14.3 Solutions
659
14.21 Phasor diagram, Fig. 14.5
Fig. 14.5
(i) X L = ωL = 2π fL = (2π)(80)(0.2) = 100.48
X C =
1
ωc
=
1
2π fc
=
1
(2π)(80)(10 × 10 −6 )
= 199.04
(ii) Z =
R 2 + (X L − X C ) 2 =
(100) 2 + (100.48 − 199.04) 2 = 140.4
(iii) I T =
V
Z
=
600
140.4
= 4.27 A(rms)
(iv) cos ϕ =
R
Z
=
100
140.4
= 0.71225 → ϕ = 44.58 ◦
(v) V R = I T R = 4.27 × 100 = 427 V rms
V C = I T X C = 4.27 × 199.04 = 850 V rms
V L = I T X L = 4.27 × 100.48 = 429 V rms
V
2
= V
2
R = (V L − V C )
2
The voltages on R, C and L are shown in the phasor diagram, Fig. 14.6.
Here the voltage lags the current as X C > X L .
(vi) ω 2 =
1
LC
Fig. 14.6
(vii) The circuit will be in resonance when X L = X C , that is, ωL =
1
ωc
or ω 2 =
1
LC
.
659
14.21 Phasor diagram, Fig. 14.5
Fig. 14.5
(i) X L = ωL = 2π fL = (2π)(80)(0.2) = 100.48
X C =
1
ωc
=
1
2π fc
=
1
(2π)(80)(10 × 10 −6 )
= 199.04
(ii) Z =
R 2 + (X L − X C ) 2 =
(100) 2 + (100.48 − 199.04) 2 = 140.4
(iii) I T =
V
Z
=
600
140.4
= 4.27 A(rms)
(iv) cos ϕ =
R
Z
=
100
140.4
= 0.71225 → ϕ = 44.58 ◦
(v) V R = I T R = 4.27 × 100 = 427 V rms
V C = I T X C = 4.27 × 199.04 = 850 V rms
V L = I T X L = 4.27 × 100.48 = 429 V rms
V
2
= V
2
R = (V L − V C )
2
The voltages on R, C and L are shown in the phasor diagram, Fig. 14.6.
Here the voltage lags the current as X C > X L .
(vi) ω 2 =
1
LC
Fig. 14.6
(vii) The circuit will be in resonance when X L = X C , that is, ωL =
1
ωc
or ω 2 =
1
LC
.