ (Wait,$I_{branch} = V_p / (R_2 + R_{34})$ is incorrect,the branch current is $I_{branch} = V_p / 70 = 0.025 \ A$ is the total current,the current through $R_2$ is $I_2 = V_p / 50 = 1.75 / 50 = 0.035 \ A$).Actually,$V_p = 3 - (I_{total} 	imes 50) = 3 - (0.025 	imes 50) = 3 - 1.25 = 1.75 \ V$.$I_2 = 1.75 / 50 = 0.035 \ A$. $P_2 = I_2^2 R_2 = (0.035)^2 	imes 50 = 0.001225 	imes 50 = 0.06125 \ W$.$I_{34} = 1.75 / 70 = 0.025 \ A$. Voltage across $R_3$ and $R_4$ is $V_{34} = I_{34} 	imes 20 = 0.025 	imes 20 = 0.5 \ V$.$P_3 = V_{34}^2 / R_3 = (0.5)^2 / 60 = 0.25 / 60 \approx 0.00417 \ W$.$P_4 = V_{34}^2 / R_4 = (0.5)^2 / 30 = 0.25 / 30 \approx 0.00833 \ W$.Comparing $P_1 = 0.03125 \ W$,$P_2 = 0.06125 \ W$,$P_3 \approx 0.00417 \ W$,$P_4 \approx 0.00833 \ W$,$R_2$ dissipates the most power.

In the circuit shown,the resistances are given in ohms and the battery is assumed ideal with $emf$ equal to $3.0 \ V$. The resistor that dissipates the most power is

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