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electromagnetic and
magnetic circuit principles
voltage and current waveforms
Example
When an instantaneous voltage of 500 Sin (314t +
π/4) is applied to a series circuit of R and L, the
current is found to be 10Sin (314t - π/6).
Calculate:
i) Peak voltage
ii) Frequency
iii) Phase angle
iv) Impedance
v) Resistance
vi) Inductance
•
Example
• Peak voltage = 500v
• Frequency
f = 314/2π = 50 Hz.
• Phase angle π/4 + π/6 = 45o + 30o = 75o
• Impedance Z = V/I = 500/10 = 50 Ω
Example
• Resistance = Z Cos φ = 50 x 0.2588 = 12.94 Ω
• Inductive reactance - XL = Z Sin φ = 50 x 0.9659 =
48.30Ω
• Inductance L = XL / 2πf
= 48.30/2π x 50
• = 0.1537 H
Circuit possessing resistance only
instantaneous value of voltage and
current
v = VmsinӨ and i = Vm/R sinӨ
i = ImsinӨ = Imsinπ2Өft
phasor diagram for the resistive circuit
VR
VR
phasor diagram for the resistive circuit
Circuit possessing inductance only:
instantaneous value of induced e.m.f.:
e = -L.di/dt = 2πfLIm
instantaneous value of applied voltage
v = 2πfLIm cos 2πft = 2πfLIm sin(2πft+π/2)
applied voltage, induced E.M.F., and current
waveforms
phasor diagram for the inductive circuit
VL leads by 90o
VL
VR
Inductive reactance
Vrms/Irms = 0.707Vm/0.707Im
= 2πfL = inductive reactance [XL]
I = V/2πfL = V/XL [ohms]
Example
•An inductor of 0.6H and negligible
resistance is connected across a 120 V a.c.
supply.
•Calculate the current when the frequency is:
•i) 30 Hz
• ii) 200 Hz
•
Example
•i) XL = 2πfL = 2π x 30 x 0.6 = 113 Ω
•IL = V / XL = 120 / 113 = 1.06 A
•ii) XL = 2πfL = 2π x 200 x 0.6 = 753 Ω
•IL = V / XL = 120 / 753 = 0.159 A
Circuit possessing capacitance only
v = Vm sin θ = Vm sin 2πft
i = C dv/dt
i = 2πfCVm cos 2πft = 2πfCVm sin(2πft+π/2)
Waveforms for capacitive circuit
phasor diagram for the capacitive circuit
VR
VC
VC lags by 90o
Capacitive reactance
Vrms/Irms = 0.707Vm/0.707Im
= 1/(2πfC) = capacitive reactance [XL]
Example
• A capacitor of 0.6 μF is connected across a 120 V
ac supply. Calculate the current when the
frequency is:
• i)30Hz
• ii) 200 Hz
•
• i) XC = 1 / 2πfC = 1 / 2π x 30 x 0.6 x 10-6 = 8842 Ω
• IC = V / XC = 120 / 8842 = 13.6 mA
• i) XC = 1 / 2πfC = 1 / 2π x 200 x 0.6 x 10-6 = 1326 Ω
• IC = V / XC = 120 / 1326 = 0.09 A
Series Resonance
• The resonance of a series RLC circuit occurs when the
inductive and capacitive reactances are equal in
magnitude but cancel each other because they are
180 degrees apart in phase. The sharp minimum in
impedance which occurs is useful in tuning
applications. The sharpness of the minimum depends
on the value of R and is characterized by the "Q" of
the circuit.
Series Circuits (R,L,C)
impedance [Z] =√{R2 +(XL -XC)2}
φ = phase angle = tan-1(XL-XC)/R),
Cos-1 R/Z, Sin-1 = (XL - XC) / Z
phasor diagram
Example
• A 10 Ω resistor and 150μF capacitor are
connected in series across a 200 Hz, 200 V ac
supply. Calculate:
• i) Circuit impedance
• ii) Current
• iii) Phase angle
•
Example
i) Circuit impedance. XC = 1 / 2πfC = 1 / 2π x 200 x 150 x
10-6 = 5.305 Ω
Z = √ R2 + XC2 = √ 102 + 5.3052 = √128.143 = 11.32 Ω
ii) Current = I = V/Z = 200 / 11.32 = 17.67 A
iii) Phase angle = tan-1 XC / R = 27.95 degrees leading
Series Resonance (R,L,C)
XL =1/XC
f = 1/{2π√(LC)}
phasor diagram
definition: acceptor circuit
graph of current and impedance plotted against Z
•
Q factor (at resonance)
Q = XL/R = 1/R √(L/C)
bandwidth - (f2-f1) - definition of half-power
points
Q = fr/(f2-f1)
Parallel Circuits (R,L,C)
supply current = √V/R + V/XL + XCV)
φ = phase angle = phase difference VS and IS
φ = tan-1 (IL - IC)/IR
phasor diagram
Parallel Resonance (R,L,C)
f = 1/(2πL) √(L/C - R2)
phasor diagram
definition: rejector circuit
dynamic impedance RD = L/CR
Q factor (at resonance) = XL/R
Terms
• Resistance is the opposition to current flow by a
resistor
• Reactance, is similar, it is the interference of a
capacitor or an Inductor to current flow
• XL is inductive reactance and XC is capacitive
reactance
• Impedance (Z) is actually the overall opposition to
current presented by the circuit
•Conductance, Susceptance, and Admittance
are the opposites to Resistance, reactance
and impedance
Impedance triangle
Impedance Z
Resistance R
Reactance
X
Admittance triangle
Conductance G
Susceptance B
Admittance Y
• Conductance [G] = R/Z2. Is 1/R when X is = 0
• admittance [Y] = 1/Z = R/Z2
• susceptance [B] = X/Z2. Is 1/X when X is = 0
•
• Y = G+ jB and tanφ = B/G
•
R and L in series
Z = R +jXL = Z < φ
admittance = Y = 1/Z = (R/Z2 - jXL/Z2) = G –
jBL = Y < -φ
R and C in series
•
Z = R -jXC = Z<-φ
•admittance = Y = 1/Z = (R/Z2 + jXC/Z2) = G +
jBC = Y<φ
•