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e17lecture3

# e17lecture3 - Engineering 17 Lecture 3 Dr L Lagerstrom U.C...

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1 Engineering 17: Lecture 3 Dr. L. Lagerstrom, U.C. Davis What about electrons? Ch. 2 review (sample problem) Resistors in series Resistors in parallel Applying series-parallel simplification Voltage division Current division What about electrons? When we first learn about voltage and current in physics, we often are concerned about which way the actual charge carriers, i.e., electrons in most cases, are flowing. Consider a wire with some voltage across it (e.g., a wire that connects the positive and negative terminals of a battery). By definition , positive charge flows from the positive terminal to the negative terminal. From http://www.allaboutcircuits.com/vol_1/chpt_1/7.html. What about electrons? Since electrons, on the other hand, are negative charges, they will flow from the negative to the positive terminal. From http://www.allaboutcircuits.com/vol_1/chpt_1/7.html. In general it doesn’t matter whether we use conventional flow notation or electron flow notation. But most engineering treatments of circuit analysis use conventional flow notation, i.e., we assign a direction to positive current flow and don’t worry about the actual direction the electrons are going. Example 2.9 The voltage and current at the terminal of the device in (a) below were measured, with results as shown in (b). Construct a circuit model for the device and use it to predict the power the device will deliver to a 10-ohm resistor. Example 2.9 The voltage and current at the terminal of the device in (a) below were measured, with results as shown in (b). Construct a circuit model for the device and use it to predict the power the device will deliver to a 10-ohm resistor. Example 2.9, cont. The voltage and current at the terminal of the device in (a) below were measured, with results as shown in (b). Construct a circuit model for the device and use it to predict the power the device will deliver to a 10-ohm resistor. t t i v 5 30 - = Plotting the data:

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2 Example 2.9, cont. t t i v 5 30 - = Remembering KVL, we see that the voltage equation above is equivalent to a 30 V voltage source in series with a 5 ohm resistor. So our circuit model is: Example 2.9, cont. To find the power this device will deliver to a 10-ohm resistor, we have: KCL tells us that the current in the 5-ohm resistor is the same as the current in the 10-ohm resistor. So using KVL (with OL) around the circuit loop, we have: i i i i 10 5 30 or , 0 10 5 30 + = = + + - So i = 2 A = the current through the 10-ohm resistor. And thus the power delivered to (or absorbed by) it is: W 40 ) 10 ( ) 2 ( 2 2 = = = R i p Resistors in Series Consider the following “series-connected” resistors: Applying KCL to each node in the circuit we find: 7 6 5 4 3 2 1 i i i i i i i i s = - = - = = = - = = Thus we can redraw the circuit with just a single current: Resistors in Series, cont.
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