HW2solutions

# HW2solutions - After doing so and labeling the resistor...

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ECE35 Win’09 Homework 2 Solutions P3.3-4 P3.3-7 All of the elements are connected in series. Replace the series voltage sources with a single equivalent voltage having voltage 12 + 12 – 18 = 6 V. Replace the series 15 Ω , 5 Ω and 20 Ω resistors by a single equivalent resistance of 15 + 5 + 20 = 40 Ω . By voltage division P3.3-8 Use voltage division to get Then The power supplied by the dependent source is given by

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P3.4-4 P3.4-9 Each of the resistors is connected between nodes a and b . The resistors are connected in parallel and the circuit can be redrawn like this: Then So P3.4-14 Using KCL Using current division P3.5-1 The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source.

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Unformatted text preview: After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get Apply KVL to the outside loop to get So And so the power supplied by each source: Source Power delivered 8-V voltage source 3-V voltage source 3-A current source 1.25-A current source P3.6-1 P3.6-15 So the circuit is equivalent to Then P3.6-28 Replace the ammeter by the equivalent short circuit and label the current measured by the meter as i m . The 10-Ω resistor at the right of the circuit is in parallel with the short circuit that replaced the ammeter so it’s voltage is zero as shown. Ohm’s law indicates that the current in that 10-Ω resistor is also zero. Applying KCL at the top node of that 10-Ω resistor shows that DP3-3...
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## This note was uploaded on 04/15/2010 for the course ECE ece35 taught by Professor Xiejiang during the Winter '10 term at UCSD.

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HW2solutions - After doing so and labeling the resistor...

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