Questions: Employing supernode/nodal analysis techniques as appropriate, determine the power dissipated by the 1 Ω resistor in the circuit of Fig. 4.51.

Solution Steps

• Identify the nodes in the circuit and assign voltage variables to each node.
• Let \( V_1 \) be the voltage at the node between the 1Ω resistor and the 2A current source.
• Let \( V_2 \) be the voltage at the node between the 3Ω resistor and the 4V voltage source.
• Let the ground node be at the bottom of the 2Ω resistor.

• For node \( V_1 \): \[ \frac{V_1 - 4}{1} + 2 + 3 = 0 \] Simplifying: \[ V_1 - 4 + 2 + 3 = 0 \implies V_1 + 1 = 0 \implies V_1 = -1 \text{ V} \]

• For node \( V_2 \): \[ \frac{V_2 - 7}{3} + \frac{V_2 - 4}{3} = 0 \] Simplifying: \[ \frac{V_2 - 7 + V_2 - 4}{3} = 0 \implies \frac{2V_2 - 11}{3} = 0 \implies 2V_2 - 11 = 0 \implies V_2 = 5.5 \text{ V} \]

• The voltage across the 1Ω resistor is \( V_1 - V_2 \): \[ V_{1Ω} = V_1 - V_2 = -1 - 5.5 = -6.5 \text{ V} \]

• Using Ohm's Law, the current through the 1Ω resistor is: \[ I_{1Ω} = \frac{V_{1Ω}}{1Ω} = -6.5 \text{ A} \]

• The power dissipated by the 1Ω resistor is: \[ P_{1Ω} = I_{1Ω}^2 \times 1Ω = (-6.5)^2 \times 1 = 42.25 \text{ W} \]

Final Answer