Section 1: Simple Circuits
In a series circuit, current follows a single path; in a parallel circuit,
current follows more than one path.
K
What I Know
W
What I Want to Find Out
L
What I Learned
series circuits
series
parallel
IA = IB = I battery
I battery = I1+I2+I3
current
VA +VB = V battery
V battery = v1=v2=v3
Voltage
R  R1  R2 
Increase the total resistance
1
1
1



R
R1 R2
Decrease the total resistance
Equivalent
resistance
Effect of
adding
new
resistance
outcomes
• 1- To calculate equivalent resistance in parallel
• 2- To calculate current through each branch of
circuit
• 3- To Compare between voltage , Req , Current
• In series and parallel connections
Parallel Circuits
•
•
•
•
A circuit in which there are several current paths is
called a parallel circuit.
In the mountain river model, such a circuit is illustrated
by multiple paths for the water over a waterfall.
• Some paths might have a large flow of water, while
others might have a small flow.
• The sum of the flows, however, is equal to the total
flow of water over the falls.
• In addition, regardless of which channel the water
flows through, the drop in height is the same.
Similarly, in a parallel electric circuit, the total current is
the sum of the currents through each path, and the
potential difference across each path is the same.
The current through each resistor in a parallel electric
circuit depends upon the individual resistances.
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Simple Circuits
Construct parallel circuit
• Go to LMS and use the link
Parallel Circuits
•
The branches of a parallel circuit are independent of
each other. The current through each resistor
depends only upon the potential difference across it
and its resistance.
V
 branch 
Rbranch
•
The total current is the sum of the current in the
branches.
  1   2 
•
The total current is also equal to the potential
difference across the source divided by the
equivalent resistance (R).
 
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V
R
Simple Circuits
Parallel Circuits
•
If the previous equations are combined, the result is a
relationship between potential difference, the individual
resistances, and the equivalent resistances.
V
V
V



R
R1 R2
•
V can be canceled to obtain the relationship between
the equivalent resistance and the individual
resistances.
Equivalent Resistance
for Resistors in Parallel
•
1
1
1



R
R1 R2
Placing more resistors in parallel always decreases the
equivalent resistance of a circuit. The resistance
decreases because each new resistor provides an
additional path for current, thereby increasing the total
current while the potential difference remains
unchanged.
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Simple Circuits
Parallel Circuits
KNOWN
Use with Example Problem 3.
Problem
Four resistors, 50.0 Ω, 40.0 Ω, 30.0 Ω, and
20.0 Ω, are connected with a parallel circuit
across a 120.0-V battery. Find the current
through each branch of the circuit, the
equivalent resistance of the circuit, and the
current through the battery.
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw and label a circuit diagram.
• List the knowns and unknowns.

A
B
C
D
RA
RB
RC
RD
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UNKNOWN
Vbattery = 120.0 V
battery = ?
RA = 50.0 Ω
A = ?
RB = 40.0 Ω
B = ?
RC = 30.0 Ω
C = ?
RD = 20.0 Ω
D = ?
Req = ?
SOLVE FOR THE UNKNOWN
• The potential difference is the same
across all four resistors.
• Use the relationship among potential
difference, current, and resistance.
I 
120 V
 2.4 A
50.0 Ω
120 V
IB 
 3.0 A
40.0 Ω
IA 
V
R
120 V
 4.0 A
30.0 Ω
120 V
ID 
 6.0 A
20.0 Ω
IC 
Simple Circuits
Parallel Circuits
Use with Example Problem 3.
KNOWN
Problem
Vbattery = 120.0 V
battery = ?
Four resistors, 50.0 Ω, 40.0 Ω, 30.0 Ω, and
20.0 Ω, are connected with a parallel circuit
across a 120.0-V battery. Find the current
through each branch of the circuit, the
equivalent resistance of the circuit, and the
current through the battery.
RA = 50.0 Ω
A = 2.4 A
RB = 40.0 Ω
B = 3.0 A
RC = 30.0 Ω
C = 4.0 A
RD = 20.0 Ω
D = 6.0 A
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw and label a circuit diagram.
• List the knowns and unknowns.

A
B
C
D
RA
RB
RC
RD
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UNKNOWN
Req = ?
SOLVE FOR THE UNKNOWN
• Determine the equivalent resistance.
1
1
1
1
1




Req RA RB RC RD
1
1
1
1



50.0 Ω 40.0 Ω 30.0 Ω 20.0 Ω
 0.1283 Ω1

Req  7.79 Ω
Simple Circuits
KNOWN
Parallel Circuits
UNKNOWN
Vbattery = 120.0 V
battery = ?
Use with Example Problem 3.
RA = 50.0 Ω
A = 2.4 A
Problem
RB = 40.0 Ω
B = 3.0 A
Four resistors, 50.0 Ω, 40.0 Ω, 30.0 Ω, and
20.0 Ω, are connected with a parallel circuit
across a 120.0-V battery. Find the current
through each branch of the circuit, the
equivalent resistance of the circuit, and the
current through the battery.
RC = 30.0 Ω
C = 4.0 A
RD = 20.0 Ω
D = 6.0 A
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw and label a circuit diagram.
• List the knowns and unknowns.

A
B
C
D
RA
RB
RC
RD
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Req = 7.79 Ω
SOLVE FOR THE UNKNOWN
• The current through the battery is equal
to the potential difference across the
battery divided by the equivalent
resistance.
 battery 
Vbattery
Req

120.0 V
 15.4 A
7.79 Ω
EVALUATE THE ANSWER
• Resistance is measured in ohms and current is
measured in amperes, so the units are correct.
• The sum of the currents in the branches equals
the current through the battery.
Simple Circuits
Self learning
• 58,59,60,64
series
parallel
IA = IB = I battery
I battery = I1+I2+I3
current
VA +VB = V battery
V battery = v1=v2=v3
Voltage
Equivalent
resistance
R  R1  R2 
1
1
1



R
R1 R2
Effect of
adding
new
resistance
Effect of
removing
bulb
The
hottest
one
Kirchhoff's Rules
•
•
•
•
Gustav Robert Kirchhoff was a German physicist who formulated two
rules that govern electric circuits:
• The loop rule
• The junction rule
The loop rule describes electric potential differences and is based on the
law of conservation of energy.
It states that the sum of increases in electric potential around a loop in
an electric circuit equals the sum of decreases in electric potential
around that loop.
For an application, picture an electric
current traveling clockwise around the
red loop in the figure.
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Simple Circuits
Kirchhoff's Rules
•
•
•
•
For an application, picture an electric current traveling clockwise around
the red loop in the figure.
Electric potential increases by 9V as this charge travels through the
battery, and electric potential drops by 5V as this charge travels through
resistor 1.
By the loop rule, the increases in electric potential around a loop must
equal the decreases in electric potential around that loop. Therefore, the
drop in electric potential across resistor 2 must be 9V – 5V = 4V.
Note that resistor 3 does not affect our answer
because resistor 3 is not a part of the loop
that includes the battery, resistor 1, and
resistor 2.
Copyright © McGraw-Hill Education
Simple Circuits
Kirchhoff's Rules
•
•
•
•
The junction rule describes currents and is based on the law of
conservation of charge.
• Recall the law of conservation of charge states that charge can
neither be created or destroyed.
• This means that, in an electric circuit, the total current into a section
of that circuit must equal the total current out of that same section.
A junction is a location where three or more
wires are connected together.
According to Kirchhoff’s rule, the sum
of currents entering a junction is equal
to the sum of currents leaving that
junction. Otherwise, charge would build
up at the junction.
In the figure, 1 = 2 + 3 at junction A,
and 2 + 3 = 1 at junction B.
Copyright © McGraw-Hill Education
Simple Circuits
Review
Essential Questions
•
•
•
What are the characteristics of series and parallel circuits?
How are currents, potential differences, and equivalent resistances in series
circuits related?
How are currents, potential differences, and equivalent resistances in parallel
circuits related?
Vocabulary
•
•
series circuit
equivalent resistance
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•
•
voltage divider
parallel circuit
Simple Circuits