ECE35WIN09 Quiz 2 Solutions

ECE35WIN09 Quiz 2 Solutions - 1-I 2 ) = 1.75A The...

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QUIZ 1 ECE35 Win’09 NAME ___ SOLUTIONS _________ SID ______________________ In the familiar circuit at right, what is the power supplied by each of the two current sources? (1) 3A source P supplied = _____________ (2) 1.25A source P supplied = _____________ As in the HW problem 3.5-1, we need to simplify the circuit to find the voltage across the current sources. Since two 8V sources are equal and opposed, they cancel; that branch is equivalent to a simple 2 Ω resistor. The current sources, I 1 (3A) and I 2 (1.25A), are also opposed and result in a net current of (I
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Unformatted text preview: 1-I 2 ) = 1.75A The equivalent circuit is two 2 Ω resistors in parallel (equivalent to a single 1 Ω resistor) driven by a 1.75A source: the voltage developed is (I 1-I 2 )R eq = 1.75V, positive at the top. The 3A source is driving current upwards in potential, so must supply P = iV = 3A x 1.75V = 5.25W. The 1.25A source is driving current to a lower potential, so absorbs power. The power it supplies is P = iV = 1.25A x (-1.75V) = -2.1875W Your answer can be expressed as a product, in case you didn’t have a calculator....
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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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