This document discusses a series RL circuit where a resistor and inductor are connected in series. It explains that the current through both components is the same, while the resistor voltage is in phase with the current and the inductor voltage leads the current by 90 degrees. It describes how to calculate circuit values like impedance, power factor, and voltages using phasor diagrams where the current is used as a reference. Changing conditions like frequency, resistance, or inductance will affect values like current, voltages, power, and power factor. Review questions are provided to test understanding.

1. RESISTOR & INDUCTORSRESISTOR & INDUCTORS
IN SERIESIN SERIES

2. This diagram shows the waveform
relationship that would be
generated when you connect a
Resistor and Inductor in series.
OhmsL
VR
IT
VL
When you place a resistor in an
AC circuit the voltage and
current are in phase.
An Inductor however, produces a
phase shift between its voltage
and current of 900
, with the
current lagging the volts

3. CIRCUIT CURRENTCIRCUIT CURRENT

4. In a DC series circuit, the current is the same value for all componentsIn a DC series circuit, the current is the same value for all components
connected in that circuit.connected in that circuit.
The series AC circuit has this same relationship.The series AC circuit has this same relationship.
So the current flowing through the Resistor and Inductor are the sameSo the current flowing through the Resistor and Inductor are the same
value.value.
This current must be supplied from the power source, so the sourceThis current must be supplied from the power source, so the source
current must also be equal to this value.current must also be equal to this value.
100Ω
OhmsL
0.320
A
IT = IR = IL

5. Again you use Ohm’s Law and componentAgain you use Ohm’s Law and component
values from either the Resistor or Inductorvalues from either the Resistor or Inductor
to find current.to find current.
IR = VR ÷ R IL = VL÷ XL
IT = IR = IL

6. CIRCUIT VOLTAGECIRCUIT VOLTAGE

7. When you use a Phasor diagram in a series circuit, theWhen you use a Phasor diagram in a series circuit, the CURRENT is
used as the REFERENCE because it is common to all components inbecause it is common to all components in
that circuit.that circuit.
The reference is always on the X axis at 00
.
VR
IT
VL
VR
VL
IT
A Resistor has a phase relationship of 0A Resistor has a phase relationship of 000
between its current andbetween its current and
voltage i.e. the are in phasevoltage i.e. the are in phase..
The Inductor phase relationship is different, with the Inductor voltageThe Inductor phase relationship is different, with the Inductor voltage
leading the Inductor current by 90leading the Inductor current by 9000
..

8. Because this is a RL circuit, theBecause this is a RL circuit, the circuit phasor diagram will also be awill also be a
combination thecombination the Resistor phasor and theand the Inductor phasor..
So the two voltage phasors are combined, with the circuit currentSo the two voltage phasors are combined, with the circuit current
phasor being used as the reference.phasor being used as the reference.
We now have two voltagesWe now have two voltages
((VR andand VL) operating at 90) operating at 9000
to each other.to each other.
VR
VL
IT
This is the phase
relationship for this circuit

9. When you have two voltages operating at 90When you have two voltages operating at 9000
, the resultant voltage, the resultant voltage
(VT, or circuit voltage) is equal to the phasor addition of theis equal to the phasor addition of the X axis
voltagevoltage (VR) and theand the Y axis voltagevoltage (VL)
22
CRT VVV +=
22
LRT VVV +=
22
LTR VVV −=
22
RTL VVV −=
VR
VL
IT
VT
You can use Pythagoras Theorem to find the
resultant (Hypotenuse)
You can also use Pythagoras Theorem to find
the X and Y values (Adjacent & Opposite)

10. Both the resistor and inductor produce aBoth the resistor and inductor produce a
voltage drop across themselvesvoltage drop across themselves
proportional to their opposition and theproportional to their opposition and the
current flowing through them.current flowing through them.
(OHMS LAW)(OHMS LAW)
VR = IR x R VL = IL x XL

11. PROBLEMS (PROBLEMS (all circuits 50Hz)all circuits 50Hz)
1. A series RL circuit has a resistor volts of 60v and an inductor
volts of 100v. The supply voltage is?
2. In a series RL circuit the supply volts is 240v and the resistor
volts is 180v. The inductor voltage is?
3. A RL circuit has a supply voltage of 1500v and the inductor has
a voltage of 560v. The resistor voltage is?
116.6 V
158.7 V
1391.6 V
4. An RL circuit has a resistance of 150Ω and the circuit current is 4.0A.
The resistor voltage is?
5. An RL circuit has a reactance of 60Ω and the circuit current is 1.5A.
The inductor voltage is?
6. An RL circuit has a current of 15A and the inductor voltage is 400V.
The inductor value is?
7. An RL circuit has a current of 7A and the resistor voltage is 150V. The
resistor value is?
600 V
90 V
84.9mH
21.48 Ω

12. CIRCUIT IMPEDANCECIRCUIT IMPEDANCE

13. Resistors and Inductors both provide opposition to circuit current flow.
Resistors have Resistance measured in Ohms.
Inductors have Inductive Reactance (XL) also in Ohms.
VL
IT R
XL
VR
This diagram shows the phase relationship that
exists between the Resistance and the Inductive
Reactance.
It is the same phase relationship that the component
voltages have.

14. Both components provide opposition to the circuit current.Both components provide opposition to the circuit current.
The issue is the Resistance is 90The issue is the Resistance is 9000
out of phase with the Inductiveout of phase with the Inductive
reactance (Xreactance (XLL).).
Because of this phase angle, you can only find the total oppositionBecause of this phase angle, you can only find the total opposition
by the phasor addition of R and Xby the phasor addition of R and XLL
R
XL

15. When any AC circuit has a combination of different components connected in it,
the circuit total opposition is called IMPEDANCE.
R
XL
So this circuit has some value of R (due to the Resistor) and
some value of XL (due to the Inductor), but the total circuit
opposition is the phasor addition of R and XL.
IMPEDANCE (Z) measured in Ohms.
Z

16. The circuit Impedance is found by theThe circuit Impedance is found by the
phasor addition of these two values.phasor addition of these two values.
22
LXRZ +=
22
LXZR −=
22
RZX L −=
R
XL
Z
As with the voltage phasor you can also use
Pythagoras Theorem to find the X and Y values
(Adjacent & Opposite)

17. The circuit impedance can also be found byThe circuit impedance can also be found by
Z = VT ÷ IT
so
IT = VT ÷ Z
And
VT = Z x IT
The value of the Resistance and Inductive Reactance can be found byThe value of the Resistance and Inductive Reactance can be found by
using component values.using component values.
R = VR ÷ IR XL = VL ÷ IL

18. PROBLEMS (PROBLEMS (all circuits 50Hz)all circuits 50Hz)
1. A series RL circuit has a resistor of 60 Ω and a reactance of 100Ω.
The circuit impedance is?
2. In a series RL circuit the impedance is 240Ω and the resistor is 180Ω.
The inductor value is?
3. An RL circuit has an impedance of 1500Ω and the inductor has a
reactance of 560Ω. The resistor value is?
116.6 Ω
505 mH
1391.6 Ω
4. An RL circuit has an impedance of 150Ω and the circuit voltage is
400V. The circuit current is?
5. An RL circuit has a circuit current of 15A and the circuit voltage is
200V. The circuit impedance is?
6. An RL circuit has a resistance of 180Ω and the resistor voltage is 40V.
The circuit current is?
7. An RL circuit has a circuit current of 7.5A. The circuit voltage is 400V and
the resistor voltage is 200V . The value of the resistor and inductor are?
2.67 A
0.22 A
R=26.7 Ω
L=147mH
13.33 Ω

19. PHASE ANGLEPHASE ANGLE

20. The phase angle in any circuit is determined by the value of theThe phase angle in any circuit is determined by the value of the
components in the circuit.components in the circuit.
When you use Phasors to determine valuesWhen you use Phasors to determine values in a series circuit, the
phase angle in the Voltage Phasor is the same angle in the
Impedance Phasor.
VR
VL
IT
VT
R
XL
Z
Phase angle
anywhere between
near 00
and near 900
Phase angle
anywhere between
near 00
and near 900

21. The phase angle can vary anywhere betweenThe phase angle can vary anywhere between ≈≈ 0 degrees (large R0 degrees (large R
value) andvalue) and ≈≈ 90 degrees (large inductor value)90 degrees (large inductor value) depending on thedepending on the
component values in the circuit.component values in the circuit.
Phase angle
anywhere between
near 00
and near 900
VR
VL
VT
IT
Note:- φ is the same angle for the impedance and voltage phasor.

22. If (VT or Z) and (VL or XL) are known.
We can use Sin θ
Sin-1 is used to determine the angle in degrees from the known sin value.
Sin θ = VL ÷ VT
VL
VT
IT
XL
Z
θ
θ
Sin θ = XL ÷ Z

23. If (VT or Z) and (VR or R) are known.
We can use CosWe can use Cos θθ
Cos-1 is used to determine the angle in degrees from the known Cos value.
Cos θ = VR ÷ VT
VR
VT
IT R
Z
θ
θ
Cos θ = R ÷ Z

24. If (If (VR or R) and (VL or XL) are known.are known.
We can use TanWe can use Tan θθ
Tan-1 is used to determine the angle in degrees from the known Tan value.
Tan θ = VL ÷ VR
VR
VL
IT R
XL
θ
θ
Tan θ = XL ÷ R

25. POWER FACTORPOWER FACTOR

26. Power factor indicatesPower factor indicates how efficient the circuit current is converted tohow efficient the circuit current is converted to
true power and is determined bytrue power and is determined by the phase relationship that existsthe phase relationship that exists
between the supply voltage and current in a circuit.between the supply voltage and current in a circuit.
Power factor = Cos φ
Phase angle
anywhere between
near 00
and near 900
VR
VL
VT
IT
Remember, voltage is always used as the reference

27. The phase angle can vary anywhere betweenThe phase angle can vary anywhere between ≈≈ 0 degrees0 degrees
(large R) and(large R) and ≈≈ 90 degrees (large X90 degrees (large XLL value).value).
The component values in the circuit determine this.The component values in the circuit determine this.
Therefore the power factor can be betweenTherefore the power factor can be between ≈≈ 0 and0 and ≈≈ 1(unity)1(unity)
lagging.lagging.
The word lagging tells use that the circuit current is lagging the circuit voltage.
VR
VL
VT
IT
Power factor
near 1(unity)
Power factor
near 0 lag So the power factor can
be between these limits

28. POWERPOWER

29. The power dissipated by a combined circuit is determined only by the
resistive component in the circuit.
Pure Inductors consume no power.
P = IT x VT x COS φ (Circuit values)
Because the power of a circuit is determined by the resistive
component only, the phasor representation for power is drawn on
the X AXIS.
Power

30. The power can also be found by using
resistor component values.
P = IR
2
x R
P = VR
2
÷ R
P = IR x VR

31. REACTIVE POWERREACTIVE POWER

32. Reactive power is the wattless power
taken by the reactive components
measured in var.
The reactive power is drawn by the
inductor, therefore the phasor
representation for Q is drawn on the
Y AXIS.
QL
The inductor is a reactive component which
consumes energy to establish its magnetic field
during the first quarter cycle but returns the
energy back to the supply in the next quarter
cycle when the inductor discharges.
The power required to achieve this is called
Reactive Power.

33. The power equations you have been using can be usedThe power equations you have been using can be used
to determine VAR by substituted P with Q and R beingto determine VAR by substituted P with Q and R being
substituted withsubstituted with
XXLL
QL = IL
2
x XL ; QL = VL
2
÷ XL ; QL = IL x VL
(Note:- these formula are used with component values only)
QT = IT x VT x sinφ
(Note:- this formula is used with circuit values only)

34. APPARENT POWERAPPARENT POWER

35. When an inductor and resistor are connected into a circuit,When an inductor and resistor are connected into a circuit,
there will be Power and VAR developed in the circuit.there will be Power and VAR developed in the circuit.
The phase relationship is such that the power is always onThe phase relationship is such that the power is always on
the X axis while the VAR is on the Y axis.the X axis while the VAR is on the Y axis.
Power
Reactive
power

36. We now have an X and Y axisWe now have an X and Y axis
components.components.
When we have this relationship, we canWhen we have this relationship, we can
find the Phasor sum of these tofind the Phasor sum of these to
determine the Total power.determine the Total power.
In AC circuits, the phasor addition of theseIn AC circuits, the phasor addition of these
Powers is called thePowers is called the
APPARENT POWERAPPARENT POWER
(S)(S)
measured in VA (volt x amps).measured in VA (volt x amps). P
QL
True Power
ReactivePower
S
Apparent Power

37. 22
LQPS +=
22
LQSP −=
22
PSQL −=

38. The Apparent power in any circuit can byThe Apparent power in any circuit can by
also found byalso found by
S = IS = ITT x Vx VTT

39. Note:-Note:- φφ is the same value for theis the same value for the
Impedance, Power and Voltage phasorImpedance, Power and Voltage phasor..
Phase angle
anywhere between
near 00
and near 900
R Values
L Values
Circuit Values
IT

40. EFFECTS OF CHANGINGEFFECTS OF CHANGING
CIRCUIT CONDITIONSCIRCUIT CONDITIONS

41. The frequency of the supply will determine the value of
inductive reactance in the circuit.
Therefore if the inductive reactance changes with
frequency, then the circuit impedance, power, var,
phase angle, current and voltage drops will also
change.

42. For a RL series connected circuitFor a RL series connected circuit
Indicate what will happen to these if the frequency is reduced?
Current VR
VL
Power Power
Factor
Indicate what will happen to these if the resistance is increased?
Current VR
VL
Power Power
Factor
Indicate what will happen to these if the inductor value is increased?
Current VR
VL
Power Power
Factor

43. For Resistor and Inductors connected in series,For Resistor and Inductors connected in series,
The current value is the same through all componentsThe current value is the same through all components
The Resistor voltage is in phase with the circuit current
The Inductor voltage is leading the circuit current by 900
The supply voltage is found by the phasor addition of VR and VL
The Resistance value is always drawn on the X axis.
The Inductive Reactance is at 900
to the resistance and is drawn
on the Y axis.
The total opposition is called Impedance and is measured in Ohms.
The Impedance is found by the phasor addition of R and XL.

44. REVIEW QUESTIONSREVIEW QUESTIONS
((all circuits 50Hz)all circuits 50Hz)

45. 2.. A circuit draws a current of 6.8 amps when connected to a 400v supply. If theA circuit draws a current of 6.8 amps when connected to a 400v supply. If the
reactive power of the circuit is 2.2 kVAR the power factor is ?reactive power of the circuit is 2.2 kVAR the power factor is ?
3. A circuit develops 5kVAR when connected to a 200v supply with a power factor ofA circuit develops 5kVAR when connected to a 200v supply with a power factor of
0.75 lag. The current drawn by this circuit is?0.75 lag. The current drawn by this circuit is?
4. A circuit has a resistance of 5 ohms and a reactance of 6 ohms when a 50hzA circuit has a resistance of 5 ohms and a reactance of 6 ohms when a 50hz
supply is connected. Determine circuit impedance and angle by which load currentsupply is connected. Determine circuit impedance and angle by which load current
would lag the voltage.would lag the voltage.
5. A winding has 50 ohms of resistance and 0.2h of inductance. It is connectedA winding has 50 ohms of resistance and 0.2h of inductance. It is connected
across a 240v, 50hz supply. Calculate the:across a 240v, 50hz supply. Calculate the:
(a) inductive reactance(a) inductive reactance
(b) impedance(b) impedance
(c) circuit current(c) circuit current
(d) angle by which the current lags the voltage.(d) angle by which the current lags the voltage.
(e) power consumed(e) power consumed
(f) power factor(f) power factor
5184 VAR
0.59 Lag
37.88 A
7.8 Ω @ 50.20
1. A circuit is connected to a 240v 50hz supply and draws a current of 27 amps.
If the power factor is 0.6 lagging ,what is the VAR of the circuit?
80.3 Ω
51.480
62.83 Ω
2.99 A
447 W
0.623 Lag

46. 47.17 Ω
55.86 Ω
4.3 A
57.520
553.8 W
2.25 Ω
7.9 mH
6. A winding has 30 ohms of resistance and 0.15h of inductance. It is to be. A winding has 30 ohms of resistance and 0.15h of inductance. It is to be
connected to a 240v, 50hz supply. Calculate the:connected to a 240v, 50hz supply. Calculate the:
(a) inductive reactance(a) inductive reactance
(b) impedance(b) impedance
(c) circuit current(c) circuit current
(d) angle by which the current lags the voltage.(d) angle by which the current lags the voltage.
(e) power(e) power
7. A RL series circuit takes 2000w of power and has a power factor of 0.67lag. A RL series circuit takes 2000w of power and has a power factor of 0.67lag
when it is connected to a 100V, 50 Hz supply. Calculate.when it is connected to a 100V, 50 Hz supply. Calculate.
(a) the value of the resistor(a) the value of the resistor
(b) the value of the inductor(b) the value of the inductor

47. ENDEND