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The Passive Sign Convention 1 Choose an arbitrary direction of current ow 2 Label polarities of all active elements (voltage and current sources)
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3 Assign polarities to all passive elements (resistors and other loads); for passive elements, current always ows into the positive terminal 4 Compute the power dissipated by each element according to the following rule: If positive current ows into the positive terminal of an element, then the power dissipated is positive (ie, the element absorbs power); if the current leaves the positive terminal of an element, then the power dissipated is negative (ie, the element delivers power)
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EXAMPLE 24 Use of the Passive Sign Convention
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Apply the passive sign convention to the circuit of Figure 214
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Known Quantities: Voltages across each circuit element; current in circuit Find: Power dissipated or generated by each element Schematics, Diagrams, Circuits, and Given Data: Figure 215(a) and (b) The voltage drop across Load 1 is 8 V, that across Load 2 is 4 V; the current in the circuit is 01 A Assumptions: None Analysis: Following the passive sign convention, we rst select an arbitrary direction for the current in the circuit; the example will be repeated for both possible directions of current ow to demonstrate that the methodology is sound
+ + vB
Load 1 Load 2 v1 + v2 Load 1 Load 2 i vB = 12 V i = 01 A (a) v1 + v2 + v1 = 8 V v2 = 4 V Load 1 + i vB = 12 V i = 01 A (b) v1 = 8 V v2 = 4 V Load 2 + vB
1 Assume clockwise direction of current ow, as shown in Figure 215(a) 2 Label polarity of voltage source, as shown in Figure 215(a); since the arbitrarily chosen direction of the current is consistent with the true polarity of the voltage source, the source voltage will be a positive quantity 3 Assign polarity to each passive element, as shown in Figure 215(a) 4 Compute the power dissipated by each element: Since current ows from to + through the battery, the power dissipated by this element will be a negative quantity: PB = vB i = (12 V) (01 A) = 12 W that is, the battery generates 12 W The power dissipated by the two loads will be a positive quantity in both cases, since current ows from + to : P1 = v1 i = (8 V) (01 A) = 08 W P2 = v2 i = (4 V) (01 A) = 04 W Next, we repeat the analysis assuming counterclockwise current direction 1 Assume counterclockwise direction of current ow, as shown in Figure 215(b) 2 Label polarity of voltage source, as shown in Figure 215(b); since the arbitrarily chosen direction of the current is not consistent with the true polarity of the voltage source, the source voltage will be a negative quantity
2
Fundamentals of Electric Circuits
3 Assign polarity to each passive element, as shown in Figure 215(b) 4 Compute the power dissipated by each element: Since current ows from + to through the battery, the power dissipated by this element will be a positive quantity; however, the source voltage is a negative quantity: PB = vB i = ( 12 V) (01 A) = 12 W that is, the battery generates 12 W, as in the previous case The power dissipated by the two loads will be a positive quantity in both cases, since current ows from + to : P1 = v1 i = (8 V) (01 A) = 08 W P2 = v2 i = (4 V) (01 A) = 04 W
Comments: It should be apparent that the most important step in the example is the
correct assignment of source voltage; passive elements will always result in positive power dissipation Note also that energy is conserved, as the sum of the power dissipated by source and loads is zero In other words: Power supplied always equals power dissipated
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