Lecture 23. R-L and L-C Circuits.
Outline:
“Charging” and “discharging” an inductor in R-L circuits.
Reactance of L-C-R circuits.
1
Iclicker Question
A steady current flows through an inductor. If the current is
doubled while the inductance remains constant, the amount of
energy stored in the inductor
A. increases by a factor of
2.
B. increases by a factor of 2.
C. increases by a factor of 4.
D. increases by a factor that depends on
the geometry of the inductor.
E. none of the above
2
Iclicker Question
A steady current flows through an inductor. If the current is
doubled while the inductance remains constant, the amount of
energy stored in the inductor
A. increases by a factor of
2.
B. increases by a factor of 2.
C. increases by a factor of 4.
D. increases by a factor that depends on
the geometry of the inductor.
E. none of the above
1 2
𝐵2
𝑈 = 𝐿𝐼 =
∙ 𝑣𝑣𝑣𝑣𝑣𝑣
2
2𝜇0
3
“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
4
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.
5
Iclicker Question
A coil with a self-inductance of 1 H is connected in parallel to a
10-Ω light bulb and a 5-V battery. Very roughly how long would it
keep the light bulb lighted if the battery were disconnected from
the circuit? (Choose the closest answer).
A. 10s
B. 2s
C. 0.5s
D. 0.05s
E. 0.001s
6
Iclicker Question
A coil with a self-inductance of 1 H is connected in parallel to a
10-Ω light bulb and a 5-V battery. Very roughly how long would it
keep the light bulb lighted if the battery were disconnected from
the circuit? (Choose the closest answer).
A. 10s
B. 2s
C. 0.5s
D. 0.05s
E. 0.001s
𝑡
𝑖 𝑡 = 𝑖0 𝑒𝑒𝑒 −
𝐿/𝑅
𝑖 𝑡
2
= 𝑖0
2
2𝑡
𝑒𝑒𝑒 −
𝐿/𝑅
𝐿
= 0.05𝑠
2𝑅
7
“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
ℰ
𝑖 𝑡 =
𝑅
𝐿
𝜏=
𝑅
8
Iclicker Question
An inductance L and a resistance
R are connected to a source of
emf as shown. When switch S1 is
closed, a current begins to flow.
The final value of the current is
A. directly proportional to RL.
B. directly proportional to R/L.
C. directly proportional to L/R.
D. directly proportional to 1/(RL).
E. independent of L.
9
Iclicker Question
An inductance L and a resistance
R are connected to a source of
emf as shown. When switch S1 is
closed, a current begins to flow.
The final value of the current is
A. directly proportional to RL.
B. directly proportional to R/L.
C. directly proportional to L/R.
D. directly proportional to 1/(RL).
E. independent of L.
10
Iclicker Question
An inductance L and a resistance
R are connected to a source of
emf as shown. When switch S1 is
closed, a current begins to flow.
The time required for the
current to reach one-half its final
value is
A. directly proportional to RL.
B. directly proportional to R/L.
C. directly proportional to L/R.
D. directly proportional to 1/(RL).
E. independent of L.
11
Iclicker Question
An inductance L and a resistance
R are connected to a source of
emf as shown. When switch S1 is
closed, a current begins to flow.
The time required for the
current to reach one-half its final
value is
A. directly proportional to RL.
B. directly proportional to R/L.
C. directly proportional to L/R.
D. directly proportional to 1/(RL).
E. independent of L.
12
Iclicker Question
10𝑉
Switch S is closed at t = 0.
1Ω
2𝜇𝜇
1Ω
1𝜇𝜇
𝐼
Find current 𝐼 at t = 1s.
A. 𝐼 = 0𝐴
B. 𝐼 = 5𝐴
C. 𝐼 = 10𝐴
D. 𝐼 = 15𝐴
E. 𝐼 = 20𝐴
13
Iclicker Question
10𝑉
Switch S is closed at t = 0.
1Ω
2𝜇𝜇
1Ω
1𝜇𝜇
𝐼
Find current 𝐼 at t = 1s.
A. 𝐼 = 0𝐴
B. 𝐼 = 5𝐴
C. 𝐼 = 10𝐴
D. 𝐼 = 15𝐴
E. 𝐼 = 20𝐴
At 𝑡 ≫ 𝑅𝑅, 𝐿/𝑅 we can
neglect the e.m.f.
generated by the inductor,
as well as the current
through the capacitor.
ℇ
Current 𝐼 = = 10𝐴.
𝑅1
14
Response of L-C-R Circuits to Alternating Currents
So far we have considered transient processes in R-C and R-L
circuits: the approach to the stationary (time-independent)
state after some perturbation (switch on / off).
Now we switch to the discussion of the AC (cos 𝜔𝜔 + 𝜑 )
driven circuits, e.g. the circuits driven by an AC voltage
source.
We disregard all transient processes and instead consider the
steady-state AC currents: currents and voltages vary with time
as 𝑐𝑐𝑐 𝜔𝜔 + 𝜑0 , but their amplitudes are t-independent.
15
Amplitudes, rms Values, and Average Power in AC Circuits
𝐼 𝑡
𝑉 𝑡
RLC
Instantaneous 𝑉 𝑡 = 𝑉0 cos 𝜔𝜔
values:
𝐼 𝑡 = 𝐼0 cos 𝜔𝜔 + 𝜙
Instantaneous
power:
𝑃 𝑡 =𝑉 𝑡 𝐼 𝑡
Currents and voltages are
NOT necessarily in phase,
𝜙 is the phase shift
between V and I (the
“phase angle”).
1
𝑃 𝑡 = 𝑉0 cos 𝜔𝜔 ∙ 𝐼0 cos 𝜔𝜔 + 𝜙 = 𝑉0 𝐼0 cos 2𝜔𝜔 + 𝜙 + cos 𝜙
2
Average
power:
𝑉 𝑡
𝐼 𝑡
𝑃𝑎𝑎 ≡ 𝑃 𝑡
1
= 𝑉0 𝐼0 cos 𝜙
2
R
16
rms Values
𝑃𝑎𝑎 ≡ 𝑃 𝑡
Root mean square (rms): the
square root of the average of
the square of the quantity:
𝑎𝑟𝑟𝑟 = 𝑎2 𝑡
=
1
𝑉0 𝐼0 cos 𝜙
2
𝑉𝑟𝑟𝑟 =
𝑉0
2
𝐼𝑟𝑟𝑟 =
𝐼0
𝒄𝒄𝒄 𝝓 the power factor
𝑃𝑎𝑎 = 𝑉𝑟𝑟𝑟 ∙ 𝐼𝑟𝑟𝑟 cos 𝜙
2
Example:
“120V” wall outlet: 𝑓 = 60𝐻𝐻, 𝜔 = 2𝜋 ∙ 60
𝑉𝑟𝑟𝑟 = 120𝑉,
𝑟𝑟𝑟
𝑠
= 377𝑟𝑟𝑟/𝑠, 𝑇 =
𝑉0 ≈ 170𝑉
1
𝑓
≈ 17𝑚𝑚
17
Phasors / Complex Number Representation
Problem: To find current/voltage in R-L-C circuits, we need to solve differential equations.
Solution: The use of complex numbers / phasors allows us to replace linear differential
equations with algebraic ones, and reduce trigonometry to algebra .
phasor
Instantaneous
value
𝜙1 − 𝜙2
We represent voltages and currents in the R-L-C
circuits as the phase vectors (phasors) on the 2D
plane. Quantity: 𝐴 𝑡 = 𝐴0 cos 𝜔𝑡. Corresponding
phasor: a vector of length 𝐴0 rotating counterclockwise with the angular frequency 𝜔 .
Instantaneous value of 𝐴 𝑡 is the projection of the
phasor onto the horizontal axis.
If all the quantities oscillate with the same 𝜔, we can get
rid of the term 𝜔𝜔 by using the rotating (merry-goaround) reference frame.
We’ll consider the steady state of AC circuits, when all amplitudes (the phasor lengths)
are t-independent, and the only time dependence remaining is in the single frequency
sinusoidal oscillation of voltages and currents. The angle between different phasors
represents their relative (t-independent) phase.
18
Complex Numbers, Phasors
𝑍 ≡ 𝑎 + 𝑖𝑖
= 𝑅𝑅 𝑍 + 𝑖 𝐼𝐼 𝑍
𝑏
𝜙
−𝜙
−𝑏
𝑅𝑅 𝑍 = 𝑟 𝑐𝑐𝑐𝜙
𝐼𝐼 𝑍 = 𝑟 𝑠𝑠𝑠𝜙
𝑎
𝑍 = 𝑟𝑒 𝑖𝜙
𝑍 ∗ ≡ 𝑎 − 𝑖𝑖
= 𝑅𝑅 𝑍 − 𝑖 𝐼𝐼 𝑍
complex conjugate of 𝑍
Imaginary
unit:
𝑖 2 = −1
𝑖 ≡ −1
𝑒 𝑖𝜙 = cos 𝜙 + 𝑖 sin 𝜙
𝑒
𝜋
𝑖2
=𝑖
𝑒
𝜋
−𝑖 2
𝐴 𝑐𝑐𝑐 𝜔𝑡 + 𝜙 = 𝐴
𝐴 𝑠𝑠𝑠 𝜔𝑡 + 𝜙 = 𝐴
= −𝑖
𝑒𝑖
𝑒𝑖
𝜔𝜔+𝜙
𝜔𝜔+𝜙
1
= −𝑖
𝑖
Euler’s
relationship
+ 𝑒 −𝑖 𝜔𝜔+𝜙
= 𝑅𝑅 𝐴 𝑒 𝑖
2
− 𝑒 −𝑖 𝜔𝜔+𝜙
= 𝐼𝐼 𝐴 𝑒 𝑖
2𝑖
𝜔𝜔+𝜙
𝜔𝜔+𝜙
The absolute value (or modulus or magnitude):
𝑍 ≡ 𝑍 ∙ 𝑍∗
𝑍 =
𝑟 𝑐𝑐𝑐𝑐 + 𝑖 𝑠𝑠𝑠𝑠 𝑟 𝑐𝑐𝑐𝑐 − 𝑖 𝑠𝑠𝑠𝑠
= 𝑟 2 𝑐𝑐𝑐 2 𝜙 + 𝑠𝑠𝑠2 𝜙 = 𝑟
Phasor: refer to either 𝐴 𝑒 𝑖 𝜔𝜔+𝜙 or just 𝐴 𝑒 𝑖 𝜙 . In the latter case, it is understood
to be a shorthand notation, encoding the amplitude and phase of an underlying
sinusoid.
19
Complex Numbers, Phasors (cont’d)
The use of complex numbers / phasors allows us to replace linear differential
equations with algebraic ones, and reduce trigonometry to algebra:
Addition:
𝑎 + 𝑖𝑖 + 𝑐 + 𝑖𝑖 = 𝑎 + 𝑐 + 𝑖 𝑏 + 𝑑
Multiplication:
Differentiation:
𝐴𝑒 𝑖𝜔1 𝑡 ∙ 𝐵𝑒 𝑖𝜔2𝑡 = 𝐴𝐴𝑒 𝑖
𝑑
𝐴𝐴 𝑖𝑖𝑖 = 𝑖𝜔𝐴𝐴 𝑖𝑖𝑖
𝑑𝑑
𝜔1 +𝜔2 𝑡
20
Complex Power, Active Power and Reactive Power
𝑉 𝑡
𝐼 𝑡
1
𝑃 𝑡 = 𝑉0 cos 𝜔𝜔 ∙ 𝐼0 cos 𝜔𝜔 + 𝜙 = 𝑉0 𝐼0 cos 2𝜔𝜔 + 𝜙 + cos 𝜙
2
RLC
𝑃𝑎𝑎 = 𝑉𝑟𝑟𝑟 ∙ 𝐼𝑟𝑟𝑟 cos 𝜙
Complex power:
𝑺 = 𝑉 𝑡 𝐼 ∗ 𝑡 = 𝑉𝑟𝑟𝑟 𝑒 𝑖𝑖𝑖 ∙ 𝐼𝑟𝑟𝑟 𝑒 −𝑖
= 𝑷 + 𝒊𝒊
Average (active,
real) power
𝜔𝜔+𝜙
= 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟 𝑒 −𝑖𝑖 = 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟 cos 𝜙 − 𝑖 sin 𝜙
Reactive
power
𝑷 = 𝑹𝑹 𝑺 = 𝑽𝒓𝒓𝒓 𝑰𝒓𝒓𝒓 𝒄𝒄𝒄 𝝓
Apparent power:
cos 𝜙 =
𝑆 =
𝑃
𝑆
𝑃2 + 𝑄2
21
Resistor
𝑉 𝑡
𝐼 𝑡
AC current through a resistor and AC voltage across
the resistor are always in phase.
Power dissipated in a resistor:
𝑃𝑎𝑎
Active (average)
power:
1
1
= 𝑉0 𝐼0 𝑐𝑐𝑐𝑐 = 𝑉0 𝐼0 = 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟
2
2
𝜙=0
𝑃 = 𝑅𝑅 𝑆 = 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟 cos 𝜙 = 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟
𝜙=0
𝑄 = 𝐼𝐼 𝑆 = 0
If the load is purely resistive, both the
current and voltage reverse their polarity at
the same time. At every instant the product
of voltage and current is positive or zero,
with the result that the direction of energy
flow does not reverse: only the active
power is transferred from the power source
to the load.
22
Resistor Reactance
𝑉 𝑡
𝐼 𝑡
Reactance (X) is the opposition of a circuit element to a change of
electric current due to that element's L , C, or R.
Units: Ohms.
Resistor:
𝑉 = 𝐼𝑅 = 𝐼𝑋𝑅
𝑿𝑹 = 𝑹
Reactance for a resistor is the same as its resistance,
it is frequency-independent.
23
Capacitor
𝑉 𝑡
𝐼 𝑡
𝝅
𝒊(𝝎𝝎+ 𝟐 )
𝝎𝝎𝟎 𝒆
𝑉 𝑡 −
𝑄 𝑡
=0
𝐶
𝑉 𝑡 = 𝑉0 𝑒 𝑖𝑖𝑖
𝐼 𝑡 = 𝐼0 𝑒 𝑖
𝜔𝜔+𝜙
𝐼 𝑡 =
𝑑𝑄 𝑡
𝑑𝑑
1 𝑑𝑄 𝑡
1
𝑑𝑑 𝑡
=
= 𝐼 𝑡
𝐶 𝑑𝑑
𝐶
𝑑𝑑
𝑑𝑉 𝑡
= 𝑖𝑖 𝑉0 𝑒 𝑖𝑖𝑖
𝑑𝑑
𝑰𝟎 𝒊(𝝎𝝎+𝝓)
= 𝒆
𝑪
𝝓 = 𝝅/2
𝑽 𝒕
𝑖𝑖𝑖0 𝑒 𝑖𝑖𝑖 =
𝐼0 𝑖(𝜔𝜔+𝜙)
𝑒
𝐶
𝑖=𝑒
𝜋
𝑖2
For a capacitor, voltage
LAGS current by 900.
𝑰 𝒕
Current (reference phasor)
24
Capacitor Reactance
𝑉 𝑡
𝐼 𝑡
Capacitor
𝑑𝑑 𝑡
1
= 𝐼 𝑡
𝑑𝑑
𝐶
𝑉0 =
𝐼0
𝜔𝜔
𝟏
𝑿𝑪 =
𝝎𝝎
𝑖𝑖𝑉0
𝑉𝑟𝑟𝑟 =
𝑒 𝑖𝑖𝑖
𝐼𝑟𝑟𝑟
𝜔𝜔
1
𝑖
= 𝐼0 𝑒
𝐶
𝜋
𝜔𝜔+ 2
𝑉𝑟𝑟𝑟 = 𝐼𝑟𝑟𝑟 𝑋𝐶
𝑽 𝒕 =𝑰 𝒕
−𝒊 𝑿𝑪
25
Capacitor (cont’d)
𝑉 𝑡
𝐼 𝑡
Active (average)
power:
𝑽 𝒕
𝑰 𝒕
Current (reference
phasor)
𝑃 = 𝑅𝑅 𝑆 = 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟 cos 𝜙 = 0
𝜙=
𝜋
𝑰
𝑽
𝑷
2𝜋 phase 𝜔𝑡
𝜋
2
If the loads are purely reactive (capacitors and/or
inductors, no resistive elements), then the voltage
and current are 900 out of phase. For half of each
cycle, V⋅I is positive, but on the other half of the
cycle, the product is negative - on average, exactly
as much energy flows toward the load as flows
back. There is no net transfer of energy to the
load - only reactive power flows back and forth.
26
Inductor
𝐼 𝑡
𝑉 𝑡
𝑉 𝑡 −𝐿
𝑉 𝑡 = 𝑉0 𝑒 𝑖𝑖𝑖
𝐼 𝑡 = 𝐼0 𝑒 𝑖
𝑽𝟎
𝒆𝒊𝒊𝒊
=𝝎
𝝅
𝒊 𝝎𝝎+𝝓+ 𝟐
𝑳𝑳𝟎 𝒆
𝜔𝜔+𝜙
𝑑𝑑 𝑡
=0
𝑑𝑑
𝑉 𝑡 =𝐿
𝑑𝑑 𝑡
= 𝑖𝑖 𝐼0 𝑒 𝑖
𝑑𝑑
𝝓 = −𝝅/2
𝜔𝜔+𝜙
𝑑𝑑 𝑡
𝑑𝑑
𝑉0 𝑒 𝑖𝑖𝑖 = 𝑖𝜔𝐿𝐼0 𝑒 𝑖
𝑖=𝑒
𝑖
𝜋
2
𝜔𝜔+𝜙
In an inductor, current LAGS
voltage by 900
27
Inductor Reactance
𝑉 𝑡
𝐼 𝑡
Inductor:
𝑑𝑑 𝑡
𝑉 𝑡 =𝐿
𝑑𝑑
𝑉0
𝑒 𝑖𝑖𝑖
=𝐿
𝜋
𝑖 𝜔𝜔− 2
𝑖𝜔𝐼0 𝑒
𝑉0 = 𝜔𝐿 𝐼0 𝑉𝑟𝑟𝑟 = 𝜔𝜔 𝐼𝑟𝑟𝑟 𝑉𝑟𝑟𝑟 = 𝐼𝑟𝑟𝑟 𝑋𝐿
𝑿𝑳 = 𝝎𝑳
𝑽 𝒕 = 𝑰 𝒕 𝒊𝑿𝑳
28
Inductor (cont’d)
𝑉 𝑡
𝐼 𝑡
Active (average)
power:
𝜋
𝑰
𝑽
−𝑷
𝑃 = 𝑅𝑅 𝑆 = 𝑉𝑟𝑟𝑟 𝐼𝑟𝑟𝑟 cos 𝜙 = 0
2𝜋 time t
phase 𝜔𝑡
𝜋
𝜙=−
2
Again, the power IS NOT dissipated in an
inductor: it is stored in the magnetic field of
the inductor for half a period, and returned
to the power source for another half.
29
Reactance - AC (cos(𝜔𝑡+𝜑)) driven circuits!
Resistor
Capacitor
Inductor
𝑽 = 𝑰𝑿𝑹
𝟏
𝑿𝑪 =
𝝎𝝎
𝑿𝑳 = 𝝎𝑳
𝑿𝑹 = 𝑹
𝑽 𝒕 =𝑰 𝒕
−𝒊 𝑿𝑪
𝑽 𝒕 = 𝑰 𝒕 𝒊𝑿𝑳
Major differences between reactance and resistance: the
reactance for L and C changes with frequency, it reflects
(being combined with ±𝑖) the phase shift between V and I,
and it dissipates no energy.
30
Resonance in the L-C circuit (L24 in detail)
𝐼
𝑰(𝒊𝒊𝑳 )
𝑰𝑰
𝐼
𝑰(−𝒊𝒊𝑪 )
Condition for the Resonance: reactances for L and C
are of the same magnitude:
𝑋𝐶 =
1
𝜔𝜔
1
= 𝜔0 𝐿
𝜔0 𝐶
=
𝑋𝐿 = 𝜔𝐿
𝝎𝟎 =
𝟏
𝑳𝑳
𝜔0
𝑓0 =
2𝜋
31
Next time: Lecture 24. Impedance of AC circuits,
§§ 31.3 - 6
32
Appendix 1: “Discharging” an Inductor – energy transformation
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
33
Appendix 2: 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
34