(II) In the LRC circuit or Fig. 30–19, suppose I = I 0 sin ωt and V − V 0 sin( ωt + ϕ ). Determine the instantaneous power dissipated in the circuit from P = IV using these equations and show that on the average, P ¯ = 1 2 V 0 I 0 cos ϕ , which confirms Eq. 30–30. FIGURE 30-19 An LRC circuit. We multiply the instantaneous current by the instantaneous voltage to calculate the instantaneous power. Then using the trigonometric identity for the summation of sine arguments (inside back cover of text) we can simplify the result. We integrate the power over a full period and divide the result by the period to calculate the average power. P = IV = ( I 0 sin ωt ) V 0 sin( ωt + ϕ ) = I 0 V 0 sin ωt cos ϕ +sin ϕ cos ωt ) = I 0 V 0 (sin 2 ωt cos ϕ +sin ωt cos ωt sin ϕ P ¯ = 1 T ∫ 0 T P d T = ω 2 π ∫ 0 2 π ω I 0 V 0 ( sin 2 + ω t cos ϕ + sin ω t cos ω t sin ϕ ) d t = ω 2 π I 0 V 0 cos ϕ ∫ 0 2 π ω sin 2 ω t d t + ω 2 π I 0 V 0 sin ϕ ∫ 0 2 π ω sin ω t cos ω t d t = ω 2 π I 0 V 0 cos ϕ ( 1 2 2 π ω ) + ω 2 π I 0 V 0 sin ϕ ( 1 ω sin 2 ω t ∫ 0 2 π ω ) = 1 2 I 0 V 0 cos ϕ
(II) In the LRC circuit or Fig. 30–19, suppose I = I 0 sin ωt and V − V 0 sin( ωt + ϕ ). Determine the instantaneous power dissipated in the circuit from P = IV using these equations and show that on the average, P ¯ = 1 2 V 0 I 0 cos ϕ , which confirms Eq. 30–30. FIGURE 30-19 An LRC circuit. We multiply the instantaneous current by the instantaneous voltage to calculate the instantaneous power. Then using the trigonometric identity for the summation of sine arguments (inside back cover of text) we can simplify the result. We integrate the power over a full period and divide the result by the period to calculate the average power. P = IV = ( I 0 sin ωt ) V 0 sin( ωt + ϕ ) = I 0 V 0 sin ωt cos ϕ +sin ϕ cos ωt ) = I 0 V 0 (sin 2 ωt cos ϕ +sin ωt cos ωt sin ϕ P ¯ = 1 T ∫ 0 T P d T = ω 2 π ∫ 0 2 π ω I 0 V 0 ( sin 2 + ω t cos ϕ + sin ω t cos ω t sin ϕ ) d t = ω 2 π I 0 V 0 cos ϕ ∫ 0 2 π ω sin 2 ω t d t + ω 2 π I 0 V 0 sin ϕ ∫ 0 2 π ω sin ω t cos ω t d t = ω 2 π I 0 V 0 cos ϕ ( 1 2 2 π ω ) + ω 2 π I 0 V 0 sin ϕ ( 1 ω sin 2 ω t ∫ 0 2 π ω ) = 1 2 I 0 V 0 cos ϕ
(II) In the LRC circuit or Fig. 30–19, suppose I = I0 sin ωt and V − V0 sin(ωt + ϕ). Determine the instantaneous power dissipated in the circuit from P = IV using these equations and show that on the average,
P
¯
=
1
2
V
0
I
0
cos
ϕ
, which confirms Eq. 30–30.
FIGURE 30-19 An LRC circuit.
We multiply the instantaneous current by the instantaneous voltage to calculate the instantaneous power. Then using the trigonometric identity for the summation of sine arguments (inside back cover of text) we can simplify the result. We integrate the power over a full period and divide the result by the period to calculate the average power.
P = IV = (I0 sin ωt)V0 sin(ωt + ϕ) = I0V0 sin ωt cosϕ +sinϕ cos ωt)
= I0V0 (sin2ωt cosϕ +sin ωt cos ωt sinϕ
P
¯
=
1
T
∫
0
T
P
d
T
=
ω
2
π
∫
0
2
π
ω
I
0
V
0
(
sin
2
+
ω
t
cos
ϕ
+
sin
ω
t
cos
ω
t
sin
ϕ
)
d
t
=
ω
2
π
I
0
V
0
cos
ϕ
∫
0
2
π
ω
sin
2
ω
t
d
t
+
ω
2
π
I
0
V
0
sin
ϕ
∫
0
2
π
ω
sin
ω
t
cos
ω
t
d
t
=
ω
2
π
I
0
V
0
cos
ϕ
(
1
2
2
π
ω
)
+
ω
2
π
I
0
V
0
sin
ϕ
(
1
ω
sin
2
ω
t
∫
0
2
π
ω
)
=
1
2
I
0
V
0
cos
ϕ
10 Figure 30-30 gives the variation
with time of the potential difference
VR across a resistor in three circuits
wired as shown in Fig. 30-16. The cir-
cuits contain the same resistance R
and emf & but differ in the induc-
tance L. Rank the circuits according
to the value of L, greatest first.
Figure 30-30 Question 10.
VR
(I) At what frequency will a 2.40μF capacitor have a reactance of 6.10 KΩ ?
Calculate volt in the given circuit, where R= 22 Q.
50 Ω
50 μF
60 sin 200f V
0.1 H
,(t)
The value of vol) is
|cos(200t) V.
Chapter 30 Solutions
Physics for Scientists and Engineers with Modern Physics
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