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Published byIzabella Cranmore Modified over 2 years ago

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electromagnetic and magnetic circuit principles

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voltage and current waveforms

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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

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Example Peak voltage = 500v Frequency f = 314/2π = 50 Hz. Phase angle π/4 + π/6 = 45 o + 30 o = 75 o Impedance Z = V/I = 500/10 = 50 Ω

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Example Resistance = Z Cos φ = 50 x = Ω Inductive reactance - X L = Z Sin φ = 50 x = 48.30Ω Inductance L = X L / 2πf = 48.30/2π x 50 = H

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Circuit possessing resistance only instantaneous value of voltage and current v = V m sinӨ and i = V m /R sinӨ i = I m sinӨ = I m sinπ2Өft

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phasor diagram for the resistive circuit VRVR VRVR

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Circuit possessing inductance only: instantaneous value of induced e.m.f.: e = -L.di/dt = 2πfLI m instantaneous value of applied voltage v = 2πfLI m cos 2πft = 2πfLI m sin(2πft+π/2)

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applied voltage, induced E.M.F., and current waveforms

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phasor diagram for the inductive circuit VRVR VLVL V L leads by 90 o

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Inductive reactance V rms /I rms = 0.707V m /0.707I m = 2πfL = inductive reactance [X L ] I = V/2πfL = V/X L [ohms]

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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

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Example i ) X L = 2πfL = 2π x 30 x 0.6 = 113 Ω I L = V / X L = 120 / 113 = 1.06 A ii) X L = 2πfL = 2π x 200 x 0.6 = 753 Ω I L = V / X L = 120 / 753 = A

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Circuit possessing capacitance only v = V m sin θ = V m sin 2πft i = C dv/dt i = 2πfCV m cos 2πft = 2πfCV m sin(2πft+π/2)

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Waveforms for capacitive circuit

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phasor diagram for the capacitive circuit VRVR VCVC V C lags by 90 o

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Capacitive reactance V rms /I rms = 0.707V m /0.707I m = 1/(2πfC) = capacitive reactance [X L ]

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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

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i) X C = 1 / 2πfC = 1 / 2π x 30 x 0.6 x = 8842 Ω I C = V / X C = 120 / 8842 = 13.6 mA i) X C = 1 / 2πfC = 1 / 2π x 200 x 0.6 x = 1326 Ω I C = V / X C = 120 / 1326 = 0.09 A

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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.

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Series Circuits (R,L,C) impedance [Z] =√{R 2 +(X L -X C ) 2 } φ = phase angle = tan -1 (X L -X C )/R), Cos -1 R/Z, Sin -1 = (X L - X C ) / Z phasor diagram

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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

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Example i) Circuit impedance. X C = 1 / 2πfC = 1 / 2π x 200 x 150 x = Ω Z = √ R 2 + X C 2 = √ = √ = Ω ii) Current = I = V/Z = 200 / = A iii) Phase angle = tan -1 X C / R = degrees leading

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Series Resonance (R,L,C) X L =1/X C f = 1/{2π√(LC)} phasor diagram definition: acceptor circuit graph of current and impedance plotted against Z

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Q factor (at resonance) Q = X L /R = 1/R √(L/C) bandwidth - (f 2 -f 1 ) - definition of half-power points Q = f r /(f 2 -f 1 )

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Parallel Circuits (R,L,C) supply current = √V/R + V/X L + X C V) φ = phase angle = phase difference V S and I S φ = tan -1 (I L - I C )/I R phasor diagram

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Parallel Resonance (R,L,C) f = 1/(2πL) √(L/C - R 2 ) phasor diagram definition: rejector circuit dynamic impedance R D = L/CR Q factor (at resonance) = X L /R

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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

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Conductance, Susceptance, and Admittance are the opposites to Resistance, reactance and impedance

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Impedance triangle Resistance R Reactance X Impedance Z

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Admittance triangle Conductance G Susceptance B Admittance Y

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Conductance [G] = R/Z 2. Is 1/R when X is = 0 admittance [Y] = 1/Z = R/Z 2 susceptance [B] = X/Z 2. Is 1/X when X is = 0 Y = G+ jB and tanφ = B/G

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R and L in series Z = R +jX L = Z < φ admittance = Y = 1/Z = (R/Z 2 - jXL/Z 2 ) = G – jB L = Y < -φ

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R and C in series Z = R -jX C = Z<-φ admittance = Y = 1/Z = (R/Z 2 + jX C /Z 2 ) = G + jB C = Y<φ

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