HW#14 Last homework set
1.
When an AC source is connected across a resistor R, the rms current in the resistor is Irms.
(a) Find the rms voltage across the resistor. (b) Find the peak voltage of the source.
(c) Find the maximum current in the resistor. (d) Find the average power delivered to the
resistor.
(a) Ohm' s Law : Vrms = I rms R
(b) Peak voltage = Vmax = 2Vrms or 2 I rms R
(c) Maximum current = 2I rms
(d) Average power delivered = I 2rns R
2. A certain lightbulb is rated at W when operating at an rms voltage of V.
(a) What is the peak voltage applied across the bulb? (b) What is the resistance of the bulb? (c)
Does a 100-W bulb have greater or less resistance than a 45-W bulb?
(a) Peak voltage = Vmax = 2 Vrms = 2 V
(b)
V2
V2
= W⇒ R =
R
W
(c) Since W =
V2
.∴ for the same V, W is less for the greater R.
R
3. In the simple AC circuit shown in the figure, Δv = ΔVmax sin ω t.
(a) If ΔvR = kΔVmax for the first time at t=t’, what is the angular frequency of the source?
k∆Vmax = ∆Vmax sin ω t' ⇒ sin ω t' = k
⇒ ω t' = sin -1 (k)
⇒ ω=
sin -1 (k)
t'
(b) What is the next value of t for which ΔvR = kΔVmax?
∆v
Let the time be t"
∆Vmax
k∆Vmax
Guided by the figure :
sin (π - ω t' ) = sin ω t' = sin ω t"
∴ t" = π - ω t' ⇒ t" = π -
sin -1 (k)
⋅ t'
t'
t
t'
t=?
4. An AC source has an output rms voltage of Vrms at a frequency of f. The source is connected
across an inductor L. (a) Find the inductive reactance of the circuit. (b) Find the rms current in
the circuit. (c) Find the maximum current in the circuit.
(a) X = X L = ω L = 2π fL
(b) I rms =
Vrms
V
= rms
| X | 2π fL
(c) I max = 2 I rms = 2 ⋅
Vrms
Vrms
=
2π fL
2π fL
5. A source delivers an AC voltage of the form Δv = Vmax sin (kπ t), where Δv is in volts and t is
in seconds, to a capacitor. The maximum current in the circuit is Imax. (a) Find the rms voltage
of the source. (b) Find the frequency of the source. (c) Find the value of the capacitance.
(a) Vrms =
Vmax
2
(b) ω = kπ ⇒ 2π f = kπ ⇒ f =
(c) X =
k
2
V
|V|
⇒ | X |=
⇒| X |=
I
|I|
Vmax
I max
1
=
ωC
Vmax
I max
⇒
⇒C=
⇒C=
1 I max
ω Vmax
1 I max
kπ Vmax
6. An AC source with ΔVmax and frequency f Hz is connected between points a and d in the
figure.
(a) Maximum voltages between points a and b = | V ⋅
R
L
C
=| V |⋅
XR
|
X R + X L + XC
ωL
| XR |
| X R + XL + XC |
R
| XR |
= Vmax
2
X R + (X L − X C )
1/(ωC)
2
R
= Vmax ⋅
1 2
)
ωC
ωCR
R 2 + (ωL −
= Vmax ⋅
ω 2 C 2 R 2 + (ω 2 LC − 1) 2
2πfCR
= Vmax ⋅
(2πf) 2 C 2 R 2 + [(2πf) 2 LC − 1]2
(a) Maximum voltages between points a and b = | V ⋅
=| V |⋅
XR
|
X R + X L + XC
| XR |
| X R + XL + XC |
= Vmax
| XR |
2
X R + (X L − X C ) 2
R
= Vmax ⋅
1 2
)
ωC
ωCR
R 2 + (ωL −
= Vmax ⋅
= Vmax ⋅
ω 2 C 2 R 2 + (ω 2 LC − 1) 2
2πfCR
(2πf) 2 C 2 R 2 + [(2πf) 2 LC − 1]2
(a) Calculate the maximum voltages between points a and b.
(b) Calculate the maximum voltages between points b and c.
(c) Calculate the maximum voltages between points c and d.
(a) Maximum voltages between points a and b = | V ⋅
=| V |⋅
XR
|
XR + XL + XC
| XR |
| XR + XL + XC |
= Vmax
| XR |
2
X R + (X L − X C ) 2
R
= Vmax ⋅
R 2 + (ωL −
= Vmax ⋅
= Vmax ⋅
(b) Maximum voltages between points b and c = | V ⋅
ω 2 C 2 R 2 + (ω 2 LC − 1) 2
2πfCR
(2πf) 2 C 2 R 2 + [(2πf) 2 LC − 1] 2
XL
|
XR + XL + XC
ωL
R 2 + (ωL −
= Vmax ⋅
)2
ωCR
= Vmax ⋅
= Vmax ⋅
1
ωC
1 2
)
ωC
ω 2 LC
ω 2 C 2 R 2 + (ω 2 LC − 1) 2
(2πf) 2 LC
(2πf) 2 C 2 R 2 + [(2πf) 2 LC − 1] 2
(c) Maximum voltages between points c and d = | V ⋅
XC
|
XR + XL + XC
1
ωC
= Vmax ⋅
R 2 + (ωL −
= Vmax ⋅
= Vmax ⋅
1 2
)
ωC
1
2
2
2
ω C R + (ω 2 LC − 1) 2
1
2
2
2
(2πf) C R + [(2πf) 2 LC − 1]2
7.
An inductor L, a capacitor C, and a resistor R are connected in series. A f-Hz AC source
produces a peak current of Imax in the circuit. (a) Calculate the required peak voltage ∆Vmax.
(b) Determine the phase angle by which the current leads or lags the applied voltage.
(a) V = IX ⇒ | V |= | I || X |
⇒ Vmax = I max | X |
⇒ Vmax = I max | X R + X L + X C |
⇒ Vmax = I max R 2 + (ωL −
1 2
)
ωC
⇒ Vmax = I max
ω 2 C 2 R 2 + (ω 2 LC − 1) 2
ωC
⇒ Vmax = I max
(2πf) 2 C 2 R 2 + [(2πf) 2 LC − 1] 2
2π fC
ωL
X
φ
R
1/(ωC)
(b) I = V/X
As can be seen from the lower figure, phase angle betweem I and V is φ .
ωL −
From the upper figure, φ = tan -1 (
1
ωC )
R
2
ω LC − 1
= tan -1 (
)
ωCR
Curret lags voltage if ω 2 LC − 1 > 1. Current leads voltatge if ω 2 LC − 1 < 1.
X
φ V
φ
I=V/X
9. A sinusoidal voltage ∆v = Vmaxsinkt, where ∆v is in volts and t is in seconds, is applied
to a series RLC circuit. (a) What is the impedance of the circuit? (b) What is the maximum
current? (c) Determine the numerical value for ω in the equation i = Imax sin (ωt −
in the equation i = Imax sin (ωt −
Determine the numerical value for
(a) | X R + X L + X C | = R 2 + (kL −
1 2
)
kC
k 2 C 2 R 2 + ( k 2 LC − 1) 2
=
kC
(b) I =
V
|V|
⇒| I |=
X
|X|
⇒ I max =
⇒ I max =
). (d)
).
ωL
R
1/(ωC)
Vmax
2
2
2
k C R + ( k 2 LC − 1) 2
kC
kCVmax
k 2 C 2 R 2 + ( k 2 LC − 1) 2
(c) ω = k
(d) As can be seen from the lower figure, phase angle betweem I and V is φ .
kL −
From the upper figure, φ = tan -1 (
1
kC )
R
k LC − 1
= tan -1 (
)
kCR
2
X
φ V
φ
I=V/X
10. A series RLC circuit has components with values L, C, and R,0 Ω, and Δv = ΔVmax sin ωt.
(a) Find the resonant frequency of the circuit.(b) Find the amplitude of the current at the
resonant frequency.(c) Find the Q of the circuit. (d) Find the amplitude of the voltage across
the inductor at resonance.
1
(a) Resonant frequency ω 0 =
(b) I =
LC
V
|V|
⇒| I |=
X
|X|
Vmax
⇒ I max =
ω 2 C 2 R 2 + (ω 2 LC − 1) 2
ωC
Vmax
⇒ I max =
R 2 + (ωL −
At resonance, ωL −
1 2
)
ωC
1
= 0 ⇒ I max =
ωC
⇒ I max =
1
(c) Q =
ω0 L
R
=
LC
R
L
=
1
R
Vmax
R 2 + 02
Vmax
R
L
C
(d) Maximum voltages across inductor = | V ⋅
XL
|
XR + XL + XC
ωL
= Vmax ⋅
R 2 + (ωL −
At resonance, ωL −
1
=0
ωC
∴ Maximum voltages across inductor = Vmax ⋅
ω0 L
R
1
LC
= Vmax ⋅
R
V
L
= max
R
C
L
1 2
)
ωC
11. A step-down transformer is used for recharging the batteries of portable electronic devices.
The turns ratio N2/N1 for a particular transformer used in a DVD player is N2 : N1. When used with
Vrms household service, the transformer draws an rms current of Irms from the house outlet. (a) Find
the rms output voltage of the transformer (b) Find the power delivered to the DVD player.
(a)
V2 N 2
N
=
⇒ V2 = 2 ⋅ V1
V1 N1
N1
⇒ V2 =
N2
⋅ Vrms
N1
(b) Power delivered = Vrms I rms
12. The RC high-pass filter shown in the figure below has a resistance R and a capacitance C.
What is the ratio of the amplitude of the output voltage to that of the input voltage for this
filter for a source frequency of f?
∆v out = ∆v R = ∆v in ⋅
⇒ | ∆v out | = | ∆v in | ⋅
⇒ | ∆v out | = | ∆v in | ⋅
⇒ | ∆v out | = | ∆v in | ⋅
XR
XR + XC
| XR |
| XR + XC |
R
2
XR + XC
2
1 2
)
ωC
ωCR
R2 + (
⇒ | ∆v out | = | ∆v in | ⋅
⇒ | ∆v out |= | ∆v in | ⋅
∴R =
R
R
(ωCR) 2 + 1
2πfCR
1 + (2πfCR) 2
| ∆v out |
2πfCR
=
| ∆v in |
1 + (2πfCR) 2
1/(ωC)
13. A current I is charging a capacitor that has square plates of length L on each side. The
plate separation is d. (a) Find the time rate of change of electric flux between the plates. (b)
Find the displacement current between the plates.
dQ
dV
=C
dt
dt
dV
dE
But V = Ed ⇒
=d
dt
dt
dQ
=I
Also,
dt
dQ
dV
dE
∴
=C
⇒ I = Cd
dt
dt
dt
2
A
L
C = ε0 = ε0
d
d
2
L
dE
dE
∴ I =ε0
⋅d⋅
⇒ I = ε 0 L2
d
dt
dt
dE
I
⇒
=
dt ε 0 L2
(a) Q = CV ⇒
(b) I d = ε 0
d ΦE
dt
d ΦE
dt
d EA
= ε0
dt
dE
= Aε 0
dt
dE
= L2ε 0
dt
I
= L2ε 0
ε 0 L2
⇒ Id = ε 0
⇒ Id
⇒ Id
⇒ Id
⇒ Id
⇒ Id = I
(from part (a))
14.
A current I is charging a capacitor that has circular plates R in radius. The plate
separation is d.
(a) What is the time rate of increase of electric field between the plates? (b)
What is the magnetic field between the plates r from the center?
dQ
dV
=C
dt
dt
dV
dE
But V = Ed ⇒
=d
dt
dt
dQ
Also,
=I
dt
dQ
dV
dE
∴
=C
⇒ I = Cd
dt
dt
dt
2
πR
A
C = ε0 = ε0
d
d
2
πR
dE
dE
∴ I =ε0
⋅d⋅
⇒ I = πε 0 R 2
d
dt
dt
dE
I
⇒
=
dt πε 0 R 2
(a) Q = CV ⇒
I
(b) I d = I
Ampere's Law ⇒
∫
r r
B ⋅ d s = µ 0 I im
⇒ B ⋅ 2π r = µ 0 I d
µ I
⇒ B = 0
2π r
r
Id
(I d = I)
I
B=?
15. In SI units, the electric field in an electromagnetic wave is described by
(ω is not given)
Ey = Emax sin(kx − ωt).
(a) Find the amplitude of the corresponding magnetic field oscillations. (b) Find the wavelength λ.
(c) Find the frequency f.
(a) E = cB ⇒ E max = cB max
⇒ B max =
(b) k =
2π
λ
⇒ λ=
E max
c
2π
k
(d) ω = ck ⇒ 2π f = ck
⇒ f=
ck
2π
16.
The speed of an electromagnetic wave traveling in a transparent nonmagnetic
substance is v = 1/ κµ 0 ε 0 where κ is the dielectric constant of the substance. Determine the
speed of light in water, which has a dielectric constant of 1.78 at optical frequencies.
Solution:
Just plug in the numbers!
17.
At one location on the Earth, the rms value of the magnetic field caused by solar radiation
is Brms. (a) Calculate the rms electric field due to solar radiation. (b) Calculate the average
energy density of the solar component of electromagnetic radiation at this location. (c)
Calculate the average magnitude of the Poynting vector for the Sun's radiation. (d) Assuming
that the average magnitude of the Poynting vector for solar radiation at the surface of the
Earth is Searth, compare your result in part (c) with this value.
(a) E = cB ⇒ E rms = cB rms
1
1
(b) u = u E + u B = ε 0 E 2 +
B2
2
2µ 0
1
1
∴< u >= ε 0 < E 2 > +
< B2 >
2
2µ 0
1
1
2
2
= ε 0 E rms +
B rms
2
2µ 0
1
1
2
= ε 0 (cB rms ) 2 +
B rms
2
2µ 0
(from part (a))
1
1
2
2
= ε 0 c 2 B rms +
B rms
2
2µ 0
1
1
1
2
2
= ε0 ⋅
B rms +
B rms
2
2µ 0
ε 0 µ0
=
=
(c =
1
ε 0 µ0
1 1
1
2
2
B rms +
B rms
2 µ0
2µ 0
1
µ0
B rms
2
r 1 r r
EB
(c) S =
E×B ⇒ S=
µ0
µ0
⇒ S=
cB 2
⇒ S avg =
c < B2 >
⇒ S avg = c
(d) The anwser is
S avg
S earth
(E = cB)
µ0
µ0
B rms
µ0
× 100%
2
( = c < u >)
)
18. A radio wave transmits W of power per unit area. A flat surface of area A is
perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect
absorber, calculate the radiation pressure on it.
Since the radiation is completely absornbed, P =
S avg
c
W
=
c
Note : The given W is just the intensity, with is the same as S avg
19.
A power W laser beam of diameter D is reflected at normal incidence by a perfectly
reflecting mirror. Calculate the radiation pressure on the mirror.
Since the radiation is reflected, P =
S avg = I =
2⋅
∴P =
W
=
A
W
4W
=
2
D
π ( ) 2 πD
2
4W
πD 2 = 8W
c
π cD 2
2S avg
c
20. What is the wavelength of electromagnetic wave in free space that have frequency f?
c = fλ ⇒ λ =
c
f
20.