() 0.002 , ≥ 0 2 , ≥0
( )= = =
0 , <0 0 , <0
() − −0.2 , ≥ 0
( )= =
0 , <0
() 0.024 −0.003 , ≥ 0 24 −0.003 , ≥0
( )= = =
0 , <0 0 , <0
() ( 7 −0.003
− 0.021 −0.003
) , ≥0 ( 7 − 0.021 ) −0.003
, ≥0
( )= = =
0 , <0 0 , <0
() 16π ×10−3 cos(2π ×1000 ) , ≥ 0 50.2655cos(2π ×1000 ) , ≥0
( )= = =
0 , <0 0 , <0
The charge q(t) entering an element can be written as
0.5 × 10−3 , 0 ≤ < 2
−3 −3
−10 + 3 × 10 , 2 ≤ < 4
( ) = 1 −3 7 −3
×10 − × 10 , 4 ≤ < 7
3 3
0,
The current through the element can be written as
1
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 0.5 × 10−3 , 0 ≤ < 2 0.5 , 0 ≤ < 2
−1 , 2 ≤ < 4
−10−3 , 2 ≤ < 4
( )
( )= = 1 = 1
3 , 4≤ <7
−3
× 10 , 4 ≤ < 7
3
0 , 0 ,
The current i(t) is shown in Figure S1.6.
Figure S1.6
5
( ) = ∫ 5 × 10 −3 = 5 × 10 −3 × 5 = 25 × 10−3
0
−0.2 5 −1
−1
5
( ) = ∫ 5 × 10 −6 −0.2
= 5 × 10 −6 0
= 5 × 10 × −6
= 1.5803 × 10 −5 = 15.803
0
−0.2 −0.2
3( − 1)
5 5 5 −0.5 5 −2.5
( ) = ∫ 3 (1 − ) = ∫3 − 3∫
−0.5 −0.5 5
=3 −3 0
= 3(5 − 0) + = 9.4925
0 0 0
0
−0.5 0.5
From integral table, we have ∫ =
( − 1)
. Thus,
2
2
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, ( −3 − 1)
−3 5
5
2 2
( ) = ∫2 −3
=2 0
= −15
( −15 − 1) − −0
( −0 − 1) ≈ = 0.2222
0
9 9 9
1
From integral table, we have ∫ sin( ) = − cos( ) . Thus,
5
π π
5
7 35 70
( ) = ∫ 7 sin = − cos = − [ cos(π ) − 1] = = 22.2817
0 5 π 5 0 π π
5
P = VI = 5 V × 2 A = 10 W, absorbing power
P = VI = 2 V × (-3 A) = -6 W, delivering power
P = VI = (-5 V) × 4 mA = -20 mW, delivering power
P = VI = (-12 V) × (-10 mA) = 120 mW, absorbing power
p(t) = v(t) i(t) = (5 V) × (2 mA) = 10 mW
p(t) = v(t) i(t) = [5 sin(2π1000t) V] × [25 cos(2π1000t) mA]
= 125 sin(2π1000t) cos(2π1000t) mW = 62.5 sin(2π2000t) mW
p(t) = v(t) i(t) = 420 e-0.15t u(t) mW
p(t) = v(t) i(t) = [3 cos(2π100t) V] × [8 cos(2π100t) mA]
= 24 cos2(2π100t) mW = [12 + 12 cos(2π200t)] mW
3
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, p(t) = v(t) i(t) = [2 sin(2π100t) V] × [6 sin(2π100t) mA]
= 12 sin2(2π100t) mW = [6 - 6 cos(2π200t)] mW
The circuit with one current source and one voltage source is shown in Figure S1.21.
Figure S1.21 Circuit with one current source and one voltage source.
The circuit with one current source and one voltage source is shown in Figure S1.22.
Figure S1.22 Circuit with one current source and one voltage source.
4
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
( )= = =
0 , <0 0 , <0
() − −0.2 , ≥ 0
( )= =
0 , <0
() 0.024 −0.003 , ≥ 0 24 −0.003 , ≥0
( )= = =
0 , <0 0 , <0
() ( 7 −0.003
− 0.021 −0.003
) , ≥0 ( 7 − 0.021 ) −0.003
, ≥0
( )= = =
0 , <0 0 , <0
() 16π ×10−3 cos(2π ×1000 ) , ≥ 0 50.2655cos(2π ×1000 ) , ≥0
( )= = =
0 , <0 0 , <0
The charge q(t) entering an element can be written as
0.5 × 10−3 , 0 ≤ < 2
−3 −3
−10 + 3 × 10 , 2 ≤ < 4
( ) = 1 −3 7 −3
×10 − × 10 , 4 ≤ < 7
3 3
0,
The current through the element can be written as
1
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 0.5 × 10−3 , 0 ≤ < 2 0.5 , 0 ≤ < 2
−1 , 2 ≤ < 4
−10−3 , 2 ≤ < 4
( )
( )= = 1 = 1
3 , 4≤ <7
−3
× 10 , 4 ≤ < 7
3
0 , 0 ,
The current i(t) is shown in Figure S1.6.
Figure S1.6
5
( ) = ∫ 5 × 10 −3 = 5 × 10 −3 × 5 = 25 × 10−3
0
−0.2 5 −1
−1
5
( ) = ∫ 5 × 10 −6 −0.2
= 5 × 10 −6 0
= 5 × 10 × −6
= 1.5803 × 10 −5 = 15.803
0
−0.2 −0.2
3( − 1)
5 5 5 −0.5 5 −2.5
( ) = ∫ 3 (1 − ) = ∫3 − 3∫
−0.5 −0.5 5
=3 −3 0
= 3(5 − 0) + = 9.4925
0 0 0
0
−0.5 0.5
From integral table, we have ∫ =
( − 1)
. Thus,
2
2
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, ( −3 − 1)
−3 5
5
2 2
( ) = ∫2 −3
=2 0
= −15
( −15 − 1) − −0
( −0 − 1) ≈ = 0.2222
0
9 9 9
1
From integral table, we have ∫ sin( ) = − cos( ) . Thus,
5
π π
5
7 35 70
( ) = ∫ 7 sin = − cos = − [ cos(π ) − 1] = = 22.2817
0 5 π 5 0 π π
5
P = VI = 5 V × 2 A = 10 W, absorbing power
P = VI = 2 V × (-3 A) = -6 W, delivering power
P = VI = (-5 V) × 4 mA = -20 mW, delivering power
P = VI = (-12 V) × (-10 mA) = 120 mW, absorbing power
p(t) = v(t) i(t) = (5 V) × (2 mA) = 10 mW
p(t) = v(t) i(t) = [5 sin(2π1000t) V] × [25 cos(2π1000t) mA]
= 125 sin(2π1000t) cos(2π1000t) mW = 62.5 sin(2π2000t) mW
p(t) = v(t) i(t) = 420 e-0.15t u(t) mW
p(t) = v(t) i(t) = [3 cos(2π100t) V] × [8 cos(2π100t) mA]
= 24 cos2(2π100t) mW = [12 + 12 cos(2π200t)] mW
3
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, p(t) = v(t) i(t) = [2 sin(2π100t) V] × [6 sin(2π100t) mA]
= 12 sin2(2π100t) mW = [6 - 6 cos(2π200t)] mW
The circuit with one current source and one voltage source is shown in Figure S1.21.
Figure S1.21 Circuit with one current source and one voltage source.
The circuit with one current source and one voltage source is shown in Figure S1.22.
Figure S1.22 Circuit with one current source and one voltage source.
4
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
