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Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition

Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition provides you with expert textbook solutions that ensure you understand every concept thoroughly.

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    Chapter 2 Solutions Section 2.1 Introduction 2.1 Current source 2.2 Voltage source 2.3 Resistor 2.4 Capacitor 2.5 Inductor Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges. 2.7 The relationship of charge and current is ( ) ( ) ( ) 0 0 t q dt t i t q t t + = ∫ so ( ) ( ) ( ) 0 0 10 sin 2 t q dt t t q t t + = ∫ π ( ) ( ) ( ) 0 10 cos 10 2 0 t q t t q t t + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = π π 2.8 The coulomb of one electron is denoted by e and ( ) ( ) ( ) 0 0 t q dt t i t q t t + = ∫ So ( ) ( ) 0 0 12 1 / ) ( t q dt t e e t q t n t t + = = ∫ If t 0 = 0 and ( ) 0 0 = t q , ( ) 2 6 t e t n = 2.9

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    ( ) ∫ = idt t q ( ) t d t q t ∫ = 0 5 ( ) t t q 5 = 2.10 ( ) [ ] ( ) ( ) 25 0 5 5 5 5 5 0 = − = = t t q Coulombs 2.11 Using the definition of current-charge relationship, the equation can be rewritten as: e t n dt dq i Δ Δ = = Thus, the current flow within t 1 and t 2 time interval is, A i 3 ) 10 6 . 1 ( 2 10 ) 2 75 . 5 ( 19 19 − = × − × − = − The negative sign shows the current flow in the opposite direction with respect to the electric charge. 2.12 Assuming the area of the metal surface is S , The mass of the nickel with depth d = 0.15mm is m = ρ × d × S Meanwhile, using the electro-chemical equivalent, the mass of the nickel can be expressed as m = k × I × t where I = σ × S. Equating the two expressions of the mass, the coating time is found: t = ρ × d / σ = 1.24 × 10 5 s ≈ 34.4 hour Section 2.3 Voltage 2.13 By the definition of voltage, when a positive charge moves from high voltage to low voltage, its potential energy decreases. So a is “+”, b is “-”. In other words, u ab =1V. 2.14 The current i ( t ) is defined as: ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ≤ < = elsewhere t t i 0 1 0 3 Therefore, the charge

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    3 3 1 0 ∫ = = dt g C The energy in Joules is given by: J C V J 15 3 5 = × = × = 2.15 1 electron = 19 10 6 . 1 − × − Coulombs. Therefore, there are 18 10 25 . 6 × electrons in a coulomb. Coulombs of 16 10 5 × electrons = 3 18 16 10 8 10 25 . 6 10 5 − × = × × Therefore, the voltage is 1875 10 8 15 3 = × = = − C J V V 2.16 ( ) dt t t q ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∫ π 2 3 sin 2 1 0 C 4244 . 0 2 3 cos 3 4 1 0 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = t π π J 122 . 2 4244 . 0 5 = × = × = C V J 2.17 C V J q 10 2 20 = = A s C dt dq i 5 . 2 4 10 = = = Section 2.4 Respective Direction of Voltage and Current 2.18 True 2.19 False 2.20 True Section 2.5 Kirchoff’s Current Law 2.21 According to KCL

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    3 7 0 2 − − = i Therefore, A i 4 2 = 2.22 Notice that 4 2 3 0 i + − = Therefore, A i 1 4 − = It can be seen that A i 2 5 − = i 6 can now be found. 6 1 2 5 0 i − + + = A i 8 6 = 2.23 i = 10mA 5k Ω 2k Ω 13k Ω + v - i 1 i 2 Figure S2.23: Circuit for Problem 2.23. According to KCL, 2 1 10 i i mA + = and 2 1 18 2 ki ki = Therefore, mA i 9 1 = , mA i 1 2 = V mA k R I V 13 1 13 = × = × = 2.24 Using the KCL method, we have

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    3 2 1 i i i + = Also, at the other end of the node, we have, 3 2 4 i i i + = Then, combine these two equations we obtain, 4 1 i i = 2.25 The algebraic sum of currents flowing into the node in Figure 2.11 is i 1 + i 2 - i 3 - i 4 where i 1 and i 2 flow in, and i 3 and i 4 flow out As current is defined as the rate of variation of the charge, i 1 + i 2 - i 3 - i 4 = dq/dt By law of the conservation of charge, charges cannot be stored in the node and charges can neither be destroyed nor created. Therefore the variation rate of the charge is zero and further the KCL law holds, i 1 + i 2 - i 3 - i 4 = 0 2.26 The resistance across the voltage source is Ω = + × + = + = 50 120 40 120 40 20 120 // 40 20 R So the current flowing through the voltage source is A V R V I 1 . 0 50 5 = Ω = = Therefore, mA i 25 1 . 0 160 40 = × = 2.27 By KCL, x I V V 2 3 5 3 0 − + − = Note that, x x I V V I 3 3 = → = Therefore, x x x I I I 2 5 3 3 0 − + − = x x I I 5 3 3 0 − − =

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    A 5 . 1 2 3 − = → − = x x I I 2.28 The voltage across node C and D (node C is positive) is U 2 - U 1 , the voltage across node C and B is U AB - U 1 , and the voltage across node D and B is U AB - U 2 The KCL equation for node C is U 1 /R = (U 2 - U 1 ) /R + (U AB - U 1 ) /2R which is 5U 1 - 2U 2 = U AB The KCL equation for node D is U 2 /2R + (U 2 - U 1 ) /R = (U AB - U 2 ) /R which is 5U 2 - 2U 1 = 2U AB Solving U 2 and U 1 , U 1 = 9U AB /21, U 2 = 12U AB /21 Thus, by the KCL equation at node A I AB = U 1 /R + U 2 /2R R AB = U AB / I AB = 7R /5 R 2 R R 2R R A + - B C D I AB U AB + - + - U 1 U 2 Figure S2.28: Circuit for Problem 2.28. Section 2.6 Kirchoff’s Voltage Law 2.29 By KVL, D C B A v v v v + − − − = 0 6 6 2 0 + + − − = A v Then V v A 10 = 2.30 Using KVL, 14 6 12 0 + − + − = R v Then 4 = R v V

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    2.31 Using KVL, 5 3 0 + + − = D v Then 8 = D v V Using KVL, 8 2 8 0 + + − − = C v Then 2 = C v V 2.32 Using KVL, C v − + − = 4 9 0 Then 5 − = C v V Using KVL, 6 5 3 5 0 − − − + − = E v Then 13 − = E v V 2.33 Using KVL, 3 4 0 − + − = D v Then 1 = D v V Using KVL, C v + + + = 1 3 1 0 Then 5 − = C v V Using KVL, 0 3 5 = − + E v Then 8 = E v V 2.34 Using KVL, 1 7 0 + − = E v Then 8 = E v V Using KVL, 8 5 0 + + − = B v Then 13 = B v V Using KVL, 7 5 0 − − = D v Then

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    2 − = D v V Using KVL 2 13 0 − + = A v Then 11 − = A v V 2.35 Using KVL, V V V H H 8 2 10 = → + = V V V V I I H 5 5 2 = → + = + V V V C C 2 6 4 = → = + V V V V D C D 1 4 5 = → + = + 2.36 V = 6V + - R 1 R 2 4 Ω 6 Ω 3 Ω i i 1 i 2 i 3 i 4 Figure S2.36: Circuit for Problem 2.36. a) R 1 =6 Ω , R 2 =3 Ω By KCL, i 1 + i = i 2 i 4 + i = i 3 Applying KVL on the loop made by 6V source, R 1 and R 2 , 6 i 1 + 3 i 2 = 6 Applying KVL on the loop made by 6V source, 3 Ω resistor and 6 Ω resistor, 3 i 3 + 6 i 4 = 6 Applying KVL on the loop made by 3 Ω resistor, 4 Ω resistor and R 1 , 3 i 3 + 4 i = 6 i 1 Now we have 5 equations for 5 variables and i can be solved using Cramer’s rule or variable elimination. Then we get i = 1/4A, i 1 = 7/12A, i 2 = 5/6A, i 3 = 5/6A, i 4 = 7/12A b) i = 0 => i 1 = i 2 , i 3 = i 4.

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    Moreover, the voltage across the 4 Ω resistor is zero. This also leads to the voltage across R 1 equals to the voltage across the 3 Ω resistor and the voltage across R 2 equals to the voltage across the 6 Ω resistor, R 1 i 1 =3 i 3 R 2 i 2 =6 i 4 Therefore, R 1 / R 2 = 3/6 = 1/2 2.37 The current from node A to node C is I AB - I 1 , the current from node D to node B is I 1 - I 2 , and the current from node C to node B is I AB - I 1 + I 2 . Then the KVL equation for loop ADC is I 1 2R + I 2 R = (I AB - I 1 )R which is 3I 1 + I 2 = I AB The KVL equation for loop DBC is (I 1 - I 2 )R = I 2 R + (I AB - I 1 + I 2 )2R which is 3I 1 - 4I 2 = 2I AB Solving I 1 and I 2 , I 1 = 2I AB /5, I 2 = -I AB /5 Thus, U AB = I 1 2R + (I 1 - I 2 )R=7I AB R /5 R AB = U AB / I AB = 7R /5 R 2 R R 2R R A + - B C D I AB U AB I 1 I 2 Figure S2.37: Circuit for Problem 2.37. Section 2.7 Ohm’s Law 2.38 By KCL, the current through the 4 Ω resistor is 3A. Using Ohm’s law, 12 3 4 = × = × = R I V V 2.39 By KVL , the voltage across the resistor is 5V. Using Ohm’s law:

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    Ω = = = 5 . 2 2 5 I V R 2.40 Using Ohm’s law: A R V I 3 5 15 = = = 2.41 The resistivity of copper is, m − Ω × = − 8 10 72 . 1 ρ The resistance of the copper wire is, Ω = × × × = = − − 70 10 2286 . 1 5 . 0 10 72 . 1 10 8 A L R ρ Using the definition of Ohm’s Law, we can obtain the maximum allowable voltage, which is: V R I V 70 70 1 max max = ⋅ = = 2.42 Using Ohm’s law, the resistance of the tissue is Ω = × = − k R 5 . 804 10 43 . 12 10 6 Because, A ρ L = R Therefore, 2 3 10 468 . 3 75 1 5 . 04 8 m L m Ω = k Ω − × × ⋅ 94 . 15 = L m 2.43 The resistance across the voltage source is 3 2 3 2 1 3 2 1 // R R R R R R R R R + ⋅ + = + = Using Ohm’s law R V I = , and 3 2 3 2 1 3 2 3 2 2 3 2 2 ) ( ) ( R R R R R R R V R R R I R R R i s + + + ⋅ ⋅ + = ⋅ + = 3 2 3 2 1 2 ) ( R R R R R R V s + + ⋅ =

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    2.44 By KCL: A 5 20 = I Therefore, by Ohm’s Law: V 100 5 20 = ⋅ = A V V 30 10 3 = ⋅ = B V V 30 15 2 = ⋅ = C V By KVL: V 130 = S V 2.45 Using Ohm’s law: 2 . 1 10 200 240 3 = × = I mA 2.46 For short circuit, v ( t ) = 0 Using ohm’s law R short = v ( t )/ i ( t ) = 0 For open circuit i ( t ) = 0 Using ohm’s law R open = v ( t )/ i ( t ) = infinity 2.47 V out = Vs(Rs /(Rs + Rc)) - Vs(Rb /(Ra+Rb)) By product rule of derivation d V out /d Rs = VsRc/(Rs+Rc) 2 When Rs = 0, the above expression is minimum. However, you cannot control over sensor resistance, but you can choose Rc to achieve the maximum slope. We differentiate d V out /d Rs with respect to Rc , d(d V out /d Rs) / d Rc =Vs(Rs-Rc) /(Rs+Rc) 3 The maximum slope is achieved when Rc = Rs. Rc can be chosen to have the same value as the nominal resistance of the sensor. 2.48 Using data pairs in the table, we get three equations in A, B, C (1/273) = A + B ln(16330) + C (ln(16330)) 3 (1/298) = A + B ln(5000) + C (ln(5000)) 3 (1/323) = A + B ln(1801) + C (ln(1801)) 3 Solving those equations, we get A = 0.001284 B = 2.364 × 10 -4 C = 9.304 × 10 -8

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    Section 2.8 Power and Energy 2.49 a) The voltage and the current are in associated direction, P = u × i = -3 × 1 = -3W => active element b) The voltage and the current are not in associated direction, P = - u × i = 2 × 2 = 4W => passive element c) The voltage and the current are not in associated direction, P = - u × i = -2 × 3 = -6W => active element 2.50 a) The voltage and the current are in associated direction, P = u × i = 20 × 2.5e -2t = 50e -2t W => active element b) The voltage and the current are not in associated direction, P = - u × i = -20 × 2sin t = -40sin t W => t > 0, active element; t < 0, passive element 2.51 The current can be found using the power formula: 5 . 0 10 5 = = = V P I A And the resistance can then be found using Ohm’s law: Ω = = = 20 5 . 0 10 I V R 2.52 2 I R I V P × = × = 500 5 20 2 = × = W 2.53 Applying KCL, the current through each resistor is 4mA. Given 2 I R P × = , ( ) mW P 80 10 4 10 5 2 3 3 1 = × × × = − ( ) mW P 224 10 4 10 14 2 3 3 2 = × × × = − ( ) mW P 512 10 4 10 32 2 3 3 3 = × × × = − 2.54 According to KVL: 2 1 3 5 . 1 ki ki = By KCL: 2 1 3 i i mA + = mA i 1 2 =

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    V k mA V 1 1 1 = Ω × = 2.55 From the problem 2.54, we have mA 2 1 = i and mA 1 2 = i . Recall R I P 2 = . Therefore, ( ) mW 6 10 5 . 1 10 2 3 2 3 5 . 1 = × ⋅ × = − k P ( ) mW 2 10 2 10 1 3 2 3 2 = × ⋅ × = − k P ( ) mW 1 10 1 10 1 3 2 3 1 = × ⋅ × = − k P 2.56 + - V I Figure S2.56: Resistor with current direction and voltage polarity. Consider Figure S2.56, the power absorbed by the resistor is P=VI By Ohm’s law, V=RI Thus P=I 2 R = V 2 /R Therefore P is always a nonnegative number when R is positive. 2.57 The voltage and the current are in associated direction, so P = u × i and the waveform of the power is shown below P (W) 1 2 0 10 -10 t (s) Figure S2.57: Power curve. The consumed energy is ∫ = 2 0 P W =0

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    which is resulted from the negative symmetry of P around t = 1. 2.58 a) Let the current through the -5V voltage source be in associated direction with the voltage, which is from the positive side to the negative side. By KCL at the node below the -5V voltage source, the current through it equals 2A+4A=6A. Therefore the power of the -5V voltage source is P = -5V × 6A =30W b) To zero the power above, the current through the -5V voltage has to be zero. So the 4A current source needs to be changed to -2A by KCL. Section 2.9 Independent and Dependent Sources 2.59 In Figure S2.59, i axis is the current through the source and u axis is the voltage across the source. u s denotes the voltage provided by the ideal voltage source and i s denotes the current provided by the ideal current source. It is shown that for ideal voltage source the voltage does not change with the current through it; for ideal current source the current does not change with the voltage. i u u s i s Ideal voltage source Ideal current source Figure S2.59: Voltage VS current curves. 2.60 Because the two current sources are in series, the currents through them are the same, i = 1A 2.61 Because the 3 Ω resistor is in series with the current source, the current through the resistor is 1A and, by Ohm’s law, its voltage is u = 3 Ω × 1A = 3V By KVL, the voltage across the current source is u + 2V = 5V The power supplied by the current source is (the current and the voltage are not in associated direction)

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    P = -5V × 1A = -5W 2.62 Using KVL: x x V V 3 10 0 + + − = Therefore, 5 . 2 = x V 2.63 Using KVL and Ohm’s law, ( ) x x I I 20 10 5 . 3 6 0 3 + × × − − = x I 3 10 48 . 3 6 × − = mA I x 724 . 1 − = 2.64 The voltage across the 2 Ω resistor is -1V, where the voltage polarity is associated with the current direction. So by Ohm’s law I = -1V/2 Ω = -0.5A Therefore, the current of the dependent source is 2 I = -1A 2.65 According to KCL, the current through R3 is x i β According to Ohm’s law, 3 0 R i v x β = i x is given by: 2 1 R R v i s x + = Therefore, 2 1 3 0 R R v R v s + = β 2.66 According to Ohm’s law, the current flow (in associated direction with the voltage) through the 5 Ω resistor is i 1 = 4.9V/5 Ω = 0.98A. Because the CCCS is in series with the 5 Ω resistor, i 1 = 0.98 i => i = 1A which is the current through the 6 Ω . By KCL, the current through the 0.1 Ω resistor is i- 0.98 i = 0.02A. So the voltage across the 0.1 Ω resistor

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