EE302Lecture8 - resistance of (3/2)R. Using the formula P =...

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2 Why? Below is a past exam problem. I g = 0.008 = I 1 + I 2 + I 3 + I 4 I g = 0.008 = 2I 1 + 10I 3 + I 3 + I 4 I g = 0.008 = 20I 4 + 10I 4 + I 4 + I 4 = 32I 4 I 4 = 0.008/32 = 0.00025 I 1 = 20 ( 0.00025 ) = .005 R 1 =Vg/I 1 =4/.005=800 Ω R 2 =Vg/I 2 =4/.0025=1600 Ω R 3 =R 4 =Vg/I 3 =4/.00025=16000 Ω
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7 What is the power in the 4 ohm resistor?
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8 1) parallel R eq of 20 and 30 ohms = = (20)(30)/(20+30) = 600/50 = 12 ohms 2) the 12 ohms is in series with 8 ohms for an equivalent resistance of 20 ohms. 3) then the 20 ohms is in parallel with the 20 ohms on the right and the combination is an equivalent resistance of 10 ohms. 4) then this 10 ohms is in series with the 7 ohms resistor and sums to a final answer of 17 ohms . The parallel R’s have a resistance of R/2 and this R/2 in series with the R connected to the 10 V supply is a total
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Unformatted text preview: resistance of (3/2)R. Using the formula P = 40 watts = V 2 /R eq = 10 2 /(3R/2) and solving for R = 5/3 ohms . 9 The following circuits will be redrawn in class to show how they are easy to solve after being redrawn: Chapter 2, Problem 44. For each of the circuits in Fig. 2.108, obtain the equivalent resistance at terminals a-b . 5 10 20 20 a b Figure 2.108 10 Chapter 2, Problem 45. Find the equivalent resistance at terminals a-b of each circuit in Fig. 2.109. 5 20 25 60 12 15 10 30 Figure 2.109 11 Chapter 3, Problem 34. Determine which of the circuits in Fig. 3.83 can be redraw with no crossing branches. Figure 3.83...
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EE302Lecture8 - resistance of (3/2)R. Using the formula P =...

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