Φ21 Fall 2006
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At
HW17 Solutions
Problem K33.22
t = 0 s,
the current
the current in the circuit in the gure is
1
2 I0 ?
I0 .
At what time is
Solution: The inductor opposes any change in the current through it.
τ = L/R.
In an LR circuit, the time constant is
The current drops
exponentially with this time constant.
I(t) =
e−tR/L
−tR/L
tR/L
t
1
I0
2
= 1/2
= ln (1/2)
Figure 1: An LR circuit for problem K33.22.
= ln 2
(
)
= L ln 2/R = 50 × 10−3 H ln 2/ (500 Ω)
=
2
I0 e−t/τ = I0 e−tR/L =
6.93 × 10−5 s
Energy within an L-C circuit
Consider an L-C circuit with capacitance
C , inductance L, and no voltage
source. As a function of time, the charge on the capacitor is Q(t) and the current through the inductor is
I(t). Assume that the circuit has no resistance and that at one time the capacitor was charged.
Part A. As a function of time, what is the energy
Solution: The energy in an inductor is
Part C. What is the total energy
inductance
stored in the inductor? (In terms of
L
and
I (t).)
stored in the capacitor? (In terms of
C
and
Q(t).)
UL (t) = 12 LI(t)2 .
Part B. As a function of time, what is the energy
Solution: The energy in a capacitor is
UL (t)
UC (t) =
Utotal
UC (t)
2
1 Q(t)
2 C .
stored in the circuit? (In terms of the maximum current
I0
and
L.)
Solution: To solve this, we have to know how the current in an L-C circuit behaves. If there is an initial
current owing somehow, it will begin charging the capacitor. The inductor will cause this current to keep
owing until the charge builds up enough to oppose the current, at which point the current will stop and
the capacitor will be charged. Then, the capacitor will force the current in the opposite direction while the
inductor tries to keep the current constant.
In the end, the current will ow back and forth sinusoidally
through the circuit, charging the opposite plates of the capacitor.
The mathematical derivation is in the
book, and the current is
I(t) = I0 cos (ωt)
Without an external voltage source, the frequency will be the resonant frequency of the capacitor and
inductor,
√
ω = 1/ LC .
This tells us
UL (t), but we need the charge on the capacitor. Either integrate the
Q(t) or use the phasor diagram. The current phasor is to the right.
Vc = XC I0 = I0 /ωC and the direction is down. The charge on the
current (because it is the derivative of
The voltage across the capacitor is
capacitor is
Q(t) = CVC (t) = C
I0
I0
cos(ωt − π/2) =
sin (ωt)
ωC
ω
1
The total energy is then
Utotal
1 I02
1 2
LI0 cos2 (ωt) +
sin2 (ωt)
2
2C ω 2
1 2
1 LCI02
LI0 cos2 (ωt) +
sin2 (ωt)
2
2 C
1 2
LI
2 0
= UL (t) + UC (t) =
=
=
Notice that with no resistance to dissipate the energy, the energy is constant.
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A Radio Tuning Circuit
A radio can be tuned into a particular station frequency by adjusting the capacitance in an L-C circuit.
Suppose that the minimum capacitance of a variable capacitor in a radio is 4.11 pF.
Part A. What is the inductance
L
of a coil connected to this capacitor if the oscillation frequency of the
L-C circuit is 1.58 MHz, corresponding to one end of the AM radio broadcast band, when the capacitor is
set to its minimum capacitance?
Solution: The inductance is found from the resonant frequency formula
2
ω 2 = (2πf ) =
1
LC
⇒
L=
1
2
(2πf ) C
=
1
2
(2π1.58 × 106 Hz) (4.11 × 10−12 F )
Part B. The frequency at the other end of the broadcast band is 0.533 MHz.
capacitance
Cmax
= 2.47 × 10−3 H
What is the maximum
of the capacitor if the oscillation frequency is adjustable over the range of the broadcast
band?
Solution: The capacitance is found from the same formula.
C=
1
2
(2πf ) L
=
1
(2π0.533 ×
106
2
Hz) (2.47 × 10−3 H)
2
= 3.61 × 10−11 F
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Problem K35.36
Evaluate
VR
in the gure at various EMF frequencies.
First, let's think about the phasor diagram.
Solution:
In the series
circuit, we know the current is constant, so we draw a current phasor
in some direction with length
I0 .
Then we draw the
same direction as the current phasor, with a value of
draw the
VC
phasor at a right angle to
VR
VR phasor
VR = I0 R.
in the
Then,
and in the trailing direction
(because voltage lags in a capacitor), which is clockwise from VR . The
I
length is VC = XC I0 = 0 . Since there is no inductance, we can use the
ωC
pythagorean theorem to nd the total voltage, which is the vector sum
Figure 2: A driven R-C circuit for
problem K35.36.
of the voltage phasors of the series components.
√
√
VR2
Vtotal =
Now that we know
I0 ,
we can nd
+
VC2
=
(
2
(I0 R) +
I0
ωC
√
)2
= I0
R2 +
1
= E0
ω2 C 2
VR .
E0 R
VR = I0 R = √
R2 + ω21C 2
Make sure you understand how this formula came about. This is the imprortant part. Now for the numbers.
Part A.
f = 100 Hz
Solution:
Part B.
= 0.994 V
ω = 300 · 2π s−1
and
VR = 2.87 V.
ω = 1000 · 2π s−1
and
VR = 7.07 V.
and
VR = 9.49 V.
f = 3000 Hz
Solution:
Part E.
1
(200π s−1 )2 (1.59×10−6 F)2
f = 1000 Hz
Solution:
Part D.
(10 V)(100 Ω)
(100 Ω)2 +
f = 300 Hz
Solution:
Part C.
ω = 2πf = 200π s−1 and VR = √
ω = 3000 · 2π s−1
f = 10, 000 Hz
Solution:
ω = 10000 · 2π s−1
and
VR = 9.95 V.
Notice that the lower frequencies are blocked by the capacitor while the high frequencies get through to
the resistor.
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