Solution of Practice problems for Test1

Solution of Practice problems for Test1 - Solution of...

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Solution of Practice problems for Test1 1. (2.11 in text) In the circuit of Fig. 1, calculate V 1 and V 2 . Fig. 1 11 15 0 6 V VV −++= ⎯⎯ →= 22 52 0 3 V −+ + = 2. (2.13 in text) For the circuit in Fig. 2 use KCL to find the branch currents I 1 to I 4 . I 2 7A I 4 1 2 3 4 4A I 1 3A I 3 Fig. 2 At node 2, 37 0 1 0 ++ = ⎯→ ⎯= II A At node 1, I I A 12 1 2 1 2 += = At node 4, 24 2 4 2 44 =+ =−= A At node 3, 77 2 5 43 3 = I A Hence, IA I A I A I A 3 4 12 10 5 2 == = = − ,, , + V 1 _ + V 2 _ + 5 V _ + 1 V 2 V 2A
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3. (2.25 in text) V 0 = 5 x 10 -3 x 10 x 10 3 = 50V Using current division, I 20 = + = ) 50 01 . 0 ( 20 5 5 x 0.1 A V 20 = 20 x 0.1 kV = 2 kV p 20 = I 20 V 20 = 0.2 kW 4. (2.39 in text) Evaluate R eq for each of the circuits shown in Fig. 4. Fig. 4 (a) We note that the top 2k-ohm resistor is actually in parallel with the first 1k-ohm resistor. This can be replaced (2/3)k-ohm resistor. This is now in series with the second 2k-ohm resistor which produces a 2.667k-ohm resistor which is now in parallel with the second 1k-ohm resistor. This now leads to, R eq = [(1x2.667)/3.667]k = 727.3 . (b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor. This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in parallel with the 4k-ohm resistor producing, R eq = [(4x12)/16]k = 3 k . 2 k Ω 2 k Ω 1 k Ω c c 1 k Ω (a) 4 k Ω 6 k Ω 12 k Ω 12 k Ω c c (b)
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5. (3.15 in text) Apply nodal analysis to find i o
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This note was uploaded on 10/12/2010 for the course ECE 2004 at Virginia Tech.

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Solution of Practice problems for Test1 - Solution of...

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