# Lecture 9/12/03 ```Write your responses to the
following questions and turn it in.
Outline
• RCL circuit(s) (L = inductance or coil)
–
–
–
–
• What activities are, do you feel, most useful
• What activities are, do you feel, useless for
• Any suggestions on any activities you want
to see during lectures?
• What do you like most about the labs?
• What do you dislike about the labs?
What’s “L?”
Ohm’s Law of L?
LC circuit
RLC circuit(?)
Physics of L
Side: How can you make L large?
• If you can create more B field for a given
current, L will be larger
– VL = L dI/dt, where L is called “self
inductance.”
– Coil (many loops compared to a single loop) is
better.
– Putting iron (or ferrite) core will help “trap” B
field.
• Ferrite (insulator; susceptibility ~ 100) is better
since it prevents Eddy currents in the core.
• Works at higher frequency (up to GHz!)
Responses to sinusoidal signal
Solving this Eq.
1K
• Ohm’s Law of capacitor
A
in
– I = C dVC /dt = C d/dt (Vin – Vout )
• Ohm’s Law of inductor
C
0.1 uF
B
L
6.8 mH
– VL = Vout = L dI/dt.
– For sinusoidal, VL = jωL I (ZL ~ jωL)
• Block high-freq signal – opposite of capacitor
• Combining the two:
• Vout = L dI/dt = LC (d/dt)2 (Vin – Vout)
• If
– Vin = Aejωt and
– Vout = Bej(ωt+ϕ ),
• The above Eq. can be re-written as
– Bej(ωt+ϕ ) = LC (jω) 2 (Aejωt – Bej(ωt+ϕ ))
– Vout = L dI/dt = LC (d/dt) 2 (Vin – Vout)
1
More arithmetic
• Bej(ωt+ϕ) = LC (jω) 2 (Aejωt – Bej(ωt+ϕ)) or
• Bejϕ = – LCω2 (A – Bejϕ)
– Single Eq for two unknowns, B and ϕ!
– Both imaginary and real parts need to be equal!
• Bcosϕ = – LCω2 (A – Bcosϕ) and
• Bsinϕ = – LCω2 (– Bsinϕ).
More arithmetic
2
• Bcosϕ = – LCω (A – Bcosϕ) and
• sinϕ = 0, cosϕ = &plusmn;1. (ϕ is either 0 or π:
both lead to the same solution!)
• B(1 – LCω2 ) = – LCω2 A.
• B = – A LCω2/(1 – LCω2)
= – A Lω/(1/Cω – Lω)
More arithmetic
• Bcosϕ = – LCω2 (A – Bcosϕ) and
• Bsinϕ = – LCω2 (– Bsinϕ).
• The 2nd Eq. holds only when Bsinϕ = 0, or
LCω2 = 1.
– B = 0 won’t satisfy Eq. 1.
– In general LCω 2 ≠ 1.
– So sinϕ = 0, cosϕ = &plusmn;1
Some interpretation &amp; observations
• B = – A Lω/(1/Cω – Lω)
• like a voltage divider, but it’s as if the resistance
(imaginary) is in the opposite directions for C and
L, so the sum = subtraction.
• When LCω 2&lt;1 B and A have opposite sign (180&deg;
out of phase), and LCω 2 &gt;1, they are in phase.
• When 1/Cω ~ Lω (or LCω2 ~1), B will be much
larger than A – amplification w/o amplifier!
• This is like forced resonant oscillation.
– Used in radio (TV) station selection. (C is variable for
a typical radio to tune to various stations.)
– If C ~ 100 pF, what should L be to tune to 1500 kHz?
Dividing by 0?
Graphically, minimal resistance
12
LCω 2=1)
• When 1/Cω = Lω (or
There is no
solution – dividing by 0!
• In reality, there is always some resistance (long
coil has large resistance unless super conductor!)
in addition to L and C, so dividing by 0 won’t
happen. i.e.
• Bejϕ = A jLω / (1/jCω + R + jLω).
– When LCω 2=1, Bejϕ = A jLRω (= A jR/Cω)
– i.e. ϕ = π /2 (90&deg; shifted: cosine in, – sine out, for
example.)
10
Output amplitude
RCω0 = 0.1
8
6
Output phase
4
2
0
-5
-4
-3
-2
-1
0
LCω 2 = 1
1
2
3
4
5
Log(ω 2 in unit
of 1/L C)
2
More resistance (damping)
Strong damping
4
3.5
RCω0 = 0.3
3.5
3
3
Output phase
2.5
2.5
2
Output amplitude
2
Output amplitude
Output phase
RCω0 = 1.0
1.5
1.5
1
1
0.5
0.5
0
-5
-4
-3
-2
-1
0
0
LCω 2
1
=1
2
3
Log(ω 2
4
5
in unit
of 1/L C)
-5
-4
-3
-2
-1
0
1
LCω 2 = 1
2
3
4
5
Log(ω 2 in unit
of 1/L C)
3
```