Unformatted text preview: UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department EECS 40/42/100, Spring 2007 Prof. ChangHasnain Homework #1 Solution
Due at 6 pm in 240 Cory on Wednesday, 1/24/07 Total Points: 100 1. Hambley, P1.8 [11 points] (a) The graph of i(t) should look something like this: (The axes labels are t(ms) and i(A)) The important points to note:  The graph is a sine wave. [1 point]  The amplitude is 10A, meaning the peaks are at 10A and 10A. [1 point]  One period ends when the argument to sin() is 2. This occurs when [1 points] 200! " t = 2! , or when (solving for t) t = 0.01s = 10ms (b) i = dq "q= dt 5 ms 0.005 # i(t )dt =
0 # 10 sin(200$ ! t )dt =
0 10 cos(200$ ! t ) 200$ 0 0.005 = 0.0318C [4 points] (c) We can repeat the same calculation replacing 5ms with 10ms, or we can notice that integrating sin() over one full period gives us 0 coulombs. [4 points] 2. Hambley, P1.17 [10 points]
p (t ) = v(t ) " i (t ) = 20e ! t W
" [4 points] [4 points] [2 points] w = # p (t )dt = ! 20e !t
0 " 0 = 20 J w > 0, so the element absorbs energy. 3. Hambley, P1.22 [9 points] The power the cell can deliver is p = vi = 0.12W. In 75 hours, the energy is W = pT = 9Whr = 0.009kWhr So the cost per kWhr is $0.50/0.009kWhr = $55.56/kWhr.
Page 1 of 4 [3 points] [3 points] [3 points] UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department This is 556 times the cost of energy from electrical utilities! 4. Hambley, P1.29 change ia=1A and id=3A [8 points] KCL at the left node: ic = ib ! ia = 3 A ! 1A = 2 A. KCL at the bottom node: ie = ic + ih = 2 A + 4 A = 6 A KCL at the top node: i f = ia + id = 1A + (!3 A) = !2 A . KCL at the right node: i g = i f ! ih = !2 A ! 4 A = !6 A. 5. Hambley, P1.34 [13 points] KCL at the top left: ib = !ia = !2A. [2 point] KCL at the bottom: ic = ia ! id = 2 A ! 1A = 1A. [2 point] KVL around the outside: ! vb + v d ! v a = 0 " vb = v d ! v a = 4V ! 10V = !6V [2 point] KVL around the right loop: v d " vc = 0 ! vc = v d = 4V [2 point] The power for each element is then: PA = v a (!ia ) = !20W [1 point] PB = vb i b = 12W [1 point]
PC = vc i c = 4W [2 points] [2 points] [2 points] [2 points] [1 point] PD = v d i d = 4W [1 point] Adding these all together: PA + PB + PC + PD = !20W + 12W + 4W = 4W = 0W so power is indeed conserved. [1 point] 6. Hambley, P1.45 [10 points]
p (t ) = v(t ) " i (t ) = v(t ) 2 / R = 2.5e !4t W
" " [5 points] w = # p (t )dt = # 2.5e ! 4t dt =
0 0 2.5 ! 4t e !4 " =
0 2.5 = 0.625 J 4 [5 points] 7. Hambley, P1.52 [10 points]
5 (a) v x + 5v x " 10V = 0V ! v x = V = 1.667V 3 v 5 (b) i x = x = V = 0.556 A 3! 9 (c) The power for the independent source is: 50 ' 5 $ P = (10V ) ( i x = (10V )% ! A " = ! W = !5.556W 9 & 9 # The power for the resistor is:
Page 2 of 4 [3 points] [3 points] [1 point] UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department & 5 #& 5 # 25 P = v x ' i x = $ V !$ A ! = W = 0.926W % 3 "% 9 " 27 The power for the dependent source is: & 5 #& 5 # 125 P = 5v x ' i x = 5$ V !$ A ! = W = 4.630W % 3 "% 9 " 27 These sum to zero, so power is conserved.
8. Hambley, P1.55 [9 points] [1 point] [1 point] [1 point] Because the two sources are in parallel, they must by definition have the same voltage. The voltage source forces this to be 10 volts. So the equivalent source is a 10 V voltage source. [9 points] Note that the current source doesn't matter since the voltage source can have any current through it. 9. Hambley, P1.57 [10 points] Ohm's law on the 4 ohm resistor: v x = 4" ! 1A = 4V [3 points] v KCL at the top: i3 = 1A + x = 3 A , where i3 is the current (from lefttoright) through the 2 3 ohm resistor. [3 points] KVL around the outside: " v s + (3!)i3 + 4V + (2!)(1A) = 15V [4 points] 10. Hambley, P2.2 [10 points] (a) Req = 7! + 12! + 2! = 21!
Page 3 of 4 [5 points] UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department (b) Req = 4! + 8! //(7! + 5!) + 5.2! + 18! //(3! + 6!) = 4! + 8! // 12! + 5.2! + 18! // 9! = 4! + 4.8! + 5.2! + 6! = 20! [5 points] Page 4 of 4 ...
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 Spring '07
 ChangHasnain
 Resistor, EECS Department

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