240_homework_4_soln

240_homework_4_soln - ECE 240 Electrical Engineering...

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ECE 240 – Electrical Engineering Fundamentals Solutions to Homework #4 1. Short circuiting the two voltage sources: i = 2 2 + 2 (4) = 2 v o ' = 2(2) = 4 Opening the current source and shorting the 12 V source: v o '' = 2 2 + 2 (24) = 12 Opening the current source and shorting the 24 V source: v o ''' = 2 2 + 2 (12) = 6 v o = 4 + 12 6 = 10 V 2. Shorting the 12 V source on the right, the top and left 2 Ω resistors are in parallel for 1 Ω and the middle and right 2 Ω resistors are in parallel for 1 Ω . Thus: v o ' = 1 1 + 1 (12) = 6 Shorting the 12 V source on the left, the middle and left 2 Ω resistors are in parallel for 1 Ω and the top and right 2 Ω resistors are in parallel for 1 Ω . Thus: v o ' = 1 1 + 1 (12) = 6 v o = 6 + 6 = 12 V 3. Shorting the 24V source, the 6A divides between the 2 Ω and 4 Ω (2 + 2 in series), thus: I o ' = 4 4 + 2 (6) = 4 . Opening the 6A source, the 24V is across a series string of 3 – 2 Ω resistors so: I o '' = 24 6 = 4 . Thus, I o = 4 + 4 = 8A 4. Replacing everything but the middle 2 Ω resistor with its Thevenin equivalent: R th is found by shorting the two voltage sources and opening the current source. Thus, R th = 2 Ω . Open circuiting the terminals across the middle 2 Ω , the 4A flows through the top 2 Ω resulting in 8V across that resistor. Thus, V oc = 8 + 24 – 12 = 20V. Placing the 2 Ω resistor back on this Thevenin circuit we obtain: v o = 2 2 + 2 (20) = 10 V 5. Replacing everything but the top 2 Ω resistor with its Thevenin equivalent: R th is found by shorting the two voltage sources. Thus all three 2
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240_homework_4_soln - ECE 240 Electrical Engineering...

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