1
Induction
We define self-inductance Land mutual inductance M as
Φ1
Φ2
= Li1
= M i1
(1)
(2)
Faraday’s law yields
dΦ1
dt
dΦ2
E2 = −
dt
di1
dt
di1
= −M
dt
E1 = −
2
= −L
(3)
(4)
Transformer
The same flux passes through both coils. Thus the rate of change of the flux is
the same too. Faraday’s law yields
v1
v2
=
(5)
N1
N2
It also yields the relationship between the self-inductance and mutual inductance:
Φ1
Φ1
= L1 i1 + M i2 = N1 Φ
= M i1 + L2 i2 = N2 Φ
(6)
(7)
Since both equation must satisfy at any moment, thie yields
L1 L2 = M 2
2.1
(8)
Circuit Equation
We assume the primary coil is connected across a power source v = V sin ωt and
the secondary coil is connected across a resistor R. Appling Kirchhoff’s loop
rule to the primary circuit and secondary circuit, we get
di2
di1
−M
dt
dt
di2
di1
− L2
i2 R − M
dt
dt
v − L1
=
0
(9)
=
0
(10)
Solving these equation under the condition L1 L2 = M 2 , we obtain
1
L2
i1 = V −
cos ωt +
sin ωt
L1 ω
L1 R
s
2
2 R
L2
1
−1
sin ωt − tan
+
= V
ωL1
L1 R
ωL2
MV
L2 V
i2 = −
sin ωt = −
sin ωt
RL1
MR
1
(11)
(12)
(13)
Note that i2 is 180◦ out of phase with the power source in the primary circuit,
and L2 /L1 = (N2 /N1 )2 for the coils with the same geometry.
2.2
Equivalent Resistance
The effective impedance of the primary circuit is
Z1 = r
1
2 2
L2
1
+
ωL1
L1 R
Also, the phase constant is written as
Req
RL1
1
R
=
=
ωL2
L2
ωL1
ωL1
(14)
(15)
Therefore, the current in the primary circuit is exactly the same as the circuit
with resistor Req = L1 R/L2 inserted parallel with L1 .
2.3
Energy Conservation
The instantaneous input power and output power are
L2
1
2
cos ωt +
sin ωt sin ωt
−
Pin = i1 v = V
L1 ω
L1 R
L2 V 2
sin2 ωt
Pout = i22 R =
L1 R
(16)
(17)
Thus for one cycle, we have
P̄in = P̄out =
L2 V 2
2L1 R
(18)
Recall P̄in and P̄out are both zero for purely inductive circuit (power factor is
zero!), and for a transformer, the all power privided into the primary circuit is
consumed in the resistor in the secondary circuit.
2