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ISU EE C.Y. Lee

Chapter 12RL Circuits

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2ISU EE C.Y. Lee

ObjectivesDescribe the relationship between current and

voltage in an RL circuitDetermine impedance andphase angle in a series

RL circuit

Analyze a series RL circuit

Determine impedance andphase angle in aparallel

RL circuit

Analyze aparallel RL circuit

Analyze series-parallel RL circuits

Determinepowerin RL circuits

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3ISU EE C.Y. Lee

Sinusoidal Response of

RL Circuits

Theinductor voltage

leads the source voltageInductance causes a phase shiftbetween voltageand current that depends on the relative values of

the resistance and the inductive reactance

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4ISU EE C.Y. Lee

Impedance and Phase Angle of

Series RL Circuits

The phase angle is the phase difference betweenthe total current and the source voltage

The impedance of a series RL circuit is

determined by the resistance (R) and the inductivereactance (XL)

(Z = R + jXL)

(ZL= j

L= jXL)(Z= R+ZL= R+jXL)

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5ISU EE C.Y. Lee

Impedance and Phase Angle of

Series RL Circuits

In the series RL circuit, the total impedance is thephasor sum of R and jXLImpedance magnitude: Z = R2 + X2LPhase angle: = tan-1(XL/R)

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6ISU EE C.Y. Lee

Impedance and Phase Angle ofSeries RL Circuits

Z = (5.6)2 + (10)2 = 11.5 k

= tan-1(10/5.6) = tan-1(1.786) = 60.8

Example: Determine the impedance and the phase angle

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7ISU EE C.Y. Lee

Analysis of Series RL Circuits

Application of Ohms Law to series RL circuitsinvolves the use of the quantities Z, V, and I as:

IZV=

Z

VI=

I

VZ=

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8ISU EE C.Y. Lee

Analysis of Series RL Circuits

Example: If the current is 0.2 mA, determine the source

voltage and the phase angle

XL = 2(10103)(10010-3) = 6.28 kZ = (10103)2 + (6.28103)2 = 11.8 k

VS = IZ = (200A)(11.8k) = 2.36 V

= tan-1(6.28k/10k) = 34.2

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9ISU EE C.Y. Lee

Relationships of I and V in a

Series RL CircuitIn a series circuit, the current is the same through

both the resistor and the inductor

The resistor voltage is in phase with the current, andthe inductor voltage leads the current by 90

I

VR VL

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KVL in a Series RL CircuitExample: Determine the source voltage and the phase angle

VS = (50)2 + (35)2 = 61 V

= tan-1(35/50) = tan-1(0.7) = 35

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12ISU EE C.Y. Lee

Variation of Impedance and

Phase Angle with Frequency

Zf

For a series RL circuit;

as frequency increases:

R remains constant XL increases

Z increases

increases

R

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13ISU EE C.Y. Lee

Impedance and Phase Angle of

Parallel RL Circuits

Total impedance in parallel RL circuit:Z = (RXL) / ( R2 +X2L)

Phase anglebetween the applied V and the total I:

= tan-1(R/XL)

( )

jRX

RXZ

jXR

RjX

Z

jXRZ

L

L

L

L

L

=

+=

+=

1

111

== jRXRX

IIZVL

L

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14ISU EE C.Y. Lee

Conductance, Susceptance and

Admittance

Conductance is the reciprocal of resistance:

G = 1/R

Inductive susceptance is the reciprocal ofinductive reactance:

BC = 1/XLAdmittance is the reciprocal of impedance:

Y = 1/Z

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15ISU EE C.Y. Lee

Ohms Law

Application of Ohms Law to parallel RL circuits

using impedance can be rewritten for admittance

(Y=1/Z):

Y

I

V=

VYI=

V

IY=

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16ISU EE C.Y. Lee

Relationships of the I and V in a

Parallel RL Circuit

The applied voltage, VS, appears across both theresistive and the inductive branches

Total current, Itot, divides at the junction into the

two branch current, IRand ILIR= V/R

Itot= V/Z

= V((XLjR)/RXL)

IL= V/(jXL)

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17ISU EE C.Y. Lee

KCL in a Parallel RL CircuitFrom KCL, Total current

(IS) is the phasor sum of thetwo branch currents

Since IRand IL are 90

out of phase with each

other, they must be added

as phasor quantities

Itot = I2R+ I

2L

= tan-1(IL/I

R)

IR= V/R

Itot= V/Z

= V((XLjR)/RXL)

IL= V/(jXL)

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18ISU EE C.Y. Lee

KCL in a Parallel RL Circuit

Example: Determine the value of each current, and describe

the phase relationship of each with the source voltage

IR= 12/220 = 54.5 mA

IC = 12/150 = 80 mA

Itot = (54.5)2 + (80)2 = 96.8 mA

= tan-1(80/54.5) = 55.7

Vs

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19ISU EE C.Y. Lee

Series-Parallel 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 (XL)

Find the impedance (Z) of the series portion and the

impedance of the parallel portion and combine them to get

the total impedance

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20ISU EE C.Y. Lee

RL Circuit as a Low-Pass 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 low-pass filter

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21ISU EE C.Y. Lee

RL Circuit as a Low-Pass Filter

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RL Circuit as a High-Pass Filter

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Summary

When a sinusoidal voltage is applied to an RL

circuit, the current and all the voltage drops arealso sine waves

Total current in an RL circuit always lags thesource voltage

The resistor voltage is always in phase with the

currentIn an ideal inductor, the voltage always leads the

current by 90

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25ISU EE C.Y. Lee

SummaryIn an RL circuit, the impedance is determined by

both the resistance and the inductive reactancecombined

The impedance of an RL circuit varies directly with

frequencyThe phase angle () if a series RL circuit variesdirectly with frequency

In an RL lag network, the output voltage lags theinput voltage in phase

In an RL lead network, the output voltage leads the

input voltage in phase