Chapter 12. RL Circuits ISU EE. C.Y. Lee


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1 Chapter 12 RL Circuits
2 Objectives Describe the relationship between current and voltage in an RL circuit Determine impedance and phase angle in a series RL circuit Analyze a series RL circuit Determine impedance and phase angle in a parallel RL circuit Analyze a parallel RL circuit Analyze seriesparallel RL circuits Determine power in RL circuits 2
3 Sinusoidal Response of RL Circuits The inductor voltage leads the source voltage Inductance causes a phase shift between voltage and current that depends on the relative values of the resistance and the inductive reactance 3
4 Impedance and Phase Angle of Series RL Circuits The phase angle is the phase difference between the total current and the source voltage The impedance of a series RL circuit is determined by the resistance (R) and the inductive reactance (X L ) (Z = R + jx L ) (Z L = j L= jx L ) (Z= R+Z L = R+jX L ) 4
5 Impedance and Phase Angle of Series RL Circuits In the series RL circuit, the total impedance is the phasor sum of R and jx L Impedance magnitude: Z = R 2 + X 2 L Phase angle: θ = tan 1 (X L /R) 5
6 Impedance and Phase Angle of Series RL Circuits Example: Determine the impedance and the phase angle Z = (5.6) 2 + (10) 2 = 11.5 kω θ = tan 1 (10/5.6) = tan 1 (1.786) =
7 Analysis of Series RL Circuits Application of Ohm s Law to series RL circuits involves the use of the quantities Z, V, and I as: V = I = Z = IZ V Z V I 7
8 Analysis of Series RL Circuits Example: If the current is 0.2 ma, determine the source voltage and the phase angle X L = 2π( )( ) = 6.28 kω Z = ( ) 2 + ( ) 2 = 11.8 kω V S = IZ = (200µA)(11.8kΩ) = 2.36 V θ = tan 1 (6.28k/10k) =
9 Relationships of I and V in a Series RL Circuit In a series circuit, the current is the same through both the resistor and the inductor The resistor voltage is in phase with the current, and the inductor voltage leads the current by 90 I V R V L 9
10 KVL in a Series RL Circuit From KVL, the sum of the voltage drops must equal the source voltage (V S ) I V R V L Since V R and V L are 90 out of phase with each other, they must be added as phasor quantities V L = I(jX L ) V S = IZ = I(R+jX L ) V S = V 2 R + V2 L θ = tan 1 (V L /V R ) I V R = IR 10
11 KVL in a Series RL Circuit Example: Determine the source voltage and the phase angle V S = (50) 2 + (35) 2 = 61 V θ = tan 1 (35/50) = tan 1 (0.7) = 35 11
12 Variation of Impedance and Phase Angle with Frequency For a series RL circuit; as frequency increases: R remains constant X L increases Z increases θ increases f Z 12 R
13 Impedance and Phase Angle of Parallel RL Circuits Total impedance in parallel RL circuit: Z = (RX L ) / ( R 2 +X 2 L ) Phase angle between the applied V and the total I: θ = tan 1 (R/X L ) 13 1 Z = 1 Z = Z = R jx jx R ( jx ) RX X L L V = IZ = I + L L R jr L RX X L jr L
14 Conductance, Susceptance and Admittance Conductance is the reciprocal of resistance: G = 1/R Inductive susceptance is the reciprocal of inductive reactance: B C = 1/X L Admittance is the reciprocal of impedance: Y = 1/Z 14
15 Ohm s Law Application of Ohm s Law to parallel RL circuits using impedance can be rewritten for admittance (Y=1/Z): V = I Y I =VY Y = I V 15
16 Relationships of the I and V in a Parallel RL Circuit The applied voltage, V S, appears across both the resistive and the inductive branches Total current, I tot, divides at the junction into the two branch current, I R and I L I R = V/R I tot = V/Z = V((X L jr)/rx L ) I L = V/(jX L ) 16
17 KCL in a Parallel RL Circuit From KCL, Total current (I S ) is the phasor sum of the two branch currents Since I R and I L are 90 out of phase with each other, they must be added as phasor quantities I R = V/R I tot = I 2 R + I2 L θ = tan 1 (I L /I R ) 17 I L = V/(jX L ) I tot = V/Z = V((X L jr)/rx L )
18 KCL in a Parallel RL Circuit Example: Determine the value of each current, and describe the phase relationship of each with the source voltage V s I R = 12/220 = 54.5 ma I C = 12/150 = 80 ma I tot = (54.5) 2 + (80) 2 = 96.8 ma θ = tan 1 (80/54.5) =
19 SeriesParallel RL Circuits An approach to analyzing circuits with combinations of both series and parallel R and L elements is to: Calculate the magnitudes of capacitive reactances (X L ) Find the impedance (Z) of the series portion and the impedance of the parallel portion and combine them to get the total impedance 19
20 RL Circuit as a LowPass Filter An inductor acts as a short to dc As the frequency is increased, so does the inductive reactance As inductive reactance increases, the output voltage across the resistor decreases A series RL circuit, where output is taken across the resistor, finds application as a lowpass filter 20
21 RL Circuit as a LowPass Filter 21
22 RL Circuit as a HighPass Filter For the case when output voltage is measured across the inductor At dc, the inductor acts a short, so the output voltage is zero As frequency increases, so does inductive reactance, resulting in more voltage being dropped across the inductor The result is a highpass filter 22
23 RL Circuit as a HighPass Filter 23
24 Summary When a sinusoidal voltage is applied to an RL circuit, the current and all the voltage drops are also sine waves Total current in an RL circuit always lags the source voltage The resistor voltage is always in phase with the current In an ideal inductor, the voltage always leads the current by 90 24
25 Summary In an RL circuit, the impedance is determined by both the resistance and the inductive reactance combined The impedance of an RL circuit varies directly with frequency The phase angle (θ) if a series RL circuit varies directly with frequency In an RL lag network, the output voltage lags the input voltage in phase In an RL lead network, the output voltage leads the input voltage in phase 25
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