Download Lecture 21. R-L and L-C Circuits.

Lecture 21. R-L and L-C Circuits.
Outline:
Mutual Inductance
“Charging” and “discharging” an inductor in R-L circuits.
Oscillations in L-C circuits
1
Self-Induction and Inductance
𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑠𝑠𝑠𝑠: ℰ = −
Variable
battery
𝐼
𝑩
The back
e.m.f. :
So: The flux depends on the current, so a wire loop
with a changing current induces an e.m.f. in itself that
opposes the changes
If we change the current by varying the voltage, the flux
changes and an “additional” e.m.f. is induced (sometimes
called the back e.m.f.). Lenz: This e.m.f. will drive a current in
the direction that opposes the change in the magnetic flux.
𝑑Φ
𝑑𝑑
ℰ =−
= −𝐿
𝑑𝑑
𝑑𝑑
Inductance:
(or self-inductance)
𝑑Φ𝐵
𝑑𝑑
Φ
𝐿≡
𝐼
𝐿 – the coefficient of proportionality between Φ
and 𝐼, purely geometrical quantity.
Units: T⋅m2/A = Henry (H)
The self-inductance
L is always positive.
This definition applies when there is a single current path
(see below for an inductance of a system of distributed currents 𝐽 𝑟 ).
Inductance as a
circuit element:
2
Inductance vs. Resistance
Inductor affects the current flow only if there
is a change in current (di/dt ≠ 0). The sign of
di/dt determines the sign of the induced
e.m.f. The higher the frequency of current
oscillations, the larger the induced e.m.f.
3
Mutual Inductance
Let’s consider two wire loops at rest. A time-dependent current in loop 1 produces a
time-dependent magnetic field B1. The magnetic flux is linked to loop 1 as well as
loop 2. Faraday’s law: the time dependent flux of B1 induces e.m.f. in both loops.
loop 2
𝑩𝟏
𝐼1
loop 1
Primitive “transformer”
The e.m.f. in loop 2 due to the time-dependent I1 in loop 1:
𝜕Φ1→2
ℰ2 = −
𝜕𝜕
Φ1→2 = 𝑀1→2 𝐼1
The flux of B1 in loop 2 is proportional to the current I1 in
loop 1. The coefficient of proportionality 𝑀1→2 (the so-called
mutual inductance) is, similar to L, a purely geometrical
quantity; its calculation requires, in general, complicated
integration. Also, it’s possible to show that
𝑀1→2 = 𝑀2→1 = 𝑀
so there is just one mutual inductance 𝑀.
𝑀 can be either positive or negative, depending on the choices made for the senses of
transversal about loops 1 and 2. Units of the (mutual) inductance – Henry (H).
ℰ2 = −𝑀
𝜕𝐼1
𝜕𝜕
𝑀=
Φ1→2
𝐼1
4
Calculation of Mutual Inductance
l
𝑀1→2 = 𝑀2→1 = 𝑀
Calculate M for a pair of coaxial solenoids
(𝑟1 = 𝑟2 = 𝑟, l 1 = l 2 = l, 𝑛1 and 𝑛2 - the
densities of turns).
𝐵1 = 𝜇0 𝑛1 𝐼1
the flux per one turn
2
𝑁2 Φ
𝜇
𝑁
𝑁
𝜋𝑟
0
1
2
𝑀=
= 𝑁2 𝜇0 𝑛1 𝜋𝑟 2 =
l
𝐼1
2 𝜋𝑟 2
𝜇
𝑁
0
𝐿 = 𝜇0 𝑛2 l 𝜋𝑟 2 =
l
Experiment with two coaxial coils.
Φ = 𝐵1 𝜋𝑟 2 = 𝜇0 𝑛1 𝐼1 𝜋𝑟 2
𝑀=
𝐿1 𝐿2
The induced e.m.f. in the secondary coil is much greater if we use a ferromagnetic core:
for a given current in the first coil the magnetic flux (and, thus the mutual inductance)
will be ~𝐾𝑚 times greater.
𝑑𝐼1
𝑑𝐼1
ℰ2 = 𝑀∗
= 𝐾𝑚 𝑀
𝑑𝑑
𝑑𝑑
no-core mutual inductance
5
“Discharging” an Inductor
𝑉
𝑖 𝑡
𝑟
−𝐿
Initially the switch is closed, the coil L conducts a certain
current (depends on the coil’s resistance 𝑟 ), and the
corresponding magnetic field energy is stored in the coil.
𝑑𝑑 𝑡
ℰ = −𝐿
>0
𝑑𝑑
𝑑𝑑 𝑡
− 𝑖 𝑡 𝑅 = 0,
𝑑𝑑
𝐼0
The switch is open at t = 0. As the current begins to
decrease, the e.m.f. is induced in the coil that opposes the
change by “pushing” the current through the coil and the
resistor 𝑅.
- 𝑖 𝑡 is the instantaneous value of the current through the coil
and through the resistor 𝑅. Let’s assume that 𝑅 ≫ 𝑟:
𝑑𝑑 𝑡
𝑅
+ 𝑖 𝑡 = 0,
𝑑𝑑
𝐿
𝑖 𝑡 = 𝐼0 𝑒𝑒𝑒 −
𝑡
𝐿/𝑅
𝐼0 = 𝑉/𝑟 - the steady current through the inductor (switch S is closed).
𝐿
𝜏=
𝑅
𝑡=𝜏
𝑑𝑑 𝑡
𝑅
= − 𝑑𝑑,
𝑖 𝑡
𝐿
- the time constant of the circuit which consists of
an inductor 𝐿 and resistor 𝑅 .
Let’s check that 𝐿/𝑅 has units of time:
𝐿 𝐿𝐼 2
𝐽
= 2→
=𝑠
𝑅 𝑅𝐼
𝐽/𝑠
6
Capacitors vs. Inductors
𝑖 𝑡 = 𝐼0 𝑒𝑒𝑒 −
𝑡
𝐿/𝑅
“current
supply”
𝐼0 was the current trough the
inductor when the switch was “ON”.
It could be unrelated to the net R of
the circuit.
The initial rate of energy dissipation:
𝑃 ≈ 𝐼0 2 𝑅 (the larger R, the faster
energy decreases) – thus, 𝜏 ∝ 1/𝑅.
𝐼0 = 𝑉0 /𝑅
𝜏 = 𝑅𝑅
“voltage
supply”
Looks similar, right? BUT the current
𝐼0 = 𝑉0 /𝑅 is inversely proportional
to R! Thus, the initial rate of energy
dissipation: 𝑃 ≈ 𝑉0 2 /𝑅 (the larger R,
the slower energy decreases) – thus,
𝜏 ∝ 𝑅.
11
“Discharging” an Inductor (cont’d)
𝑉
𝑟
Let’s check that the energy stored in the magnetic field is
100% transformed into Joule heat after switch S is open.
The initial magnetic field energy stored in the inductor:
𝑖 𝑡
1
𝑈𝐵 = 𝐿 𝐼0
2
2
1 𝑉
= 𝐿
2 𝑟
Switch S is open at t = 0. The back e.m.f.: ℰ 𝑡
Power dissipated
in the resistors:
𝐼0 = 𝑉/𝑟 - the steady current
through the inductor (switch S
is closed).
𝑑𝑑 𝑡
= 𝐿
𝑑𝑑
= 𝑅+𝑟 𝑖 𝑡
𝑉
𝑡
𝑖 𝑡 = 𝑒𝑒𝑒 −
𝑟
𝐿/ 𝑅 + 𝑟
𝑃 = 𝑅 + 𝑟 𝑖2 𝑡
Net thermal
energy released in
the resistors:
2
∞
∞
𝑉
2
� 𝑅 + 𝑟 𝑖 𝑡 𝑑𝑑 = 𝑅 + 𝑟 �
𝑟
0
= 𝑅+𝑟
𝑉
𝑟
2
0
2
−𝜏/2 𝑒 −∞/𝜏 − 𝑒 −0/𝜏
𝜏 = 𝐿/ 𝑅 + 𝑟
𝑒 −2𝑡/𝜏 𝑑𝑑
1 𝑉
= 𝐿
2 𝑟
2
14
Example
𝑉
𝑟
𝑖 𝑡
A. never
Initially the switch is closed, the coil L
conducts a certain current, and the
corresponding magnetic field energy is
stored in the coil. The switch is open at t =
0, and the back e.m.f. ℰ is induced in the
coil. Can ℰ 𝑡 = 0 be greater than V?
B. always
C. only if R+r is sufficiently small
D.
only if R+r is sufficiently large
ℰ = −𝐿
𝑑𝑑 𝑡
𝑑𝑑
𝑑
𝑡
𝑖0 𝑒𝑒𝑒 −
𝑑𝑑
𝐿/ 𝑅 + 𝑟
𝑅+𝑟
𝑡
= 𝐿𝑖0
𝑒𝑒𝑒 −
𝐿
𝐿/ 𝑅 + 𝑟
𝑉
𝑡
=
𝑅 + 𝑟 𝑒𝑒𝑒 −
𝑟
𝐿/ 𝑅 + 𝑟
= −𝐿
𝑖0 = 𝑉/𝑟
15
“Charging” an Inductor
As an inductor opposes any decrease in the current
flowing through it, it also opposes any increase in that
current.
At t = 0 the switch is closed [𝑖 𝑡 = 0 = 0]. As the current
begins to increase, the e.m.f. is induced in the coil that
opposes the change.
ℰ−𝐿
𝑑𝑑 𝑡
−𝑖 𝑡 𝑅 =0
𝑑𝑑
𝑑𝑑 𝑡
𝑅
ℰ
+ 𝑖 𝑡 = ,
𝑑𝑑
𝐿
𝐿
𝑡 ≫ 𝜏:
ℰ
𝑡
𝑖 𝑡 =
1 − 𝑒𝑒𝑒 −
𝑅
𝐿/𝑅
𝑖 𝑡 =
ℰ
𝑅
16
Oscillations in L-C Circuits
L-C circuits: the circuits with TWO elements that can store energy
(ideally, without dissipation). The energy flow back and forth
between L and C results in harmonic oscillations of 𝑞 𝑡 and 𝑖 𝑡 .
𝑞2
electric field energy
𝑈𝐸 =
in the capacitor
2𝐶
𝑞 𝑡
𝑞0 2
=
2𝐶
2𝐶
2
𝐿𝐿 2 magnetic field energy
𝑈𝐵 =
in the inductor
2
Let’s say, at t = 0 the capacitor is fully charged, 𝑖=0.
𝐿 𝑖 𝑡
+
2
2
19
Oscillations in L-C Circuits (cont’d)
𝑑𝑑 𝑡
𝑑2𝑞 𝑡
≡
,
𝑑𝑑
𝑑𝑡 2
𝑑𝑑 𝑡
𝑞 𝑡
−𝐿
−
= 0,
𝑑𝑑
𝐶
𝜔=
1
𝐿𝐿
1 2𝜋
𝑇≡ ≡
= 2𝜋 𝐿𝐿
𝑓
𝜔
Solution: 𝑞 𝑡 = 𝑞0 𝑐𝑐𝑐 𝜔𝜔 + 𝜑0
amplitude
𝑞 2 𝑞0 2
𝑞0 2
2
𝑈𝐸 =
=
𝑐𝑐𝑐 𝜔𝜔 =
1 + 𝑐𝑐𝑐 2𝜔𝜔
2𝐶 2𝐶
4𝐶
2
2𝑞
𝐿𝐿 2 𝐿𝜔2 𝑞0
𝐿𝜔
0
𝑈𝐵 =
=
𝑠𝑠𝑠2 𝜔𝜔 =
2
2
4
𝑇/2
𝑑2𝑞 𝑡
1
+
𝑞 𝑡 =0
𝑑𝑡 2
𝐿𝐿
2
𝑖 𝑡 =
angl. frequency
phase at t=0
𝑑𝑞 𝑡
= −𝜔𝑞0 𝑠𝑠𝑠 𝜔𝜔 + 𝜑0
𝑑𝑑
1 + 𝑠𝑠𝑠 2𝜔𝜔
𝑇
20
L-C-R Circuits
𝑑𝑑 𝑡
𝑞 𝑡
−𝐿
−
− 𝑖𝑖 = 0,
𝑑𝑑
𝐶
Solution for weak damping
𝑑2𝑞 𝑡
𝑅 𝑑𝑞 𝑡
1
+
+
𝑞 𝑡 =0
𝑑𝑡 2
𝐿 𝑑𝑑
𝐿𝐿
1
𝐿𝐿
𝑅2
≫ 4𝐿2:
𝑅𝑅
𝑞 𝑡 = 𝑞0 𝑒𝑒𝑒 −
𝑐𝑐𝑐 𝜔∗ 𝑡 + 𝜑0
2𝐿
weak damping (“small” R)
𝜔∗ =
1
𝑅2
−
𝐿𝐿 4𝐿2
strong damping (“large” R)
23
Next time: Lecture 22. AC circuits, Reactance.
§§ 31.1 - 2
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