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Unformatted text preview: e can redraw and label the diagram as follows: 3 Note because all elements are in parallel, we have V = 9 V. Hence the value of i can be calculated 9 using the dependent source. i = 100 = 90 mA. Apply KCL at node A. i + .100 = i1 + i2 + i3 V V + i2 + i + .100 = 50 10 9 9 .090 + .100 = + i2 + 50 10 i2 = .090 + .100 - .180 - .9 i2 = -890 mA a) V = 9 V; i = 90 mA b) In the diagram above, the current through the CCVS (i2 ) was defined in the downward direction. However, our analysis revealed i2 to be negative. Hence, i2 must flow from the negative terminal to the positive terminal (upward). c) i2 = -890 mA 4...
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- Spring '05
- Electrical Engineering