Slide 4.0CE163 - 2023 F. Sepulveda - CSEE - Essex University
Foundations of Electronics I
Lecture 4 (Week 5)
Parallel resistive circuits
Kirchhoff’s current law
Power in parallel circuits
Francisco Sepulveda
E-mail: f.sepulveda
CE163
Slide 4.1CE163 - 2023 F. Sepulveda - CSEE - Essex University
Parallel
Circuits
Required reading:
Chapter 5 in Floyd & Buchla (2014, 8th ed.)
Slide 4.2CE163 - 2023 F. Sepulveda - CSEE - Essex University
Resistors in Parallel
Resistors that are connected between the same two voltage
points (or nodes) are said to be in parallel.
Slide 4.3CE163 - 2023 F. Sepulveda - CSEE - Essex University
Resistors in Parallel
Example Show how to connect the resistors on the
breadboard in parallel.
Slide 4.4CE163 - 2023 F. Sepulveda - CSEE - Essex University
Resistors in Parallel
Solution: This is one way. Notice that one node is colored in pink;
the other is black and all resistors are between these two
nodes.
Slide 4.5CE163 - 2023 F. Sepulveda - CSEE - Essex University
Parallel Circuits
A parallel circuit is identified by the fact that it has more
than one current path (branch) connected to a common
voltage source.
Slide 4.6CE163 - 2023 F. Sepulveda - CSEE - Essex University
Parallel Circuit Rule for Voltage
Because all components are connected across the same
voltage source, the voltage across each is the same.
For example, the source voltage is +5.0 V. What will a
voltmeter read if it is placed across each of the resistors?
Slide 4.7CE163 - 2023 F. Sepulveda - CSEE - Essex University
Parallel vs. Series circuits
Series network:
 same current at all components
 different voltages at each component
 use KVL and voltage divider
Parallel network:
 same voltages at all branches
 different currents at each branch
 use KCL and current divider
Slide 4.8CE163 - 2023 F. Sepulveda - CSEE - Essex University
Parallel Resistors
The total resistance of resistors in parallel is the reciprocal
of the sum of the sum of the reciprocals of the individual
resistors:
R =
1
1
R
+
1
R
+
1
R
+. . . +
1
R
Example:
The resistors in a certain parallel circuit are 680Ω, 1.5kΩ, and
2.2kΩ. What is the total resistance? 386 W
NB: in parallel circuits, the total (or equivalent) resistance is always smaller
than any of the individual resistances.
Slide 4.9CE163 - 2023 F. Sepulveda - CSEE - Essex University
Two Parallel Resistors
The special case of resistance
of two parallel resistors can be
found by:
R =
1
1
R
+
1
R
which can be
reduced to
R =
RR
R + R
This is called the
product-over-sum
rule.
Example: Calculate the total resistance if R = 27kΩ and
R = 56kΩ.
= 18.2kΩR =
27Ω x 56Ω
27Ω + 56Ω
Slide 4.10CE163 - 2023 F. Sepulveda - CSEE - Essex University
Parallel Circuits
Tabulating current, resistance, voltage, and power is a useful way to
summarize parameters in a parallel circuit.
Continuing with the previous example, complete the parameters
listed in the Table.
I1= R1= 0.68 kW V1= P1=
I2= R2= 1.50 kW V2= P2=
I3= R3= 2.20 kW V3= P3=
IT= RT = 386 W VS= 5.0 V PT=
5.0 V
5.0 V
5.0 V
13.0 mA
2.3 mA
3.3 mA
7.4 mA 36.8 mW
16.7 mW
11.4 mW
64.8 mW
Slide 4.11CE163 - 2023 F. Sepulveda - CSEE - Essex University
Kirchhoff’s Current Law
Kirchhoff’s current law (K C L) is generally stated as:
The sum of the currents entering a node is equal to the sum of
the currents leaving the node.
Notice in the previous example that the current from the source is equal
to the sum of the branch currents.
I1= R1= 0.68 kW V1= P1=
I2= R2= 1.50 kW V2= P2=
I3= R3= 2.20 kW V3= P3=
IT= RT = 386 W VS= 5.0 V PT=
5.0 V
5.0 V
5.0 V
13.0 mA
2.3 mA
3.3 mA
7.4 mA 36.8 mW
16.7 mW
11.4 mW
64.8 mW
Slide 4.12CE163 - 2023 F. Sepulveda - CSEE - Essex University
Current Divider
When current enters a node (junction) it divides inversely
proportional to each branch resistance.
The most widely used formula for the current divider is the
two-resistor equation. For resistors R and ,
=
+
and =
+
Notice the subscripts. The resistor in the numerator is not the same as
the one for which current is found.
I
T
I
1
I
2
Slide 4.13CE163 - 2023 F. Sepulveda - CSEE - Essex University
Current Divider
Example Assume that is a 2.2kΩ resistor that is in
parallel with R, which is 4.7kΩ. If the total
current into the resistors is 8.0 mA, what is the
current in each resistor?
Solution
=
+
=
4.7 kΩ
6.9 kΩ
8.0 mA = 5.45mA
=
R
+
=
2.2 kΩ
6.9 kΩ
8.0 mA = 2.55mA
Notice that the larger resistor has the smaller current.
Slide 4.14CE163 - 2023 F. Sepulveda - CSEE - Essex University
Power in Parallel Circuits
Power in parallel resistors can be calculated with any of the standard
power formulas. Most of the time, the voltage and resistance are known,
so =
is the most convenient one to use.
The total power is the sum of the powers dissipated in each resistor.
Example: Assume 10 V is applied to the parallel combination of
= 270Ω and = 150Ω. Calculate the total power.
Solution:
=
=
10 V
270 Ω
= 0.370 W =
=
10 V
150 Ω
= 0.667 W
Alternatively,
=
+
=
270 Ω 150 Ω
270 Ω + 150 Ω
= 96.4 Ω =
=
10 V
96.4 Ω
= 1 + 2 = 1.037W
Slide 4.15CE163 - 2023 F. Sepulveda - CSEE - Essex University
Power in Parallel Circuits
Question:
Assume there are 8 wires with equal resistance that form
a rear window defroster for an automobile. The wires are
in parallel.
(a) If the defroster dissipates 90 W when connected to a
12.6 V source, what power is dissipated by each resistive
wire?
(b) What is the total resistance of the defroster?
Answer: (a) Each of the 8 wires will dissipate 1/8 of the total power or
(b) The total resistance is
Follow
up: What is the resistance of each wire?
90W
8wires
= 11.25W
=
=
12.6V
90W
= 1.76Ω
1.76Ω × 8 = 14.1Ω
Slide 4.16CE163 - 2023 F. Sepulveda - CSEE - Essex University
Key Terms
Parallel The relationship in electric circuits in which
two or more current paths are connected
between two separate points (nodes).
Branch One current path in a parallel circuit.
Kirchhoff’s
current law
A law stating the total current into a node
equals the total current out of the node.
Node A point or junction in a circuit at which
two or more components are connected.
Current divider A parallel combination of resistances in
which the input current divides in inverse
proportion to the branch resistances.