
 Background
 Magnetic field lines
 Magnetic Field Lines for Current in Coil of Wire (Fig.
1a)
 Curl the fingers of your right hand in the direction
of the current. Your thumb points in the direction
of the field along the axis of the coil.
 The field at other points is tangent to the field
lines which are dashed in Fig. 1a.
 Magnetic Field Lines for Bar Magnet (Fig. 1b)
 Magnetic field lines leave North pole and go to
South pole. The field at any point is tangent to the
field line.
 North pole of magnet acts like current carrying
coil of wire with current in counterclockwise direction
as you view that end of the coil.
 Before we knew the cause of magnetism was a net
flow of charge, we said a circular coil with current
in counterclockwise direction as you view that end
of the coil acted like a North pole.
 By experiment we find that a North Pole repels another
North pole or a coil with the current in a counterclockwise
direction.
 Magnetic Flux
 Φ_{B} =
B ^{. }A = BA cos B,A
for B constant.
 For Fig. 2a, B,A
= 90^{o}. cos 90^{o} = 0 and
Φ_{B }= 0.
 For Fig. 2b, Φ_{B}
= BA cos B,A.
 For Fig. 2c, B,A
= 0^{o}. cos 0^{o} = 1 and
Φ_{B }= BA.
 For Fig. 2d, notice that
 If B, A, or B,A
vary, Φ_{B}
= ∫ B ^{.} dA.
 The Experiment
 First we calibrate a galvanometer to detect the presence
and the direction of a current. In Fig. 3 when current
enters the terminal on the right hand side of the galvanometer,
the needle deflects to the left.
 In Fig. 4a, the bar magnet is at rest. There is no current
in coil of wire and the galvanometer does not deflect.
 In Fig. 4b, the magnet moves toward the coil and the
galvanometer deflects toward the left indicating an induced
current and associated induced magnetic field (shown by
dashed line) opposing the field of the magnet.
 In Fig. 4c, the rate of change of the magnetic flux
produced by the magnet has increased and there is a greater
current in the galvanometer.
 In Fig. 4d, the magnet is inside the coil at rest. Current
is zero.
 In Fig. 4e, the magnet moves to the right. Current
in coil reverses.
 Results of the Experiment
The induced current creates an induced magnetic field that
opposes the change produced by the external magnetic field.
In this experiment the bar magnet provided the external field.
 When the magnet is at rest there is no change of flux
or induced current (Fig. 4a above).
 When the North pole of the magnet moved toward the right
face of the coil, the induced current in the coil was
counterclockwise as viewed from its right end of the coil.
Another way of looking at this is the right end of the
coil acts like a North pole and repels the North pole
of the magnet as you move the magnet toward it. You must
exert a force through a distance doing work to induce
a current in the coil (Fig. 4b above).
 The greater rate of change of magnetic flux, the greater
induced current. As the magnet gets nearer to the coil,
the flux changes more rapidly. You can also see this by
bringing up two magnets at the same speed of one or bringing
up one more rapidly (Fig. 4c above).
 When the magnet is at rest, there is no change of flux
and no current (Fig. 4d above).
 When the North pole is moved away from the right end
of the coil the current in the coil reverses. The end
of the coil nearest the magnet acts like a South pole
and you must do work to remove the North pole of the magnet
and induce a current in the coil (Fig. 4e above).
 Faraday's Law
 Induced emf = the rate of change of magnetic flux.
 Since Φ_{B}
= ∫ B ^{.} dA, you can change
the magnetic flux by changing the magnetic field, the
area through which the field exists or the angle between
the magnetic field and the area.
 The direction of the induced current is given by Lenz's
Law: The direction of the induced current is such to oppose
the change that produced it. Lenz's Law is really a statement
of conservation of energy. In order to get an induced
emf or current, you must do work.
 Motional emf
 As conductor in Fig. 5 moves to the right, the electrons
in the conductor move with its velocity v and experience
a magnetic force down equal to qvB, where q is the charge
of the electron. They accumulate at the bottom of the
conductor leaving the top of the conductor with an excess
of positive charge.
 This separation of charge produces an electric field
from a to b that results in an electric force up on the
electron equal to qE, where q is the charge of the electron.
 When the electric force equals the magnetic force, qE
= qvB, the motion of the electron ceases and the
electric field in the conductor E = vB. The
potential difference between the ends of the conductor
V_{ab} = E=
(Bv)
 The motional emf = Bv. In
Fig. 5, B was perpendicular to v.
If the magnetic field is not perpendicular to the velocity,
the motional emf = Bv sin B,v.
 Alternating Current Generator
 In Fig. 2b above, the magnetic flux Φ_{B}
= BA cos B,A.
 Let B,A
= Θ = ωt,
where ω equals
the angular velocity of the rotating coil. Then
Φ_{B} = BA
cos ωt and
ε
=  NdΦ_{B}/dt
= N ωBA sin ωt,
where N = the number of turns of the coil.
 Figure 6a and 6b below are plots of the magnetic flux
and the electromotive force as a function of time t, respectively.
 Sample problems in 108
Problem Set for Induction: 110.
 The Production of an Electric Field by a Changing Magnetic Flux
 The electromotive force ε
equals the energy per unit charge and the electric field E
equals the force per unit charge.
In Fig. 7 above we can explain the motion of charge in the
wire as an induced current. We can also explain it as the
production of an electric field that, in turn, produces a
motion of positive charge as a current. The work done per
unit charge around the loop of length 2 πr
equals ε = E(2 πr) =  dΦ_{B }/dt.
 In general we may write E
• d
= B
• dA
 Sample problem in 108
Problem Set for Induction: 12
 Selfinductance L = NΦ_{B}/I
 Calculation of L for a solenoid
For a solenoid, B = µ_{o}nI = µ_{o}(N/)I,
where n = the number of turns per unit length = N/.
NΦ_{B} = µ_{o}(N^{2}/)I πr^{2},
where r is the radius of the solenoid.
L = NΦ_{B} /I
= µ_{o}(N^{2}/)( πr^{2}).
 Energy in Magnetic Field
 Since L = NΦ_{B}/I,
L = N (dΦ_{B}/dt)/(dI/dt)
= ε/(dI/dt).
 Energy = dU = (power)dt = εI
dt = L(dI/dt)I dt = LI dI.
U = L_{o }∫^{I } I dI = 1/2 LI^{2}.
 Using L from solenoid, the magnetic energy
U_{B} = 1/2 [(µ_{o}(N^{2}/)( πr^{2})]I^{2
} = 1/2[µ_{o}(N^{2}/)( πr^{2})][B/µ_{o}(N/)]^{2}
= (1/2µ_{0})
B^{2}[ πr^{2}]
= (1/2µ_{o})
B^{2}[volume]
 While we have derived this from L for a solenoid, in
general
U_{B} = 1/2 (B^{2}/µ_{o})A(length)
and the magnetic energy density
u_{B} = (1/2µ_{o })B^{2}.
Compare with electric energy density
u_{E} = (1/2) ε_{o
}E^{2}.
 Example
Find the magnetic energy stored for a < r < b for
the coaxial inductor shown in Fig. 8 below.
Assume that the inductor is sufficiently long ( >> b),
or that the gap is sufficiently small, that end effects
can be neglected.
From Ampere's law, we know that for a < r <
b, B = µ_{o}I/2 πr.
The volume element is the volume of a cylindrical
shell with radius r, surface area 2 πr,
and thickness dr.
dV = 2 πr
dr
U = (1/2µ_{o}) _{a}∫^{b
} B^{2} dV
= (1/2µ_{o}) _{a}∫^{b
}[µ_{o}I/2 πr]
^{2} 2 πr
dr
= [µ_{o}I^{2}/4 π]
ln (b/a)
 Sample problems in 108
Problem Set for Induction: 11, 15.
 RL Circuits
 Inductance L = ΔV_{L}/(dI/dt),
where ΔV_{L }is
the potential difference across the inductor and dI/dt is
the rate of change of current in the inductor.
 Unit of Inductance = 1 Henry (H) = 1Vs/A
 Similar to resistance and capacitance, the inductance
depends only on the geometry of the conductor.
 Build up of current
 When a source of emf is connected across a resistor
and inductor wired in series with a battery (Fig. 9b above),
the current goes from zero to some finite value in a very
short time. Since the rate of change of current is very
high, the potential difference across the inductor = V_{cb}=
L(dI/dt) is very high, while the small current makes the
potential difference across the resistor V_{ac}
= IR very small. As time passes, the current builds up
and the rate of change of current decreases. After a very
long time, it is as though the inductor were not in the
circuit and the total emf is across the resistor giving
a maximum current = ε/R.
A plot of V_{ac} and V_{cb}
(with dashes) for the build up of the current is shown
in Fig. 9a above.
 For the build up of the current,
I(t) = ( ε/R)(1
 e^{Rt/L})
V_{ac}(t) = IR = ε(1
 e^{Rt/L})
V_{cb}(t) = L dI/dt = L( ε/R)
e^{Rt/L} (R/L) = ε
e^{Rt/L}
 Decay of the Current
 When the battery is removed and a wire is placed across
the resistor and the inductor (Fig. 9b above), the potential
difference Vab = 0. As the current decreases, the inductor's
"back emf" opposes the decrease in current.
A plot of Vac and Vbc (with dashes) for the decay of the
current is shown in Fig. 9a’ above.
 For the decay of the current,
I(t) = ( ε/R)e^{Rt/L}
V_{ac}(t) = IR = ε
e^{Rt/L}
V_{cb}(t) = L dI/dt = L( ε/R)(e^{Rt/L})(R/L)
=  εe^{Rt/L}
 Sample problem in 108
Problem Set for Induction: 13
 LC Circuits
 Comparison of LCoscillator with spring

Spring 

Oscillator 

F_{net} =
ma 

V_{ab} = V_{ac}
+ V_{cb} 

 kx =
m dv/dt 

0
= q/C + L dI/dt 
Since
v = dx/dt, 
Since
I = dq/dt, 

dv/dt = d/dt(dx/dt)
= d^{2}x/dt^{2} 

dI/dt = d/dt(dq/dt)
= d^{2}q/dt^{2} 

kx = m d^{2}x/dt^{2}
or^{ } 

q/C = L d^{2}q/dt^{2}
or 

d^{2}x/dt^{2
}+ (k/m)x = 0 

d^{2}q/dt^{2
}+ (1/LC)q = 0 

ω
= 2 πf = (k/m)^{1/2} 

ω
= 2 πf = (1/LC)^{1/2} 
^{
}
 Charge, Current, and Potential
Energy as a function of time for LC circuit
 For fully charged capacitor
at time t = 0,
q(t) = Q_{0} cos ωt,
with ω = 2 πf
= (1/LC)^{1/2.
}
 I(t) = dq/dt = I(t) = ωQ_{0}
sin ωt
 Electric potential energy
U_{e} = 1/2 q^{2}/C = (1/2C) Q_{0}^{2
}cos^{2} ωt
 Magnetic potential energy
U_{m} = 1/2 LI^{2} = (L/2) ω^{2}Q_{0}^{2
}sin^{2} ωt
 Since ω=
2 π/T,
maximum electric
energy occurs at t = 0, T/2, 2T . . . maximum
magnetic energy occurs at t = T/4, 3T/4 . . .
Sample problem in 108
Problem Set for Induction: 14
 Electromagnetic Waves
 From Faraday's law we find
(Equation
1)
This equation states that a changing magnetic flux produces
an electric field.
 From Ampere's law
(Equation
2)
This equation states that a current produces a magnetic field.
 Maxwell's addition
 Since nature is amazingly
symmetrical, Maxwell believed that a changing electric
flux should produce a magnetic field.
 As an example, he pointed
out the discontinuity in the magnetic field while a capacitor
is charged. In Fig. 11 below, you see the magnetic field
lines associated with the current entering the capacitor
and leaving the capacitor, but there is apparently no
magnetic field inside the capacitor.
 While the capacitor is charging,
the electric field inside the capacitor is changing since
E = q/ ε_{o}A.
dE/dt = (dq/dt)/ ε_{o}A
or
(dq/dt) = current = ( ε_{o}A)
dE/dt.
Maxwell called this current a displacement current
I_{d} and generalized it by writing I_{d
} = ε_{o }d/dt
∫ E ^{. }dA (Equation
3)
 With this addition we rewrite
Eq. 2
(Equation
4)
Eq. 4 states that a changing
electric flux produces a magnetic field even in the absence
of a current.
 Maxwell suggested that a
changing electric field produced a magnetic field, which
if changing, produced an electric field, which if changing,
produced a magnetic field, etc. He hypothesized that these
disturbances would be transmitted as an electromagnetic
wave. Figure 12a below is a representation of an electromagnetic
wave traveling in the Xdirection with the electric vector
in the XY plane and the magnetic vector in the YZ plane.
 The equations of motion for
the E and B vectors are
E = E_{p }cos
(kx  ωt) (Equation 5)
B = B_{p }cos (kx  ωt)
(Equation
6)
where k = 2 π/λ
and ω = 2 πf.
Since λf = c, ω/k
= (2 πf)/(2 π/λ)
= λf = c.
 Evaluating paths in Fig. 12b
and 12c.
Let us evaluate Eq.1 for the path in Fig. 12b above.
You choose the direction of going around the loop by pointing
the thumb of your right hand in the direction of B.
Your fingers then curl clockwise in the direction of the
path. For this path, the E,
ds along the righthand side of the loop is 0^{o}
and along the lefthand side of the loop it is 180^{o}.
E^{ }•^{
}ds = (E + dE) ds on the right side of
length h and –E ds for the left side. For the sides
of length dx, E,
ds is 90^{o } and cos 90^{o} = 0.
There is no contribution for these lengths.
Letting B = the average value of the loop of area h dx,
d/dt ∫ B ^{.} dA = dB/dt
hdx.
Thus becomes
(E + dE)h – Eh =  dB/dt h dx or
dE/dx =  dB/dt,
or in terms of partial derivatives,
∂E / ∂x =  ∂B / ∂t
(Equation
7)
 Write Maxwell's equation in
free space, that is, assume there is
no "real" current I.
 Then (Equation
4)
becomes (Equation
4’)
 Evaluate Eq. 4’
for the path of Fig. 12c above.
You choose the direction of going around the loop
by pointing the thumb of your right hand in the direction
of E. Your fingers then curl in the direction
of the path. For this path, the B,
ds along the lefthand side of the loop is 0^{o}
and along the righthand side of the loop it is 180^{o}.
B^{ . }ds =  (B + dB)dx
on the right side of length h and B dx for the left
side. For the sides of length dx, the B,
ds = 90^{o} and there is no contribution
to those sides.
=dB dx
Letting E = the average value of the loop of area
h dx,
µ_{o} ε_{o}
d/dt ∫ E ^{.} dA = µ_{o} ε_{o
}dE/dt hdx or
Eq. 4' becomes
 dB/dx
= µ_{o} ε_{o}
dE/dt
or in terms of partial derivatives
 ∂B / ∂x = µ_{o} ε_{o}
∂E / ∂t (Equation
8)
Taking the partial derivative
of Eq. 7 with respect to x gives
∂^{2}E / ∂x = 
( ∂ / ∂x)∂B / ∂t
(Equation
9)
Taking the paratial derivative
of Eq. 8 with respect to t gives
 (∂ / ∂t)∂B / ∂x
= µ_{o} ε_{o}
(∂^{2}E / ∂t^{2)} (Equation
10)
Since the order of differentiation
does not make any difference, you can equate the left
side of Eq. 9 to the right side of Eq. 10:
∂^{2}E / ∂x^{2} =
µ_{o} ε_{o}
(∂^{2}E / ∂t^{2})
(Equation
11)
Compare Eq. 11 with wave
equation of wave along rope
∂^{2}y / ∂x^{2} =
(1 / v^{2}) (∂^{2}y / ∂t^{2}^{)}
The equations are identical
for a wave traveling with a velocity (1/µ_{o} ε_{o})^{1/2}
and changing electric field E, and a wave with a velocity
v with changing displacement of the rope y from the
equilibrium position, if c = (1/µ_{o} ε_{o})^{1/2.
}
 Since µ_{o}=
4 π x 10^{7
}N/A^{2} and
ε_{o}
= 8.85 x 10^{12} C^{2}/Nm^{2},
(1/µ_{o} ε_{o})^{1/2}
= 1/[(4 π x 10^{7
}N/A^{2})( 8.85 x 10^{12} C^{2}/Nm^{2})]^{1/2
} =
3.00 x 10^{8} m/s!
As the late Richard Feynman said in a rewrite of Genesis,
"Let there be electricity and magnetism and there
will be light."
c = (1/µ_{o} ε_{o})^{1/2
}
 If you take the derivative
of Eq. 5 with respect to x and the derivative of Eq. 6
with respect to t and substitute into Eq. 7, you find
 kE_{p}
sin(kx  ωt) =  ωB_{p}
sin(kx  ωt) or
E_{p} = (ω/k)B_{p}
= {2 πf/(2 π/ λ)B_{p}
or
E_{p} = B_{p}(f λ)
B_{p} = E_{p}/c
 The Poynting vector S
 S = 1/µ_{o}
E x B.
 S is the energy/second per
m^{2}. Since
energy/second = power = P,
S = power/area.
 Frequently you can solve
problems by checking units.
The unit of power is J/s = W.
The unit of S is W/m^{2}.
 The direction of S is
given by E x B.
In Fig. 12, E = Ej and
B = Bk.
Since j x k = i, and S =
1/µ_{o} E x B, S
is in the positive Xdirection.
 Magnitude of S = EB/µ_{o}
= E^{2}/cµ_{o}.
 Average value of S_{av}
= (E^{2})_{av}/cµ_{o} = (E_{p})^{2}[cos^{2}(kx
 ωt)]_{av}/cµ_{o}.
[cos^{2}(kx  ωt)]_{av}
= 1/2 and S_{av} = (E_{p})^{2}/2cµ_{o}.
 Power (W) = Energy/time (J/s).
 Average value of S = Intensity
(W/m^{2}) = Power (W)/Area (m^{2}).
 Sample problems in 108
Problem Set for Induction: 1620.
 Momentum and Pressure of an Electromagnetic
Wave
 Assume that an initially at rest,
positive charge q is in the path of a wave (Fig. 13 below).
The force F_{e} on it by the electric field is qE.
In time Δt, the electric
field moves the charge a distance v Δt
and does work F_{e }v Δt = (qE) v Δt.
The charge's increase in energy = ΔU = qE v Δt
or qvE = ΔU/ Δt.
 Once the charge is in motion,
the magnetic field exerts a force on the charge to the right.
Although the magnetic force does no work, it does give the
charge momentum.
The magnetic force F_{m} = qvB = qvE/c, since B =
E/c for an electromagnetic wave.
Thus F_{m} = qvB = qvE/c = ΔU/c Δt
or
F_{m} Δt
= ΔU/c.
Since F = Δmv/ Δt
= Δp, Δp
= F Δt = ΔU/c.
If a particle absorbs energy ΔU,
it acquires a momentum ΔU/c.
The momentum of the particle must have belonged to the wave!
 Pressure = P = Force/Area.
For a wave, since F Δt
= ΔU/c,
F = [( ΔU/ Δt)/c]
and
P = [( ΔU/ Δt)]/c)Area
= [( ΔU/ Δt)/area]/c.
Power = [( ΔU/ Δt)].
(Power)/Area = S_{av}.
Thus P = S_{av}/c.
 Alternating Currents
 The basic problem is an analysis
of alternating current (ac) to find expressions for the current
amplitude I_{p} and the phase angle Φ
where the current I(t) = I_{p} sin (ωt
+ Φ), when V(t) = V_{p}
sin ωt is applied to a
circuit.
 Single Circuit Elements
 Resistor R only (Fig. 14a)
 V(t) = V_{p}
sin ωt = V_{R}
= I_{R} R.
I_{R} = V_{p} sin ωt/R.
I_{R } = I_{pR} sin ωt,
where I_{pR } = V_{pR}/R.
The current and the potential difference across R
are in phase (Fig. 14c).
 The phasor diagram is
shown in Fig. 14c for a resistive circuit only. The
maximum values for the current in R and the potential
difference across R are I_{pR } and V_{pR},
respectively. The instantaneous values of the current
and the potential difference across R, I_{R}
and V_{R}, respectively, are shown for the
time indicated in Fig. 14b.
Notice that I_{R } = I_{pR} sin ωt
and V_{R } = V_{pR} sin ωt
.
 Capacitor C only (Fig. 15a)
 V(t) = V_{p} sin
ωt = V_{C}
= q/C. q = CV_{C} = CV_{pC}sin
ωt.
I_{C } = dq/dt = ωCV_{pC
}cos ωt =
ωCV_{pC }sin
(ωt + π/2)
because
(sin ωt cos π/2
+ cos ωt sin π/2)
= cos πt.
I_{C} = I_{pc} sin (ωt
+ π/2), where
I_{pc }= V_{pC}/(1/ωC).
1/ωC = X_{C.
} The capacitive reactance takes the place of
R in a total resistive circuit. The current in the
capacitive circuit leads the potential difference
across the capacitor (Fig. 15b).
 The phasor diagram for
a capacitor only circuit is shown in Fig. 15c. The
maximum values for the current for C and the potential
difference across C are I_{pC} and V_{pC},
respectively. The instantaneous values of the current
and the potential I_{C} and V_{C}
are shown at the time indicated in Fig. 15b.
Notice that I_{C } = I_{pC } sin (ωt
+ π/2). V_{C
} = V_{pC } sin ωt.
 Inductor only (Fig. 16a)
 V(t) = V_{pL}
sin ωt = V_{L}
= L dI_{L}/dt.
Separating variables,
V_{pL }sin ωt
dt = L dI_{L}
(VpL / L)
sin ωt dt =
dI_{L}
 (V_{pL} / ωL)
[cos ωt  1] =
I_{L}  I_{oL}
I_{oL} =  (V_{pL} / ωL)
I_{L} (t) =  (V_{pL} / ωL)
cos ωt
=
(V_{pL} / ωL)
sin (ωt  π/2)
(V_{pL}/ωL)
cos ωt = (V_{pL}/ωL)_{
}sin (ωt
 π/2) because
(sin ωt cos π/2
 cos ωt sin π/2)
=  cos ωt.
I_{L} = I_{pL} sin (ωt
 π/2), where
I_{ pL }= V_{pL}/(ωL).
ωL = the inductive
reactance = X_{L}.
The current for the inductive circuit lags the potential
difference across the inductance by π/2
as shown in Fig. 16b.
 The phasor diagram for
an inductor only circuit is shown in Fig. 16c. The
maximum values for the current in L and the potential
difference across are I_{pL }and V_{pL},
respectively. The instantaneous values of the current
and the potential I_{L} and V_{L},
respectively, are shown at the time indicated in Fig.
16b.
Notice in the phasor diagram I_{L } = I_{pL
} sin(ωt
 π/2).
V_{c }= V_{pC }sin ωt.
 RLC Series Circuit
 The applied potential difference
across the circuit V(t) = V_{p }sin ωt
(Fig. 17a) and the current I(t) = I_{p} sin (ωt
 Φ) is the same everywhere
in the series circuit.
The potential difference across the resistor = V_{R}(t)
= IR = [I_{p }sin (ωt  Φ)]R
is in phase with the current (Fig. 17b and Fig. 18b).
The potential difference across the inductor = V_{L}(t)
leads the current by π/2
(Fig. 17c and Fig. 18b). V_{L}(t) = [I_{p}
sin (ωt  Φ
+ π/2)]ωL.
The potential difference across the capacitor V_{C}(t)
lags the current by π/2
(Fig. 17d and Fig. 18b). V_{C}(t) = [I_{p}
sin (ωt  Φ
 π/2)](1/ωC)
 In Fig. 18c,
V_{p}^{2
} = V_{pR}^{2} + (V_{pL} –
V_{pC})^{2
} V_{p}^{2} = (I_{p}R}^{2}
+ (I_{p}ωL –
I_{p}/ωC)^{2}
I_{p } = V_{p}/[R^{2 }+(ωL
–1/ωC)]^{1/2}
= V_{p}/Z
tan Φ = (ωL
–1/ωC)]/R
 Phasor Diagrams Revisited
 While the previous phasor
diagrams allow you to take projections and find the voltage
or current at any time, you can simplify the diagrams
by drawing them as shown in Fig. 19 b and c above.
 Draw R and I along the +Xaxis,
ωL along the +Yaxis,
and 1/ωC along the
–Y axis (Fig. 19b)
 Since I is the same in the
resistance, inductance, and capacitance
(Fig. 19c),
 V_{pR} is drawn
along the +Xaxis because V_{R} is in phase
with I.
 V_{pL} is drawn
along the +Yaxis because V_{L} leads I by
π/2.
 V_{pC} is drawn
along the Yaxis because V_{C } lags I by
π/2.
 From Fig. 19b, tan Φ= (ωL –
1/ωC)/R.
 From Fig. 19c,
V_{p} =
[V_{pR}^{2} + (V_{pL}^{2}
– V_{pC}^{2})]^{1/2
} V_{p} = {(I_{p}R)^{2}
+ [(I_{p}ωL)^{2}
– (I_{p}/ωC)^{2}]}^{1/2}
or
I_{p } = V_{p}/[R^{2} + (ωL
– 1/ωC)^{2}]^{1/2}
 For the circuit of Fig. 19a,
V_{p } = 10 V, ω
= 120 πs^{1},
R = 4.0 ω, L = 0.0159
H, and C = 0.884 mF.
X_{L }= ωL = 6.0
ω
X_{C} = 1/ωL =
3.0 ω
I_{p } = V_{p}/[R^{2} + (ωL
– 1/ωC)^{2}]^{1/2}
= 10 V/[16 + (6 – 3)^{2}]^{1/2} ω= 2.0 A
V_{pR }= I_{p}R = 8.0 V
V_{pL }= I_{p }(X_{L}) = 12 V
V_{pLc}= I_{p }(X_{c}) = 6 V
 The maximum value of I_{p
}occurs for ωL =
1/ωC, or the resonant
frequency ω_{o}
= 1/(LC)^{1/2}. At resonance, I_{p }
= V_{p}/[R^{2} + (ωL
– 1/ωC)^{2}]^{1/2}
= V_{p}/R. The resonance curve for the circuit of
Fig. 19a is shown in Fig. 20 below.
 Power = IV
= [I_{p} sin
(ωt  Φ)][V_{p}
sin ωt}
= I_{p}V_{p }[sin ωt
cos Φ  cos ωt
sin Φ)[sin ωt]
= I_{p}V_{p }[sin^{2} ωt
 cos ωt sin ωt]
= I_{p}V_{p }[sin^{2} ωt
cos Φ – 1/2(sin
2ωt](sin Φ)
The average value of sin^{2}
ωt is 1/2 and the average
value of sin 2ωt is 0.
P_{av} = (I_{p}V_{p }/2) cos Φ.
Because cos Φ =
R/[R^{2}_{ }+ (ωL
– 1/ωC)^{2}]^{1/2},
P_{av} = (I_{p}^{2}/2)R = I^{2}_{rms
}R, where the rootmeansquare value of the current is
defined as I_{p}/(2)^{1/2 }and the rootmeansquare
value of the voltage is defined as V_{p}/(2)^{1/2.
}
 Sample problems in 108
Problem Set for Induction: 2226.

