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) Impedancev) 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

VRVR

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

VR

VL

VL leads by 90o

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 + XC

2 = √ 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

Resistance R

ReactanceX

Impedance Z

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

•