homework 1 solutions

# homework 1 solutions - ,Berkeley EE42/100 Spring2013...

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University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 1 Solution 1. 2. (a) a b R eq 32 Ω 8 Ω 8 Ω 16 Ω 4 Ω a b R eq 32 Ω 16 Ω 4 Ω 16 Ω a b R eq 32 Ω 8 Ω 4 Ω a b R eq 40 Ω 4 Ω a b R eq 4 Ω // 40 Ω = 3.63 Ω

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(b) 3. 4. Start by arbitrarily labeling the all the currents in the branches and all the voltages across each resistor
a) Applying KVL to Loop 1: 10 – V1 + V2 +V3 +15 = 0 Furthermore, by using Ohm’s Law, the voltages can be expressed in terms of the associated currents and resistances. The KVL Loop1 equation can be re written as: 10 – (I 1 * 5) + (I 2 * 3) + (I 2 * 3) + 15 = 0 – (5 * I 1 ) + (6 * I 2 ) + 5 = 0 (1) Applying KVL to Loop 2: V3 – V2 + V5 + V4 = 0 ( I 2 *3) – (I 2 *3) + ( I X * 12) + ( I X * 6)= 0 ; Note the negative sign associated with I X 6*I 2 – 18*I X = 0 (2) Applying KCL at node C: I 1 + I 2 – I X = 0

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I X = I 1 + I 2 (3) Substitute (3) into (2): 6*I 2 – 18*(I 1 + I 2 ) = 0 18*I 1 – 24*I 2 = 0 (4) Simultaneously solve (4) and (1): I 1 = ଵ଴ ଵଽ = 0.5263 A Plug I 1 back into (4) yields: I 2 = 0.3947 A Using the values of I 1 and I
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