Problem 4.36 Part A
Use the mesh current method to find the power dissipated in the 2 Ohm resistor in the following circuit.
Express your answer with appropriate units.
Solution: There are four meshes in this circuit. We arbitrarily define all the mesh currents clockwise Ia is the clockwise mesh current in the mesh containing the 230V source. Ib is the clockwise mesh current in the resistive mesh at the top of the circuit. Ic is the clockwise mesh current in the resistive mesh in the middle of the circuit. Id is the clockwise mesh current in the resistive mesh at the bottom of the circuit. All the currents in the branch 1 ohm resistors (those in intersection meshes) are either defined going downward or left to right. For example, the mesh currents in the intersection meshes for the top right most mesh are: I (1 ohm) = Ia – Ib (because the net current is flowing downward) I (1 ohm) = Ic-­‐ Ib (because the net current is flowing left to right. Writing KVL using these four mesh currents gives: 230V – 1(Ia – Ib) -­‐ 1 (Ia – Ic) – 1 (Ia – Id) = 0 1 (Ia – Ib) – 1 Ib – 5 Ib + 1 (Ic – Ib) = 0 1 (Ia – Ic) – 1 (Ic – Ib) – 2 Ic + 1 (Id – Ic) = 0 1 (Ia – Id) – 1 (Id – Ic) – 5 Id -­‐ 1 Id = 0 Solving for the four mesh currents gives: Ia = 95A Ib = 15A Ic = 25A Id = 15A Then the power dissipated in the 2 ohm resistor is simply: Ic2 * R = 252 * 2 = 1250 W Part B
Find the power supplied by the 230V Source
Solution: 230*Ia = 21,850 W