1
Notes on Inductance and Circuit Transients
Joe Wolfe, Physics UNSW
Circuits with R and C
q
-q
o
What happens when you close the
switch?
(close switch at t = 0)
R
C
o
Current flows off capacitor, so
dq
i = −
dt
i
q
-q
R
C
q
q
Across capacitor: VC = C
C ≡
V
Across resistor:
V R = iR = Vc
dq
∴
dt = − i
V
= −R
q
= − RC
dq
1
so
q = − RC dt
dq
1
∫ q = − RC ∫ dt
t
ln q = − RC + const.
t = 0, q = qo ∴ ln qo = const.
t
a
ln a − ln b = ln b
ln q − ln qo = − RC
q
t
ln q = − RC
o
∴
q = qo e −t/RC
V = q/c ∴
V = Voe−t/RC
(1)
τ
= RC = time constant
V C
s
Units: ΩF = .
A V = C . C=s
dq
i = dt
qo - t/RC
V
=
e
= o e - t/RC
RC
R
i = io e - t/τ
V
Vo
Vo /e
τ
Vo /e2
t
2τ
i
io
i o/e
τ
i o/e 2
2τ
t
2
Example
A real C uses dielectric with dielectric const. 4.0 and
conductivity σ = 2 10-15 Ω -1m -1. It is charged then left
disconnected for 1 hour. What fraction of charge remains?
I
q
A
R
=
K, σ
d
C
-q
εA
KεoA
C = d =
d
ρL
d
R = A = Aσ
K ε oA K εo
d
RC = Aσ .
d = σ
4.0 x 8.9 10 -12 Fm - 1
=
= 1.78 x 104 s
2.0 10 -15 Ω -1 m -1
V = Voe - t/RC
V
Q
= V = e−t/RC = e−3600/τ = 0.82
Qo
o
∴
Example Electronic appliance requires 20 W at 40 V DC,
with max variation of 10%. What capacitance needed to store
energy between cycles?
+
Vm
C R
90%*Vm
AC
−
t
AC has 50 cycles per second, ∴ turns off 100 times/second, ie
every 10 ms.
V
−t/RC ≅ 0.90
Vm = e
t = 10 ms, C = ?, what is R?
R is the effective resistance of the circuit.
V2
We know V and P, so use P = R
R = V2/P = 80 Ω.
∴
e −t/RC ≅ 0.90
t
≅ − ln (0.90)
RC
− 10 ms
C ≅
80 Ω * ln . 9
= 13000 µF
Example the membrane of a "pacemaker" neurone of the
heart has a capacitance of 10 pF and a conductance G =
3.0 pΩ−1. Initially the potential difference across this
membrane is −80 mV. The neurone begins a pulse when the
potential difference reaches −60 mV. How long before it
begins a pulse? Neglecting the duration of the electrical impulse
itself, what is the pulse rate?
V
t
(1) →
ln
Vo = − RC
V
10 10 −1 2 -60
t = − RC ln
= −
ln  
Vo
3 10 −12
 -80
= 0.96 s
The "leakiness" of the membrane is now suddenly increased by
a factor of 2. How long between electrical impulses? and what
is the approximate pulse rate, again neglecting the duration of
the inidividual impulses.
V
RC
V
t = - R2 C ln V == 2 ln Vo = ...
o
= 0.48 s
(σ ≡ 1/ρ)
3
RC charging
R
ε
What happens when you close the
switch?
(close switch at t = 0,
q = 0, V = 0)
C
Current flows onto capacitor, so
dq
I = + dt
R
i
q
ε
C
-q
q
= iR + C
dq
take derivatives, use i = + dt
di
i
∴
0 = Rdt + C
di
dt
rearrange:
i = - RC
t
∴
ln i = - RC + const
initial condition ⇒ ln io = const
ε
Kirchoff
i = ioe-t/τ =
where io =
ε/R
ε
R
e-t/τ
(2)
τ = RC
and
i
io
i o/e
i o/e 2
τ
t
2τ
V
V
∞
V∞ (1 − 1/e)
τ
V∞ (1 − /e 2)
t
2τ
Inductance
Self-induction
B
i
flux through each turn of coil ≈ φB
d(Nφ Β )
Faraday ε = −
dt
NΦB
Inductance, L
L ≡
i
(3)
(L is the total flux linkage per unit current)
So
Note:
cf
Units:
ε
di
= − L dt
(4)
ε in coil
∝ rate of change in i
V across C
∝ integral of i
1 henry = 1 Volt second/amp
4
Lenz's law:
ε acts in direction to maintain current (and ∴
−
+
+
−
i
i
if i increases
if i decreases
∴ L resists change in i
ε
←
ε
→
B)
Example Automobile ignition coil (schematic)
(not in syllabus)
iron
contact
iron
core
spring
spring moves iron right to close contact. i flows which causes B
B attracts iron to left, which opens circuit
di
ε = − L dt
i → 0 quickly so large ε
Ideal Solenoid
A
i
∴
l
n turns/m
NΦB = n l BA
but B = µoni
NΦ B
nlµoniA
L = i
=
i
L = µon2lA
(5)
Note:
for non-magnetic materials, µ ≈ µο = 4π 10 -7 Hm-1
magnetic materials, µ >> µo
∴ iron cores for large L.
Example Coil in kettle has R = 60 Ω. It is 20 cm long, has
1 turn/mm, and is 4 mm in diameter.
i) what is L?
ii) it is connected to 240 V rms. Switch has breakdown
potential of 1000 V and is turned off at peak of cycle (V = 340
V). Estimate the duration of spark. (Neglect other L).
Vp
340
Peak i = R = 60 = 5.7A.
(5) → L = µο n2 l A
= 4π 10-7 .(103 )2 . (0.2) . π(2 10-3)2 SI
= 3.2 µH.
di
εL = - L dt ~ − 1000 V during spark
di
1000 V
∴ dt ~ const. =
L
i
5.7 x 3.2 10 - 6
∴ ∆t ≅ di =
= 18 ns.
1000 V
dt
(In practice ∆t is larger because of other L's.)
5
R - L Transients
di
Recall εL = - L
(Lenz's law: back emf)
dt
This prevents sudden change in i,
∴ i is continuous.
i) switch at a. After a long time,
R
a
di
ε
= 0, i = ε c/R
dt
b
L
ii) switch to b at t = 0
i
∴ at t = 0, i = εc /R
di
Kirchoff: - L dt = iR
R L
di
R
∴ i = i
L dt
Rt
∫ →
∴ ln i = - L + ln io
i = io e -Rt/L =
∴
where
τ
εc
R
e-t/τ
= L/R
(6)
H V A
Ω = A/s . V = s
Units:
i
io
io /e
i o/e 2
t
τ
2τ
iii) switch at b for long time (i = 0). Switch to a at t′ = 0.
di
εc - L dt = i R
solve → i =
where
εc
- t′/τ
)
R (1 - e
τ = L/R
i
i
∞
2
i∞(1 − 1/e)
τ
i∞ (1 − /e )
t'
2τ
Example
ε=
L
ε
R
r
6V
L = 264 H
r = 13k (≡ 13 kΩ)
R = 47k
Switch closed for long time. Open at t = 0.
What does the CRO show?
di
Switch closed,
after a long time dt = 0
Kirchoff's law:
V = i1 r = i 2 R
and iT = i1 + i2
CRO shows V = i2R = ε = 6 V
(7)
6
Switch opened
L forbids abrupt change in i
∴ at t = 0 i1(0) = ε /r
t → ∞ , i1 → 0
total series resistance is r+R
L
R
r
i1
 ε
- t(r + R)/L
so i1 =   e
r 
L
here τ = r+R
i2
 εR
so VCRO = i2 R = - i1 R = -  r 
 
i 2 = - i1
 εR
VCRO = -  r  e-t(r+R)/L
V
6V
t
- 22 V
Energy in B of L
~
Power supplied to L = − εL i
dUB
di
i.e. dt = L dt i


i.
U B = ∫ dUB = ∫ Li' di'
o
UB = 12 Li2
c.f. UE =
(8)
1
CV2
2
Energy density
1
2
UB
UB
2 Li
=
=
vol
Al
Al
(5) → L = µo n2 lA
(5)
(from mag.) → B = µoin
UB
B2
∴
= 12
(9)
(c.f. UE =
vol
µο
1
ε E 2)
2 o
Mutual Induction
N
1
B
N2
N 2 φ2
ε2
i1
1
G
2
Mutual Inductance M21 (of 2 w.r.t. 1)
N 2 Φ 21
c . f . L = N Φ 
M 21 ≡

i 
i1
so M21 i1 = N2 Φ21
di1
dΦ21
(Faraday's law)
M21 dt = N2 dt = - ε 2
(long proof) ⇒
so:
ε2
M21 = M12 = M
di1
di2
= − M dt , ε 1 = − M dt
(10)
(S.I. unit henry = Vs A - 1)
7
Example
Two long concentric solenoids.
M = ?
r
xx xxxx 1r
2
xxxxxx
M
B1
∴
M
M
n1 / unit length and n2 / unit length
N 2 Φ 21
n2 l Φ 2 1
= M21 =
=
i1
i1
= µo i n, and Φ 21 = b1A2
n2 l
=
(µο i1n1) π R22
i1
= π R2 2 l µ o n1 n2
(11)
LC oscillations
(assume ideal L:
rC = ∞)
rL = 0, ideal C:
Kirchoff's law:
di
q
-L
dt = C
dq
d2i
1 dq
i = dt ,
so - L 2 = C dt =
dt
d2i
1
∴
= - LC i = - ω 2 i
say
dt2
solve:
i = const sin (ωt + const)
i = im sin (ωt + φ)
1
where ω =
LC
√

i
C
(12)
di
V = - L dt = - L ω im cos (ωt + φ)
V = - Vm cos (ωt + φ)
Energy
C
UE. = 21 CV2 = 2 [....cos....]2
L
1
2
U B = 2 Li2 = 2 [....sin....]
U T = U E + UB
= 12 im 2L [LC ω 2 cos2 (
= 12 im 2 L
)]
) + sin2 (
[= constant]
also Vm = im ωL
2
1 Vm
1
so UT = 2 ω 2L 2 L = 2 CVm2
Example Front end of a radio:
Cmax (3 in //)
ii)
 εοA 
= 3 d 
Cmax
= 3
ε oπr2
d
Cmin = 1.1 pF
= .... = 16.7 pF
If we want to tune the AM band (54 - 160 kHz) what values of L and C1? How do we make the L?
√

1
1
1
fmax = 2π ω max = 2π
L(C min + C 1 )
fmax2
C max + C 1
∴ f 2 = ..... = C
min
min + C 1
2
2
∴ C1 (fmax - fmin ) = Cmax fmin2 - Cmin fmax2
∴ C1 = .... = 0.9 pF
√

1
1
1
1
f = 2π
∴ L = C . 4π2f2
LC
= 490 mH
(5) L = µο n2 lA
fmin =
1
= 2π
√
1
L(C max + C 1 )
1
1
= C
.
4qp 2 fmin2
max + C 1
say n = 2000/m , A = 5 cm2, l = ? l =
L
µon2A
= 200 m. Oops, try iron core:
µr =
µ
~ 2000
µo