CHAPTER FOUR
MUTUAL INDUCTANCE
4.1 Inductance
4.2 Capacitance
4.3 Serial-Parallel Combination
4.4 Mutual Inductance
4.1 Inductance
Inductance (L) in Henry is the circuit parameter used to
describe an inductor. The symbol a coil wire is remainder
that inductance is a consequence of a conductor linking a
magnetic field.
L = (N2µA)/l
L-inductance (H), N-kebolehtelapan magnet, A-area,l-length
Voltage drop across inductor , v = L di//dt
where, v in volts, L in Henry, i in ampere and t in second.
Voltage across the terminal is proportional to the time rate of
change of the current in the inductor.
If current is constant, voltage is zero, thus inductor behave
as a short circuit .
If current cannot change instantaneously in an inductor, that
is, the current cannot change by a finite amount in zero time.
Arcing is when switch is open but current initially continue to
flow in the air across the switch and voltage surge must be
controlled to prevent equipment damage.
EX : Current source, generate zero for t < 0 and pulse 10te-5tA
for t > 0.
i1
a.
b.
L1 100 mH
Waveform
di/dt = 10(-5te-5t + e-5t) = 10e-5t(1 – 5t)A/s
di/dt = 0 when t = 1/5s
c. VL = Ldi/dt = 0.1(10e-5t)(1 – 5t) for t > 0
VL = 0 for t < 0
i(A)
v(V)
0.73
1.0
0.6
0.2
t(s)
0.2
t(s)
The voltage across the terminals of an inductor as a
function of the current in the inductor. To find I as the
function of v ;
multiply both side with dt : vdt = L x (di/dt) x dt
next integrate both side : L ∫i(t) i(t0)dx = ∫t t0d‫ז‬
i(t) = I/L ∫t t0vd‫ ז‬+ i(t0) or normally i(t) = I/L ∫t 0vd‫ ז‬+ i(0)
EX : v(t) = 20te-10tV for t > 0
V
a.
b.
c.
L1 100 mH
Waveform
i = 1/0.1∫t0 20‫ז‬e-10‫ז‬d‫ ז‬+ 0
= 200[(-e-10‫)ז‬/100(10‫ ז‬+ 1)]t0
=2(1 – 10te-10t – e-10t) A
Current as function of time
iA)
v(V)
0.73
0.1
t(s)
t(s)
Power and Energy in the Inductor
It can be derived directly from the current and voltage
relationship. If the current reference is in direction of the
voltage drop across the terminal of the inductor,
the power p = vi, but v = Ldi/dt, so p = Lidi/dt
or p = v[I/L ∫t t0vd‫ ז‬+ i(t0)]
Power is the time rate of expending energy,
so, p = dw/dt = Lidi/dt
multiply both with dt, dw = Lidi
integrated with ref that zero energy is zero current in the
inductor, so ∫w 0dx = L ∫w 0ydy
energy, w = 1/2Li2 assume w =0 at t = 0
i(t) = 1/L ∫t t0vd‫ ז‬+ i(t0) or normally i(t) = 1/L ∫t 0vd‫ ז‬+ i(0)
w (t ) − w ( t 0 ) =
∫
t
t0
p (τ ) dτ
1 2
1 2
= Li (t ) − Li (t0 )
2
2
Ex : The current through 0.1H inductor is i(t) = 10te-5tA. Find
the voltage across the inductor and the energy stored in it.
Since v = Ldi/dt and L = 0.1H
V = 01.d/dt(10te-5t) = e-5t + t(-5)e-5t = e-5t(1 – 5t) V
Energy stored is = 1/2Li2 = ½(0.1)100t2e-10t = 5t2e-10t J
Ex : The current through 1mH inductor is i(t) = 20cos100t mA.
Find the terminal voltage and the energy stored.
4.2 Capacitance
Capacitor consists of two conducting plate separated by an
insulator (or dielectric). Capacitance occurs whenever
electrical conductor are separate by a dielectric or
insulating material. This condition applies that electric
charge electric charge is to transport through the capacitor.
Practical capacitor is in µF and pF.
Although applying a voltage to the terminal of the capacitor
cannot move a charge through the dielectric, it can
displace a charge within the dielectric. As the voltage
varies with time, the displacement of charge also varies
with time, and causing the displacement current.
charge, q(culomb) = c(F) x v(volt)
Capacitance, C = (€A)/d
A – surface area of the plate, the larger the area, the greater
the capacitance.
d – spacing between plates, the smaller the spacing the
greater the capacitance.
€ - permittivity of the material, the higher, the greater the
capacitance.
There are two types, fix and variable. The fix type are
polyester which light in weight, stable and their change with
temperature is predictable. Other dielectric material such as
mica and polystyrene may be used. Film capacitor are rolled
and house in metal or plastic film. Electrolytic capacitor
produce very high capacitance. Capacitance of a trimmer (or
padder) capacitor or a glass piston capacitor is varied by
turning the screw. Usually parallel with another capacitor so
the equ capacitance can be varied slightly.
capacitance of variable air capacitor (meshed plates) is
varied by turning the shaft. Variable capacitor are used in
radio. Capacitor are used to block dc, pass ac, shift phase,
store energy, start motor and suppress noise.
q(t) = Cv(t)
d/dtq(t) = d/dtCv(t)
The current is proportional to the rate at which the voltage
across the capacitor varies with times, i = dq/dt = Cdv/dt.
i(t) = Cdv(t)/dt ; i(‫ = )ז‬Cdv(‫)ז‬
Voltage cannot change instantaneously across the terminals
of a capacitor, coz such change would produce current
infinite.
If the voltage across the terminal is constant, the capacitor
current is zero, coz a conduction current cannot be
established in the dielectric material of the capacitor. Only a
time-varying voltage can produce a displacement current.
Thus a capacitor behaves as an open circuit in the presence
of a constant voltage.
Expressing the voltage as current by multiply both with dt
idt = Cdv or ∫v(t) v(t0)dx = 1/C∫t t0id‫ז‬
Integration left hand of equation v(t) = 1/C∫t t0i (‫)ז‬d‫ ז‬+v(t0)
So voltage capacitor, v(t) = 1/C∫t -∞i (‫)ז‬d‫ז‬
Power absorb by capacitor, p = v(t)i(t) = v(t)xCdv(t)/dt
or p = i[1/C∫t t0id‫ ז‬+v(t0)]
So by combine the definition of energy, dw = Cvdv
∫w0dx = C∫t 0ydy or w = 1/2Cv2
Energy store by capacitor from t to t0
w ( t ) −
w ( t
0
=
=
=
=
t
∫
=
)
∫
C
t
t
t
p (τ ) d τ
0
d τ
0
∫
v ( t )
v ( t
0
)
(τ )
dv
v (τ ) C
v ( τ ) dv
d τ
(τ )
v ( t )
1
2
C
1
2
C
v (τ )
2
v ( t
v ( t )
2
−
0
1
2
)
C
v ( t
so capacitor energy, w(t) = 1/2Cv(t)2
0
)
2
Capacitor is derive form it’s capacity to store energy in an
electric field ;
1. When the voltage across capacitor is not changing with
time (i.e.dc voltage) current through capacitor is zero, so
capacitor is an open circuit to dc but if a battery connected
across a capacitor, the capacitor charge.
2. Voltage on capacitor must continuous where it cannot
change abruptly, but current through capacitor can change
instantaneously.
3. Ideal capacitor does not dissipate energy. It take power
from the circuit when storing energy in its filed and returns
previously stored energy when delivering power to the
circuit.
4. A real no ideal capacitor has a parallel model leakage
resistance, as high as 100MΩ and can be neglected for most
practical application.
Ex : Calculate the charge store on a 3-pF capacitor with 20V
across it and find energy stored in the capacitor.
Q = Cv = 3 x10-12 x 20 = 60pC
W= 1/2Cv2 = ½ x 3 x 10-12 x 400 = 600 pJ
Ex: Voltage across 5 µF capacitor is v(t) = 10cos600t,
determine the current through capacitor.
i(t) = Cdv/dt = 5 x 10-6 d/dt(10cos6000t)
= -5 x 10-6 x 6000 x 10sin600t = -0.3sin6000t A
Ex: Determine the voltage across 2 µF capacitor if current
through it is i(t) = 6e-300t mA
Since v = 1/C∫t0idt + v(0) and v(0) = 0
v = 1/(2 x 10-6)∫t06e-3000tdt.10-3
= (3 x 103)/-3000 x e-3000t|t =(1 – e-3000t) V
0
Ex : Find the energy stored in each capacitor under dc
condition.
2mF
under dc
i = 3/(3=2+4) x 6mA = 2 mA
2kΩ
3kΩ
6mA
v1 = 2000i =4V
v2 = 4000i = 8V
v1
w1 = 1/2C1v12 = 16 mJ
w2 = 1/2C2V22 = 128 mJ
6mA
2kΩ
3kΩ
5kΩ
4mF
5kΩ
v2
4kΩ
4kΩ
Ex : Find the energy stored in each capacitor under dc
condition.
1kΩ
10V
20uF
10uF
6kΩ
4.3 Serial-Parallel Combination
Inductors in Series
The inductor are forced to carry the same current; thus only
one current for series combination.
v1 = L1di/dt
v2 = L2di/dt v3 = L3di/dt
v = v1 + v2 + v3 = (L1 + L2 +L3) di/dt = Leqdi/dt
Leq = L1 + L2 + L3 + ….+ LN
Inductors in Parallel
The inductor in parallel have the same terminal voltage. In
the equivalent circuit, the current in each inductor is a
function of the terminal voltage and the initial current in the
indicator.
i1 = 1/L1∫t t0vd‫ ז‬+ i1(t0) i2 = 1/L2∫t t0vd‫ ז‬+ i2(t0) i3 = 1/L3∫t t0vd‫ ז‬+ i3(t0)
i = i1 + i 2 + i 3
So I = (1/L1 + 1/L2 + 1/L3)∫t t0vd‫ ז‬+ i1(t0) + i2(t0) + i3(t0)
or I = 1/Leq ∫t t0vd‫ ז‬+ i(t0)
where 1/Leq = 1/L1 + 1/L2 + 1/L3
i(t0) = i1(t0) + i2(t0) + i3(t0)
Ex : Find Leq.
4H
20H
Leq = 18H
Leq
7H
8H
12H
10H
Ex : Find Leq.
20mH
Leq = 25mH
50mH
Leq
100mH
40mH
40mH
30mH
20mH
Ex : Find vc and the energy stored in capacitor and inductor
under dc condition.
Under dc condition, capacitor is
1Ω
5Ω
4Ω
12V
Open and inductor is short.
1F
i = iL = 12/(1 + 5) = 2A
vc equal to voltage across 5Ω
1Ω
5Ω
vc = 5i = 10V
4Ω
12V
Energy in capacitor, wc = 1/2Cvc2
=1/2(1)(102) = 50 J
Energy in inductor, wL = 1/2LiL2
=1/2(2)(22) = 4 J
2H
Ex : If i(t) = 4(2 – e-10t) mA and i2(0) = -1 mA.
Find v and i.
4H
-10t
From i(t) = 4(2 – e ) mA
i v1
i1
i(0) = 4(2 – 1) mA = 4mA
4H
Leq
Coz i = i1 + 12
v2
Leq = 5H
So v(t) = Ldi/dt = 5(4)(-1)(-10)e-10t mV = 200e-10t mV
v1(t) = Ldi/dt = 2(-4)(-1)(-10)e-10t mV = 80e-10t mV
v2(t) = v(t) – V(t) = 120e-10t mV
i1(t) = ∫t0v2dt +i1(0) = 120/4∫t0e-10tdt + 5 mA
=-3e-10t|t0 + 5mA = -3e-10t + 3 + 5 = 8 - 3e-10t mA
i2(t) = ∫t0v2dt +i2(0) = 120/12∫t0e-10tdt -1mA
=-e-10t|t0 - 1mA = -e-10t + 1 - 1 = e-10t mA
i2
12H
Capacitors in Series
Capacitor connected in series can be reduced to a single
equivalent capacitor. If each capacitor carries its own initial
voltage, the initial voltage on the individual capacitors is the
algebraic sum of the initial voltages on the individual
capacitors.
1/Ceq = 1/C1 + 1/C2 +….+1/CN
v(t0) = v1(t0) + v2(t0) +….+ vn(t0)
v(t0) = i/Ceq∫t-∞i(‫)ז‬d‫ז‬
Capacitor in Parallel
The equivalent capacitance of capacitor connected in
parallel is simply the sum of the capacitance of the individual
capacitors;
Ceq = C1 + C2 + ….+CN
Capacitor connected in parallel must carry the same voltage.
Therefore if there is an initial voltage appear across the
original parallel capacitor, this same initial voltage appears
across the equivalent capacitance, Ceq.
Ex : Find the Ceq
20µF series with 5µF = 4µF
4µF || 6µF|| 20µF = 30µF
30µF series with 60µF = 20µF
60uF
5uF
20uF
6uF
20uF
Ceq
Ex : Find the Ceq
50uF
60uF
Ceq
70uF
20uF
120uF
Ex : Find the voltage across each capacitor.
Ceq = 10mF
The total charge,
q = vCeq = 10 x 10-3 x 30 = 0.3C
Since i = dq/dt, so charge on
20mF and 30mF is 0.3C
So v1 = q/C1 = 0.3/(20 x 10-3) = 15V
v2 = q/C2 = 0.3/(20 x 10-3) = 15V
v3 = 30 – v1 – v2 = 5V
v3 = q/C = 0.3/60 = 5V
20mF 30mF
30V
v1
v2
40mF
v3
20mF
Ex : Find the voltage across each capacitor.
60V
40F
60µF
v1
v3
20µF
v2
30µF
v4
Summary
Relation
Resistor(R)
Capacitor (C)
Inductor (L)
v-i
v = iR
v = 1/C∫tt0idt + v(t0)
v = 1/2Ldi/dt
i-v
i = v/R
i = Cdv/dt
i = 1/L∫tt0idt + i(t0)
p or w
p = i2R = v2/R
w = 1/2Cv2
w = 1/2Li2
Series
Req = R1 + R2
Ceq =
(C1C2)/(C1+C2)
Leq = L1 + L2
Parallel
Req
Ceq = C1 + C2
= (R1R2)/(R1+R2)
Leq = (L1L2)/(L1+L2)
At dc
Same
Open circuit
Short circuit
v
i
Circuit variable Not applicable
cannot change
abruptly
4.4 Mutual Inductance
When two circuit are linked by a magnetic field, voltage
induced in second circuit can be related to time-varying
current in first circuit by parameter known as mutual
inductance.
The self inductance of two coil is L1 and L2 and the mutual
inductance is M and the easier to analyze is using mesh
current. Choose for direction for i1 and i2, sum the voltage
around each closed path. Coz of M, there will be two voltage
across each coil, namely self-induced voltage and a
mutually induced voltage. Self induced voltage is the
product of the self-inductance of the coil and the first
R
derivatative current in other coil.
1
Mutual induced voltage is the
product of mutual inductance of Vg
L M L
R
i1
i
2
2
1
2
coil and first derivatative current
in other coil.
From the circuit the self-induced voltage across this coil is
L1(di1/dt) and the mutually induced voltage across this coil is
M(di2/dt).
When the ref direction for a current enters the dotted
terminal of a coil, the ref polarity of the voltage that is
induces in the other coil is positive at its dotted terminal and
the other way around.
The dot convention rule indicates that the reference polarity
for the voltage induced in coil 1 by the current i2 is negative
at the dotted terminal of coil 1. This voltage (Mdi2/dt) is a
voltage rise with respect to i1. The voltage induced in coil2
by the current i1 is Mdi2/dt and its ref polarity is positive at
the dotted terminal of coil 2. This voltage is a voltage rise in
direction of i2. The sum voltage around each closed loop ;
-vg + i1R1 + L1di1/dt – Mdi2/dt = 0
i2R2 + L2di2/dt – Mdi1/dt = 0