Online Homework 11 Solution
Problem 30.90
A voltage V = 0.95 sin 754t is applied to an LRC circuit (I is in amperes, t is
in seconds, V is in volts, and the "angle" is in radians) which has
L = 27.0 mH, R = 30.0 kΩ, and C = 0.49 µF.
Part A
Sol:)
What is the impedance and phase angle?
X L = ωL = (754 rad/s )(0.0270 H ) = 20.36 Ω
XC =
1
1
=
= 2707 Ω
ωC (754 rad/s ) 0.49 × 10 -6 F
(
)
(30 × 10 Ω) + (20.36 Ω − 2707 Ω )
Z = R 2 + (X L − X C ) =
2
φ = tan −1
2
3
2
= 3.01 × 10 4 Ω
XL − XC
20.36 Ω − 2707 Ω
= tan −1
= −5.12°
R
30 × 10 3 Ω
Part B How much power is dissipated in the circuit?
Express your answer using two significant figures.
Sol:)
2
2
Vrms
V02
(
0.95V )
P = I rmsVrms cos φ =
cos φ =
cos φ =
cos(− 5.12° ) = 1.5 × 10 −5 W
4
Z
2Z
2 3.01 × 10 Ω
(
)
Part C What is the rms current and voltage across each element?
Express your answer using two significant figures.
Sol:)
V0
2
0.95 2
= 2.2 × 10 −5 A
4
Z
3.01 × 10 Ω
VR = I rms R = 2.2 × 10 −5 A 30 × 10 3 Ω = 0.67 V
I rms =
=
(
(
= (2.2 × 10
)(
)
)
A )(20.36 Ω ) = 4.5 × 10
VC = I rms X C = 2.2 × 10 −5 A (2707 Ω ) = 6 × 10 − 2 V
VL = I rms X L
−5
1
−4
V
A Series L-R-C Circuit: The Phasor Approach
In this problem, you will consider a series L-R-C circuit, containing a
resistor of resistance R, an inductor of inductance L, and a capacitor of
capacitance C, all connected in series to an AC source providing an
alternating voltage V(t) = V0cos(ωt).
You may have solved a number of problems in which you had to find the
effective resistance of a circuit containing multiple resistors. Finding the
overall resistance of a circuit is often of practical interest. In this problem,
we will start our analysis of
this L-R-C circuit by finding
its effective overall resistance,
or impedance. The
impedance Z is defined by
Z=
V0
, where V0 and I0 are
I0
the amplitudes of the voltage
across the entire circuit and
the current, respectively.
Part A Find the impedance Z of the circuit using the phasor diagram shown.
Notice that in this series circuit, the current is same for all elements of the
circuit:
Express your answer in terms of R, L, C and ω.
Sol:)
(
V0 = VR20 + VL20 − VC20
Z=
)
2
= I 0 R 2 + (X L − X C )
2
V0
1 

2
= R 2 + ( X L − X C ) = R 2 +  ωL −

I0
ωC 

2
Part B Now find the tangent of the phase angle ψ between the current and
the overall voltage in this circuit.
Express tan(ψ) in terms of R, L, C and ω.
Sol:)
V − VC 0 I 0 ( X L − X C ) X L − X C
tan φ = L 0
=
=
=
VR 0
I0R
R
2
ωL −
R
1
ωC
Part C Imagine that the parameters R, L, C and the amplitude of the voltage
V0 are fixed, but the frequency of the voltage source is changeable. If the
frequency of the source is changed from a very low one to a very high one, the
current amplitude I0 will also change. The frequency at which I0 is at a
maximum is called resonance. Find the frequency ω0 at which the circuit
reaches resonance.
Express your answer in terms of any or all of R, L, and C.
Sol:)
At resonance
X L = X C ⇒ ωL =
1
1
⇒ ω2 =
LC
ωC
1
⇒ ω0 =
LC
Part D What is the phase angle ψ between the voltage and the current when
resonance is reached?
Sol:)
Use the right figure
At resonance
XL = XC
Z is at minimum
ψ=0
Part E Now imagine that the parameters R, L, ω, and the amplitude of the
voltage V0 are fixed but that the value of C can be changed. This is one of the
easiest parameters to change when "tuning" such (radio frequency) circuits in
order to make them resonate. This is because the capacitance can be changed
just by moving the capacitor plates closer or farther apart. Find the capacitance
C0at resonance.
Express your answer in terms of L and ω.
Sol:)
At resonance
XL = XC
⇒ C0 =
1
ω2L
3
Problem 30.101
To detect vehicles at traffic lights, wire loops with dimensions on the order
of 2m are often buried horizontally under roadways. Assume the
self-inductance of such a loop is L = 5.0 mH and that it is part of an LRC
circuit as shown in the figure with C = 0.10 µF and R = 45 Ω. The ac
voltage has frequency f and rms voltage Vrms.
Part A The frequency f is chosen to match the resonant frequency f0 of the
circuit. Find f0. Express your answer to two significant figures and include the
appropriate units.
Sol:).
at resonance ⇒ ωL −
f =
1
2π LC
=
2π
1
= 0 ⇒ ω0 =
ωC
1
1
LC
(0.005H )(0.1×10 −6 F)
= 7.1kHz
Part B Determine what the rms voltage (VR)rms across the resistor will be
when f = f0 . Express your answer in terms of Vrms.
Sol:).
(VR )rms
= Vrms
Part C Assume that f, C and R never change, but that, when a car is located
above the buried loop, the loop's self-inductance decreases by 10% (due to
induced eddy currents in the car's metal parts). Determine by what factor the
voltage (VR)rms decreases in this situation in comparison to no car above the
loop. [Monitoring (VR)rms detects the presence of a car.] Express your answer
using two significant figures.
Determine what the rms voltage
(VR)rms across the resistor will
be when f = f0. Express your
answer in terms of Vrms.
Sol:).
1
1
=
= 223.6Ω
2πfC 2π (7118Hz ) 0.1 × 10 −6 F
X L = 2πfL = 2π (7118Hz )(0.9 )(0.005H ) = 201.3Ω
XC =
(
Z = R 2 + (X L − X C ) =
2
(VR )rms
)
(45Ω )2 + (201.3Ω − 223.6Ω )2
R
 45Ω 
=  Vrms = 
Vrms = 0.9Vrms
Z
 50.24Ω 
4
= 50.24Ω
The Ampère-Maxwell Law
The objective of this example is to introduce the displacement current,
show how to calculate it, and then to show that
the displacement current Idisplacement(t) is
identical to the conduction current Icharge(t).
Assume that the capacitor has plate area A and
an electric field E(t) between the plates. Take µ0
to be the permeability of free space and ε0 to be
the permittivity of free space.
Part A
First find
r r
r
B
•
d
l
,
the
line
integral
of
B
around a loop of radius R
∫
R
located just outside the left capacitor plate. This can be found from the usual
current due to moving charge in Ampère's law, that is, without the
displacement current. Find an expression for this integral involving the current
I(t) and any needed constants given in the introduction.
Sol:)
r r
B
∫ • dl = µ 0 I (t )
R
Part B
Now find an expression for
r r
r
B
∫ • dl , the same line integral of B
R
around the same loop of radius R located just outside the left capacitor plate as
before. Use the surface that passes between the
plates of the capacitor, where there is no conduction
current. This should be found by evaluating the
amount of displacement current in the
Ampère-Maxwell law above. Express your answer
in terms of the electric field between the plates E(t),
dE(t)/dt, the plate area A, and any needed constants
given in the introduction.
Sol:)
r
r
Φ E = ∫ E • dA ⇒ Φ E (t ) = AE (t )
I displacemennt = ε 0
dΦ E
dt
r r
dΦ E (t )
B
∫R • dl = µ 0 I = µ 0ε 0 dt
r r
dE (t )
B
∫R • dl = µ 0ε 0 A dt
5
A necessary consistency check
Part C We now have two quite different expressions for the line integral of
the magnetic field around the same loop. The point here is to see that they both
are intimately related to the charge q(t) on the left capacitor plate. First find the
displacement current Idisplacement(t) in terms of q(t).
Express your answer in terms of q(t), dq(t)/dt, and any needed constants given
in the introduction.
Sol:)
r q
r
Φ E = ∫ E • dA = encl
Φ E (t ) =
q (t )
ε0
I displacemennt (t ) = ε 0
ε0
dΦ E (t ) dq(t )
=
dt
dt
Part D Now express the normal current Icharge(t) in terms of the charge on the
capacitor plate q(t).
Express your answer in terms of q(t), dq(t)/dt, and any needed constants given
in the introduction.
Sol:)
I charge (t ) =
dq (t )
dt
6
Problem 31.6
Suppose an air-gap capacitor has circular plates of radius R = 2.3 cm and
separation d = 1.8 mm. A 78.0 Hz emf, ε = ε0cos(ωt), is applied to the
capacitor. The maximum displacement current is 37 µA.
Part A Determine the maximum conduction current I.
Express your answer to two significant figures and include the appropriate
units.
Sol:)
I = I displacemennt = 37µA
Part B Determine the value of ε0.
Express your answer to two significant figures and include the appropriate
units.
Sol:)
Q = CV = Cε 0 cos ωt
dQ
− ωCε 0 sin ωt
dt
I max = ωCε 0
I=
I max
A
, C = ε0
d
ωC
I d
I d
37 × 10 −6 A 1.8 × 10 −3 m
∴ ε 0 = max = max =
= 9200V
ωε 0 A 2πfε 0 A 2π (78Hz ) 8.85 × 10 −12 C 2 N ⋅ m 2 π (0.023m )2
⇒ ε0 =
(
(
)(
)
)
Part C Determine the maximum value of dΦE/dt between the plates. Neglect
fringing.
Express your answer using two significant figures.
Sol:)
dΦ E
dt
(I D )max
37 × 10 − 6 A
 dΦ E 
⇒
=
= 4.2 × 10 6 V ⋅ m s
 =
2
2
−12
ε0
8.85 × 10 C N ⋅ m
 dt  max
ID = ε0
7