662
14 Electromagnetism II
From (1) we have L 2 =
L 1 C 1
C 2
(5)
Substituting (5) in (4) we get on simplification LC = L 1 C 1 =
1
ω 2
It follows that ω =
1
√
LC
14.26 Z c =
1
ωc
=
1
2π fc
∴ C =
1
2π fz c
=
1
2π × 1000 × 500
= 3.18 × 10
−7 F = 0.318 μF
Z L = ωL = 2π fL
∴ L =
Z L
2π f
=
100
2π × 5000
= 3.18 × 10
−3 H = 3.18 mH
14.27 The maximum stored energy in the capacitor must equal the maximum stored
energy in the inductor, from the principle of energy conservation.
∴
1
2
q 2
m
C
=
1
2
Li
2
m
(1)
where i m is the maximum current and q m is the maximum charge. Substituting C V 0 for q m and solving for i m in (1)
i m = V 0
C
L
= 100
0.01 × 10 −6
10 × 10 −3 = 0.1 A
14.28 For resonance
ω =
1
√
LC
∴ C =
1
4π 2 f 2 L
=
1
4π 2 (5 × 10 5 ) 2 × 10 −3 = 1.013 × 10
−9 F
Quality factor
Q =
ωL
R
∴ R =
2π fL
Q
=
2π(500 × 10 3 )(10 −3 )
150
= 20.944
∴ Resistance to be included in series is 20.944 − 5.0 = 15.944 .
14 Electromagnetism II
From (1) we have L 2 =
L 1 C 1
C 2
(5)
Substituting (5) in (4) we get on simplification LC = L 1 C 1 =
1
ω 2
It follows that ω =
1
√
LC
14.26 Z c =
1
ωc
=
1
2π fc
∴ C =
1
2π fz c
=
1
2π × 1000 × 500
= 3.18 × 10
−7 F = 0.318 μF
Z L = ωL = 2π fL
∴ L =
Z L
2π f
=
100
2π × 5000
= 3.18 × 10
−3 H = 3.18 mH
14.27 The maximum stored energy in the capacitor must equal the maximum stored
energy in the inductor, from the principle of energy conservation.
∴
1
2
q 2
m
C
=
1
2
Li
2
m
(1)
where i m is the maximum current and q m is the maximum charge. Substituting C V 0 for q m and solving for i m in (1)
i m = V 0
C
L
= 100
0.01 × 10 −6
10 × 10 −3 = 0.1 A
14.28 For resonance
ω =
1
√
LC
∴ C =
1
4π 2 f 2 L
=
1
4π 2 (5 × 10 5 ) 2 × 10 −3 = 1.013 × 10
−9 F
Quality factor
Q =
ωL
R
∴ R =
2π fL
Q
=
2π(500 × 10 3 )(10 −3 )
150
= 20.944
∴ Resistance to be included in series is 20.944 − 5.0 = 15.944 .