Thank you very much as we will learn later because of

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Unformatted text preview: ollowing circuit. Your answer should be in terms of is , R1 , R2 and R3 . R1 10 R2 20 R3 50 is 1A 15.18 Solution Steps 1 and 2. Assume currents and voltage drops for all circuit elements. Follow the passive and active sign conventions for passive elements and sources respectively (optional). We have done this in the figure. Notice that the current source is connected in parallel with R2 , so its voltage drop should be the same as in R2 . You don’t need to assume an additional one. Step 3. Apply KCL and KVL The “useful” nodes to apply KCL are: B: is i1 i2 C: i1 i3 (You should have seen this since R1 and R3 are in series). Although we can apply KVL in any closed loop we want, it is simpler to apply it in the two “simple” loops “L1” and “L2” as noted on the figure: L1 (start from A): V2 V2 0 . Thank you very much! As we will learn later because of this trivial result, a single element in parallel with a current source is not considered a loop. L2 (start from B): V1 V3 V2 0 Now it’s just a matter of solving these equations. Using Ohm’s law the L2 equation beco...
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This note was uploaded on 03/06/2014 for the course ECE 201 taught by Professor All during the Spring '08 term at Purdue University-West Lafayette.

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