The power dissipated in the 3 Omega resista | vedclass

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Question

(A) First, we identify the circuit structure. The $ 3 \Omega $ and $ 6 \Omega $ resistors are in parallel. Their equivalent resistance $ R_p $ is give

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) First, we identify the circuit structure. The $ 3 \Omega $ and $ 6 \Omega $ resistors are in parallel. Their equivalent resistance $ R_p $ is given by: $ R_p = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \ \Omega $. This $ R_p $ is in series with the $ 2 \ \Omega $ resistor. So, the resistance of this branch is $ R_{branch} = 2 + 2 = 4 \ \Omega $. This branch is in parallel with the $ 4 \ \Omega $ resistor. The total equivalent resistance of the external circuit $ R_{eq} $ is: $ R_{eq} = \frac{4 	imes 4}{4 + 4} = 2 \ \Omega $. Including the internal resistance $ r = 1 \ \Omega $, the total resistance of the circuit is $ R_{total} = R_{eq} + r = 2 + 1 = 3 \ \Omega $. The total current $ I $ from the battery is $ I = \frac{V}{R_{total}} = \frac{4.5}{3} = 1.5 \ A $. The voltage across the parallel combination of the $ 4 \ \Omega $ resistor and the $ 4 \ \Omega $ branch is $ V_{parallel} = I 	imes R_{eq} = 1.5 	imes 2 = 3 \ V $. Since the $ 3 \ \Omega $ and $ 6 \ \Omega $ resistors are in the $ 4 \ \Omega $ branch, the voltage across them is $ 3 \ V $. The current through the $ 3 \ \Omega $ resistor is $ I_3 = \frac{V_{parallel}}{3} = \frac{3}{3} = 1 \ A $. The power dissipated in the $ 3 \ \Omega $ resistor is $ P = I_3^2 	imes R = (1)^2 	imes 3 = 3 \ W $. Wait, re-evaluating the circuit: The $ 3 \ \Omega $ and $ 6 \ \Omega $ are in series with the $ 2 \ \Omega $ resistor? No, looking at the diagram, the $ 3 \ \Omega $ and $ 6 \ \Omega $ are in parallel, and that combination is in series with the $ 2 \ \Omega $ resistor. The total branch resistance is $ 4 \ \Omega $. This branch is in parallel with the $ 4 \ \Omega $ resistor. The voltage across the $ 4 \ \Omega $ branch is $ 3 \ V $. The current through the $ 3 \ \Omega $ resistor branch is $ I_{branch} = \frac{3 \ V}{4 \ \Omega} = 0.75 \ A $. The voltage across the $ 3 \ \Omega $ resistor is $ V_3 = I_{branch} 	imes 3 = 0.75 	imes 3 = 2.25 \ V $. The power dissipated is $ P = \frac{V_3^2}{3} = \frac{2.25^2}{3} = \frac{5.0625}{3} = 1.6875 \ W $. Correction: Let's use current division. $ I_{branch} = 0.75 \ A $. The current through the $ 3 \ \Omega $ resistor is $ I_3 = I_{branch} 	imes \frac{6}{3+6} = 0.75 	imes \frac{6}{9} = 0.75 	imes \frac{2}{3} = 0.5 \ A $. Power $ P = I_3^2 	imes 3 = (0.5)^2 	imes 3 = 0.25 	imes 3 = 0.75 \ W $.

The power dissipated in the $ 3 \Omega $ resistance in the following circuit is: (in $W$)

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