Ch01 - solution of 11th edition PDF

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solution of 11th edition...

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1 Irwin, Engineering Circuit Analysis, 11e ISV

SOLUTION:

Chapter 1: Basic Concepts

2 Irwin, Engineering Circuit Analysis, 11e ISV

SOLUTION:

𝑄 𝑡 = 150C/15 x 60s = 1/6A P = VI 1Whr/15min = V x 1/6 V = 24V 𝐼 =

Chapter 1: Basic Concepts

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Chapter 1: Basic Concepts

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SOLUTION:

𝑑𝑞 𝑑𝑡 = 10e-tmA Now, P = VI Thus V = 0.2e--tV is the voltage across the element. E = ∫ 𝑝(𝑡)𝑑𝑡 𝐼 =

0.2

= ∫0 2𝑒 -tdt = 0.362 mJ

Chapter 1: Basic Concepts

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SOLUTION: The charge passing through the circuit is q

102

= ∫99 𝑖(𝑡)𝑑𝑡 102

=∫99 (−3 + 𝑡 )𝑑𝑡 = (−3 × 102 + =292.5C

Chapter 1: Basic Concepts

1022 ) 2

− (− 3 × 99 +

992 2

)

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SOLUTION:

I=

𝑑𝑞 𝑑𝑡

I = 20e-2tmA Now E = ∫ 𝑉𝐼𝑑𝑡 20

= ∫0 2.4e-4tdt

= 0.011 J

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SOLUTION:

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SOLUTION:

i(t)

=

𝑑𝑞 𝑑𝑡

= -3t2 + 12t - 10 = -3 x 52 + 12 x 5 - 10 = -25A

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SOLUTION:

a. i(0.25) = e-4x0.25 = 0.368A 0.25 b. Average value of current = ∫−0.25 𝑖(𝑡)𝑑𝑡 = 0.632A c. Charge passed = Average current x Time = 0.318A

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SOLUTION: The charge through the element can be found by finding the area under the curve from 5 to 15 seconds. Thus charge flown = 5 x (10-5) + ½ x 5 x (15-10) = 37.5mC

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SOLUTION:

P=VI = 57.5cos(-2π/3) = -28.75W

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SOLUTION:

Power supplied by 10V battery = 10V x -8A = -80W Power supplied by 6A current source = 6A x 5V = 30V

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SOLUTION:

a. V(1ms) = 9.69V b. V(8ms) = -4.16V c. Work done = q x V = 4C x V(4ms) = -21.6J

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SOLUTION:

Power delivered

Chapter 1: Basic Concepts

=VxI = 5 x (0.8-2)2V x -5A = -34.68W

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SOLUTION: (See Next page)

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Chapter 1: Basic Concepts...