DeCarlo Ch1 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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PROBLEM SOLUTIONS CHAPTER 1. Solution 1.1. (a) Charge on one electron: -1.6019 × 10 -19 C. This means that charge on 10 13 electrons is: -1.6019 × 10 -6 C. Net charge on sphere is: 1.6019 × 10 -6 C (POSITIVE). Solution 1.2. (a) 1 atom -4.646 × 10 -18 C. By proportionality, 64 g NA atoms. 3.1 g ? atoms 3.1 g 3.1 NA 64 atoms. Total Charge = - 4.646 × 10 - 18 C atom × 3.1 × 6.023 × 10 23 64 atoms =- 1.3554 × 10 5 C (b) Total charge per atom is -4.646 × 10 -18 C. Total charge per electron is –1.6019 × 10 -19 C. Therefore, there are 29 electrons per atom of copper. (c) 0.91 A 0.91 C/s. i = Q t t = Q i = 1.36 × 10 5 0.91 = 1.49 × 10 5 sec. (d) We know there are 3.1 NA 64 = 2.9174 × 10 22 atoms in the penny. Removing 1 electron from 0.05 × 3.1 NA 64 atoms means removing 0.05 × 3.1 NA 64 electrons. Therefore, Net charge = 0.05 × 3.1 NA 64 × 1.6019 × 10 - 19 = 234 C Solution 1.3 (a) 7.573 × 10 17 × - 1.6019 × 10 - 19 ( 29 = - 0.1213 C (b) Current = 0.1213 10 - 3 = 121.3A flowing from right to left. (c) Again, use proportionality: 10 A = x × 1.6019 × 10 - 19 60sec x = 10 × 60 1.6019 × 10 - 19 = 3.75 × 10 21 (d) i t ( 29 = dq dt = 1 - e - 5 t A. This is an exponential evolution with an initial value of 0, a final value of 1, and a time-constant of 1/5 (signal reaches ~63% of it’s final value in one time-constant).
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0.2 time in sec i(t) 1 (e) Current is the slope of the charge waveform. Therefore, by inspection: Solution 1.4 (a) 6.023 × 10 23 × (-1.6019 × 10 -19 ) = –9.65 × 10 4 C. (b) Current flows from right to left (opposite electrons), and: I = 9.65 × 10 4 10 - 3 = 9.65 × 10 7 A
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(c) Using proportionality: 5 A = x × 1.6019 × 10 - 19 60sec x = 5 × 60 1.6019 × 10 - 19 = 1.87 × 10 21 (d) i t ( 29 = dq dt = 1 + 0.5 π cos π t ( 29 i 1sec ( 29 = 1 - 1.57 = - 0.57 A . Current flows from left to right. Solution 1.5 (a) i t ( 29 = 1 - 4 e - 2 t + 3 e - 3 t t 0. Then q t ( 29 = i t ( 29 dt 0 t = 1 - 4 e - 2 τ + 3 e - 3 τ ( 29 d 0 t τ = τ ] -∞ t - 4 e - 2 τ d τ 0 t + 3 e - 3 τ d τ 0 t = t - 4 - 0.5 e - 2 τ [ ] 0 t + 3 - 0.333 e - 3 τ [ ] 0 t = t + 2 e - 2 t - e - 3 t - 1 (b) By inspection: (c) q t ( 29 = 120cos 120 π t ( 29 . Hence i t ( 29 = dq dt = - 120 π× 120sin 120 π t ( 29 =- 14400 π sin 120 π t ( 29 A Solution 1.6. (a) i ( t ) = 1 - cos( π t ) A. Hence q ( t ) = i ( τ ) d τ 0 t = 1 - cos( πτ ) ( 29 d τ 0 t = t - 1 π sin( πτ ) 0 t = t - 1 π sin( π t ) C (b) Charge is integral of current. Graphically, the charge at time t is the area under the current curve up to time t: (note the quadratic nature between 2 and 4 seconds)
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Solution 1.7 Again, Q is the running area under the current curve. Between 0 and 3 seconds, current decreases linearly until zero. So, Q tot = 7.5 C. From 0 to 6: Q tot = 7.5 + Q 3_6 = 7.5 -1/1 × 0.5 + -1/1 × 0.5 + -1 × 1 = 5.5 C, where the curve from 3 to 6 was divided into two triangular sections and one rectangular one. Solution 1.8 Charge is the area under the current curve. Thus, Q = 0.1*4 – 0.1*2 = 0.2 C. Solution 1.9 Calculate the change in energy for the electron: E = Q V = 3.218 × 10 -15 . Equate this to kinetic energy: 3.218 × 10 - 15 = 1 2 mv 2 v = 8.4 × 10 7 m / s where the mass of an electron, 9.1066 × 10 -31 has been substituted. Solution 1.10 P = VI. Hence I = P/V = 2 × 10 3 /120 = 16.6667 A PROBLEM Solution 1.11 (a) It is necessary to integrate the i(t) curve to obtain q(t). We do this interval by interval: (i) 0 t 1 ms, q ( t ) = 0 + d τ 0 t = t μ C (ii) 1 ms t 2 ms, q ( t ) = 1 - 2 d τ 1 t = 3 - 2 t μ C (iii) 2 ms t 3 ms, q ( t ) = - 1 + d τ 2 t = - 3 + t μ C
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(iv) 3 ms t 5 ms, q ( t ) = 0 + 8 - 2 τ ( 29 d τ 3 t
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