Physics 1c practical, 2015
Homework 5 Solutions
Serway 33.42
L = 20.0 mH, C = 10−7 F , R = 20Ω, ∆Vmax = 100V
a
The resonant frequency for a series RLC circuit is f =
1
2π
√
1
LC
= 3.56 kHz
b
At resonance Imax =
∆Vmax
R
=5A
c
Q=
ω0 L
R
= 22.4
d
∆VL,max = XL Imax = ω0 LImax = 2.24 kV
Serway 33.49
(∆Vout )max
=
(∆Vout )rms
=
N2
2000
(∆Vin )max =
170 = 971V
N1
350
971
√ = 687V
2
Serway 33.77
(2 points)
The circuit is simply a voltage divider.
ZR
R
Vout
=
=√
2
Vin
ZRLC
R + (ωL − 1/ωC)2
Setting this equal to 1/2, squaring both sides, and rearranging terms:
(ωL − 1/ωC)2 = 3R2
Homework 5 Solutions – 2015
2
√
ωL − 1/ωC = ± 3R
√
2πf L − 1/2πf C = ± 3R
This equation must hold for f1 = 200 Hz and f2 = 4 kHz.
2πf1 L − 1/2πf1 C
=
2πf2 L − 1/2πf2 C
=
√
s1 3R
√
s2 3R
Where s = +1 or − 1 to replace the ±; the sign will be determined below. A simple change of
variables A = 1/C shows that this is a pair of linear equations in L and A. Solving (I used Cramer’s
Rule):
√
(
(
)
)
√
s2 f1 − s1 f2
3R s2 /f1 − s1 /f2
1
L=
= 2 3πR
;
A=
2π
f2 /f1 − f1 /f2
C
f2 /f1 − f1 /f2
Unfortunately, the signs of s1 and s2 are ambiguous, but if we restrict ourselves to positive values of
L and C (negative values are meaningless), then the only possible values are s1 = −1 and s2 = +1.
√
(
)
(
)
√
3R
1/f1 + 1/f2
1
f1 + f2
= 2 3πR
L=
;
2π
f2 /f1 − f1 /f2
C
f2 /f1 − f1 /f2
Numerically,
√
(
)
3(8 Ω)
1/(200 Hz) + 1/(4 kHz)
L=
= 580 µH.
2π
(4 kHz)/(200 Hz) − (200 Hz)/(4 kHz)
(
)
1
(4 kHz)/(200 Hz) − (200 Hz)/(4 kHz)
√
C=
= 54.6 µF.
(4 kHz) + (200 Hz)
2 3π(8 Ω)
(0 points)
C is given in part (a). Points for part (b) are included in (a).
(0.5 point)
Referring to the expression at the beginning of part a, Vout /Vin will be maximum when the denominator is minimum. This will occur when ω0 L − 1/ω0 C = 0. In that case, Vout /Vin = 1
(0.5 point)
Solving for ω0 in part c,
√
ω0 = 1/ LC;
1
f0 =
2π
√
f0 = ω0 /2π = 1/2π LC
√[
(
)] [
(
)]
√
2π
f2 /f1 − f1 /f2
f1 + f2
√
2 3πR
1/f1 + 1/f2
f2 /f1 − f1 /f2
3R
√
√
f0 = f1 f2 = (200 Hz)(4 kHz) = 894 Hz.
The resonance frequency is the geometric mean of the two half-amplitude frequencies. The circuit
actually has the more general property that the resonant frequency is the geometric mean of any
two frequencies with equal voltage amplitudes.
Homework 5 Solutions – 2015
3
(1 point)
The phase for an arbitrary frequency f is given by:
(
)
(
)
XL (f ) − XC (f )
2πf L − 1/2πf C
−1
−1
ϕ(f ) = tan
= tan
R
R
(
[
]
[
])
√
√ 1
1/f1 + 1/f2
f1 + f2
ϕ(f ) = tan−1
3f
− 3
f2 /f1 − f1 /f2
f f2 /f1 − f1 /f2
( √
[
])
3
f1 f2
−1
ϕ(f ) = tan
f−
f2 − f1
f
For the values we are interested in:
( √ )
ϕ(f1 ) = tan−1 − 3 = −π/3 radians,
−1.05 radians,
or
− 60◦ .
( √ )
ϕ(f2 ) = tan−1 + 3 = +π/3 radians, +1.05 radians, or + 60◦ .
√
From part c, f0 = f1 f2 , giving
( √
[
])
√
f1 f2
3
−1
ϕ(f0 ) = tan
f1 f2 − √
= tan−1 0 = 0.
f2 − f1
f1 f2
(0.5 point)
The average power across the resistor (for voltages given as amplitude) is
2
P = Vout
(f )/2R
By the statement of the problem, Vout (f1 ) = Vout (f2 ) = Vin /2, so
P (f1 ) = P (f2 ) =
2
1 Vin
1 (10.0 V)2
=
= 1.56 W.
8 R
8 8.00 Ω
By parts b and c, at resonance Vout (f0 ) = Vin , giving
P (f0 ) =
2
1 Vin
1 (10.0 V)2
=
= 6.25 W.
2 R
2 8.00 Ω
(0.5 point)
From the discussion in the text:
Q=
ω0 L
2πf0 L
2π(894 Hz)(580 µH)
=
=
= 0.407
R
R
8.00 Ω
This is a very poor Q because the pass band is extremely wide relative to the resonance frequency.
QP9
RMS quantities below are denoted by a tilde, for example: Ve .
Homework 5 Solutions – 2015
4
(1.5 points)
Because of Kirchhoff’s current rule, the current through this circuit will be the same for each part
of the loop. Looking particularly at the capacitor:
VeY Z
VeY Z
Ie =
=
= (2πf ) C VeY Z
XC
1/ωC
Ie = (2π)(1.00 kHz)(1.00 µF)(15.5 V) = 97.4 mA.
(2 points)
The total impedance of the circuit is
Z=
Ohms Law reads:
√
R2 + (ωL − 1/ωC)2
√
e = Ie R2 + (ωL − 1/ωC)2
VeXZ = IZ
Solving for L:
v(
)2
u
1
1u
VeXZ
t
L= 2 ±
− R2
ω C
ω
Ie
√(
)2
10.1 V
1
1
L=
±
− (35.0 Ω)2
((2π) 1.00 kHz)2 1.00 µF (2π) 1.00 kHz
97.4 mA
L = 40.8 mH
or
9.8 mH
It may help to draw a phasor diagram to see where the two possible values of L come from. The
magnitude of VXZ and the component along VR are fixed. The value of L is chosen so that the
phasors VC + VR + VL = VXZ . One possible diagram follows below. (The diagram is not necessary
for full credit.)
(1.5 points)
The RMS voltage across XY will be
√
e = Ie R2 + ω 2 L2
VeXY = IZ
For L = 40.8 mH:
VeXY = (97.4 mA)
√
[35.0 Ω]2 + [(2π)(1.00 kHz)]2 [40.8 mH]2 = 25.2 V.
For L = 40.8 mH:
√
VeXY = (97.4 mA) [35.0 Ω]2 + [(2π)(1.00 kHz)]2 [9.8 mH]2 = 6.9 V.
Homework 5 Solutions – 2015
5
QP13
Since Rb burns 100 Watts at 120 V and Pb = I 2 Rb , to reduce Pb to 25 Watts, the current I has to
be reduced by 1/2. Therefore, Rtotal = 2Rb and Radj = Rb . Now that the two resistors have the
same resistance, Ptotal = 2Pb = 50 Watts. The resistance of Rb is found from
2
VRM
S
Rb
⇒ Rb = 144 Ω.
100 =
In the case when Ladj is used, we can look at Pb = Vb2 /Rb . This implies the voltage across the bulb
Vb has to be reduced by 1/2.
Vb = VAC √
Rb2
Rb
+ (ωLadj )2
Rb
1/2 = √ 2
Rb + (ωLadj )2
(ωLadj /Rb )2 = 3
Ladj ≈ 0.66 H
Since there is no RMS power in the adjustable Ladj , the total power is 25 Watts.
QP28
a
A capacitor resists sudden voltage changes. So at t = 0, VC = 0.
b
at t → ∞ there will be no current across the capacitor. So we can effectively remove capacitor from
2
the original circuit and calculate VC . We get VC = R2R+R
VB
1
Homework 5 Solutions – 2015
6
c
VC =
∫
1 t
Q
=
IC dt
C
C 0
IC + I2 = I1
(1)
(2)
d
Kirchoff’s law for battery, R1 , R2 loop, and Capacitor, R2 loop are:
VB − I1 R1 − I2 R2 = 0
∫
1 t
Ic dt − I2 R2 = 0
C 0
(3)
(4)
e
solving (2) and (3) for I2 we get I2 =
VB −R1 IC
R1 +R2 .
Substituting this value in (4) we get
∫
VB − R1 Ic
1 t
IC dt −
R2 = 0
C 0
R1 + R2
∫ t
R1 + R2 1
VB
⇒
IC dt −
+ IC = 0
R1 R2 C 0
R1
(5)
Now using (1) we get
d
VC +
dt
f
time constant =
R1 R2
R1 +R2 C
(
R1 + R2
R1 R2
)
1
VB
VC =
C
CR1
(6)