Unit 19
Resistive-Inductive Parallel Circuits
Unit 19
Resistive-Inductive Parallel
Circuits
ET = 240 V
ER = 240 V
EL = 240 V
Parallel circuit components have the same voltage
drop.
The voltage drops are in phase.
Unit 19
Resistive-Inductive Parallel Circuits
Unit 19
Resistive-Inductive Parallel Circuits
Objectives:
• Discuss the operation of a parallel circuit
containing resistance and inductance.
• Compute circuit values of an R-L parallel
circuit.
• Compute power factor values of an R-L
parallel circuit.
Unit 19
Resistive-Inductive Parallel Circuits
A parallel circuit with an inductor and resistor.
The currents through the inductor and resistor are
90out of phase.
Current flow through the resistor is in phase with
the voltage.
Unit 19
Resistive-Inductive Parallel Circuits
Current flow through the inductor lags the voltage
by 90(out of phase).
Unit 19
Resistive-Inductive Parallel Circuits
ET = 240 V
R = 15 Ω
Unit 19
Resistive-Inductive Parallel Circuits
XL = 20 Ω
IT = 20 A
A sample parallel circuit problem with an inductor
and resistor.
The resistance, inductive reactance, and source
voltage are the only given values.
Unit 19
Resistive-Inductive Parallel Circuits
ET = 240 V
ER = 240 V
R = 15 Ω
EL = 240 V
XL = 20 Ω
The volt-drops across the components are equal.
ET = ER = EL = 240 V
Unit 19
Resistive-Inductive Parallel Circuits
ET = 240 V
ER = 240 V
R = 15 Ω
IR = 16 A
EL = 240 V
XL = 20 Ω
IL = 12 A
Next, use Ohm’s law to calculate the current
through each component.
IR = ER / R = 16 A and IL = EL / XL = 12 A
IL = 12 A
IR = 16 A
Vector addition is required to determine the total
circuit current.
IT2 = IR2 + IL2 = 400 A  IT = 20 A
Unit 19
Resistive-Inductive Parallel Circuits
ET = 240 V
IT = 20 A
Z = 12 Ω
ER = 240 V
R = 15 Ω
IR = 16 A
EL = 240 V
XL = 20 Ω
IL = 12 A
Now the total circuit resistance can be
found using the IT and Ohm’s law.
RT = ET / IT = 240 / 20 = 12 Ω
Unit 19
Resistive-Inductive Parallel Circuits
ET = 240 V
IT = 20 A
Z = 12 Ω
VA = 4800
ER = 240 V
R = 15 Ω
IR = 16 A
EL = 240 V
XL = 20 Ω
IL = 12 A
The total apparent power can be found using
Ohm’s law.
VA = ET x IT = 240 x 20 = 4800
Unit 19
Resistive-Inductive Parallel Circuits
ET = 240 V
IT = 20 A
Z = 12 Ω
VA = 4800
ER
R
IR
P
= 240 V
= 15 Ω
= 16 A
= 3840 W
EL = 240 V
XL = 20 Ω
IL = 12 A
The total true power can be derived using Ohm’s
law.
Watts = ER x IR = 240 x 16 = 3840 W
Unit 19 Resistive-Inductive
Parallel Circuits
ET = 240 V
IT = 20 A
Z = 12 Ω
VA = 4800
ER
R
IR
P
= 240 V
= 15 Ω
= 16 A
= 3840 W
Unit 19
Resistive-Inductive Parallel Circuits
Vector relationships of component powers and
angle theta (θ).
Unit 19
Resistive-Inductive Parallel Circuits
EL = 240 V
XL = 20 Ω
IL = 12 A
VARsL = 2880
The total reactive power can also be derived using
Ohm’s law.
VARsL = EL x IL = 240 x 12 = 2880 VARsL
Unit 19
Resistive-Inductive Parallel Circuits
Power Factor is equal to the cosine of angle theta (θ).
PF = COS θ = COS 36.87° = 0.80
Unit 19
Resistive-Inductive Parallel Circuits
Review:
ET = 240 V
IT = 20 A
Z = 12 Ω
VA = 4800
PF = 80%
ER
R
IR
P
= 240 V
= 15 Ω
= 16 A
= 3840 W
EL = 240 V
XL = 20 Ω
IL = 12 A
VARsL = 2880
The power factor can be calculated.
PF = Watts / VA = 3840 / 4800 = .80 or 80%
1. The voltage applied across components in a
parallel circuit must be the same.
2. The current flowing through resistive parts of
the circuit will be in phase with the voltage.
3. The current flowing through inductive parts of
the circuit will lag the voltage by 90.
Unit 19
Resistive-Inductive Parallel Circuits
Review:
4. The total current in a parallel circuit is equal
to the sum of the individual currents. Vector
addition must be used because the current
through the resistive parts or the circuit is
90 out of phase with current flowing
through the inductive parts.
Unit 19
Resistive-Inductive Parallel Circuits
Review:
5. The impedance of an R-L parallel circuit can be
computed by using vector addition to add the
reciprocals of the resistance and inductive
reactance.
6. Apparent power, true power, and reactive
power add in a circuit. Vector addition is used
because true power and reactive power are 90
out of phase.