Lesson 3
Simple Circuits
Focus Question
How can you build a simple circuit to fit
the needs of a given situation?
New Vocabulary
series circuit
equivalent resistance
voltage divider
parallel circuit
Review Vocabulary
resistance: the measure of how strongly an
object or material impedes the flow of electric
charge produced by a potential difference; equal
to the potential difference divided by the current
A River Model
You can use a mountain river to model an electric circuit.
Some of the similarities are shown in the table below.
Mountain River
Electric Circuit
Water flows downhill.
Positive charges move from high
potential to low potential.
Can have one or more streams
Can have one or more current paths
Large rocks and other obstacles that hinder
water flow
Resistors
Solar energy evaporates groundwater,
causing clouds to form; clouds precipitate
on the mountaintops.
Battery moves charges from low
potential to high potential.
Series Circuits
A circuit in which there is only one path for the
current is called a series circuit.
Series Circuits
• The current is the same through the entire series
circuit.
• In an electric circuit, the increase in voltage
provided by the energy source (ΔVsource) is equal
to the sum of voltage drops across resistors.
ΔVsource = ΔV1 + ΔV2 +… = IR1 + IR2 + …
• Thus, the current through the circuit is:
I=
DVsource
R1 + R2 + ...
Series Circuits
• The same current would exist in the circuit with a
singe resistor (R) that has a resistance equal to the
sum of the resistances of the individual resistors.
• Such a resistance is called the equivalent resistance
of the circuit. The equivalent resistance of resistors
in series equals the sum of the individual
resistances of the resistors.
Equivalent Resistance
R =R1 +R2 +...
for Resistors in Series
• Notice that the equivalent resistance is greater than
that of any individual resistor.
Series Circuits
A voltage divider is a
series circuit used to
produce a source of
potential difference that
is less than the potential
difference across the
battery.
Series Circuits
• Voltage dividers often are used with sensors, such as
photoresistors.
• The resistance of a photoresistor depends upon the
amount of light that strikes it and can be used in a
light meter.
• In this device, the
potential difference
is converted to a
measurement of
illuminance.
KNOWN
Series Circuits
Use with Example Problem 1.
Problem A
Suppose 15 V are applied across three
resistors in series (15.0 Ω, 22.0 Ω, and 47.0
Ω). Determine the current in the circuit and
the potential difference across the 47.0-Ω
resistor.
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw a circuit diagram.
• List the knowns and unknowns.
Vsource = 15 V
R2 = 22.0 Ω
I=?
R1 = 15.0 Ω
R3 = 47.0 Ω
V3 = ?
SOLVE FOR THE UNKNOWN
• Determine the equivalent resistance.
R  R1  R2  R3
 15.0   22.0   47.0   84.0 
•
Use the relationship among potential difference,
current, and total resistance to find the current
through the circuit and the potential difference
across R3.
I 
R1
I
R2
R3
UNKNOWN
Vsource
15 V

 0.18 A
R
84.0 
V3  I R3  0.18 A(47.0 )  8.5 V
EVALUATE THE ANSWER
• Current is in amperes and potential difference is in
volts, so the units are correct.
Series Circuits
KNOWN
Use with Example Problem 1.
Problem B
Suppose 15 V are applied across three
resistors in series (15.0 Ω, 22.0 Ω, and
47.0 Ω), as in the previous problem.
Demonstrate that the total power is equal to
the sum of the individual power amounts
used in the three resistors.
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw a circuit diagram.
• List the knowns and unknowns. I comes
from the previous problem.
UNKNOWN
Vsource = 15 V
R1 = 15.0 Ω
P=?
P2 = ?
I = 0.18 A
R2 = 22.0 Ω
P1 = ?
P3 = ?
R3 = 47.0 Ω
SOLVE FOR THE UNKNOWN
• Use the relationship among power, potential
difference, and current to find the power for the
whole circuit.
P  IV  0.18 A15.0 V  2.7 W
•
Use the relationship among power, current, and
resistance for each resistor.
P  I 2R
P1  0.18 A 15.0   0.49 W
2
R1
I
R2
R3
P2  0.18 A 22.0   0.71 W
2
P3  0.18 A 47.0   1.5 W
2
KNOWN
Series Circuits
UNKNOWN
Vsource = 15 V
R1 = 15.0 Ω
P=?
P2 = ?
I = 0.18 A
R2 = 22.0 Ω
P1 = ?
P3 = ?
Use with Example Problem 1.
R3 = 47.0 Ω
Problem B
Suppose 15 V are applied across three
resistors in series (15.0 Ω, 22.0 Ω, and 47.0
Ω), as in the previous problem. Demonstrate
that the total power is equal to the sum of
the individual power amounts used in the
three resistors.
SOLVE FOR THE UNKNOWN
• Use the relationship among power, potential
difference, and current to find the power for the
whole circuit.
P  IV  0.18 A15.0 V  2.7 W
•
Response
Use the relationship among power, current, and
resistance for each resistor.
SKETCH AND ANALYZE THE PROBLEM
• Draw a circuit diagram.
• List the knowns and unknowns. I comes
from the previous problem.
P1  0.49 W, P2  0.71 W, P3  1.5 W
Add the individual powers.
Ptotal  P1  P2  P3
R1
I
R2
R3
•
 0.49 W  0.71 W  1.5 W  2.7 W
EVALUATE THE ANSWER
• The total power equals the sum of the power for
the individual resistors, as expected.
KNOWN
Series Circuits
Use with Example Problem 2.
UNKNOWN
Vsource = 12.0 V
V1, before = ?
V2, before = ?
R1 = R2 = 1.5×106 Ω
V1, after = ?
V2, after = ?
RV = 1.0×107 Ω
SOLVE FOR THE UNKNOWN
Problem
• Before the voltmeter is connected, the potential
A voltage divider consisting of two 1.5-MΩ
difference across each resistor will be half the supply
resistors is connected to a 12.0-V source.
potential difference, or 6.0 V.
Determine the potential difference across each
• When the voltmeter is connected, it acts as a parallel
resistor before and after a voltmeter is
resistance:
connected, assuming the voltmeter’s
resistance is 1.0×107 Ω.
1
1
1


R
1.5106  1.0107 
Response
SKETCH AND ANALYZE THE PROBLEM
R  1.3106 
• Draw a circuit diagram. • Find the voltage drop across the parallel
• List the knowns and
combination.
unknowns.
 V

V  I R   source R
R  R 
 1

12.0 V1.3 M  5.6 V

 1.5 M  1.3 M
KNOWN
Series Circuits
UNKNOWN
Vsource = 12.0 V
V1, before = ?
V2, before = ?
R1 = R2 = 1.5×106 Ω
V1, after = ?
V2, after = ?
Use with Example Problem 2.
RV = 1.0×107 Ω
SOLVE FOR THE UNKNOWN
Problem
• Before the voltmeter is connected, the potential
A voltage divider consisting of two 1.5-MΩ
difference across each resistor will be half the supply
resistors is connected to a 12.0-V source.
potential difference, or 6.0 V.
Determine the potential difference across each
• When the voltmeter is connected, it acts as a parallel
resistor before and after a voltmeter is
resistance.
connected, assuming the voltmeter’s
resistance is 1.0×107 Ω.
R||  1.3106 
• Find the potential difference across the parallel
Response
combination.
SKETCH AND ANALYZE THE PROBLEM
V||  5.6 V
• Draw a circuit diagram.
• List the knowns and
• Use the loop rule to find the potential difference
unknowns.
across the first resistor.
V1  Vsource  V||  12.0 V  5.6 V  6.4 V
EVALUATE THE ANSWER
• Potential difference is in volts, so the units are
correct.
Parallel Circuits
Search about Parallel circuit and make a summary about it in a
creative way then upload your work in LMS
You should include:
1. The parallel definition
2. Current equation
3. Voltage equation
4. Equivalent resistance equation
5. A picture for a parallel circuit
Parallel Circuits
A circuit in which there are
several current paths is called a
parallel circuit.
Parallel Circuits
• 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.
Parallel Circuits
• 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.
• 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.
Ibranch =
V
Rbranch
Parallel Circuits
• The total current is the sum of the current in the
branches.
I = I1 + I 2 + …
• The total current is also equal to the potential
difference across the source divided by the equivalent
resistance (R).
V
I=
R
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
Parallel Circuits
• 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.
Real Life Application
• Usages of parallel and series 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.
I
UNKNOWN
Vbattery = 120.0 V
Ibattery = ?
RA = 50.0 Ω
IA = ?
RB = 40.0 Ω
IB = ?
RC = 30.0 Ω
IC = ?
RD = 20.0 Ω
ID = ?
SOLVE FOR THE UNKNOWN
• The potential difference is the same across all
four resistors.
• Use the relationship among potential
difference, current, and resistance.
I 
IA
IB
IC
ID
RA
RB
RC
RD
Req = ?
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 
Parallel Circuits
Use with Example Problem 3.
KNOWN
Problem
Vbattery = 120.0 V
Ibattery = ?
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
IA = 2.4 A
RB = 40.0 Ω
IB = 3.0 A
RC = 30.0 Ω
IC = 4.0 A
RD = 20.0 Ω
ID = 6.0 A
Req = ?
SOLVE FOR THE UNKNOWN
• Determine the equivalent resistance.
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw and label a circuit diagram.
• List the knowns and unknowns.
I
UNKNOWN
1
1
1
1
1




Req RA RB RC RD
IA
IB
IC
ID
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
KNOWN
Parallel Circuits
UNKNOWN
Vbattery = 120.0 V
Ibattery = ?
Use with Example Problem 3.
RA = 50.0 Ω
IA = 2.4 A
Problem
RB = 40.0 Ω
IB = 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
IC = 4.0 A
RD = 20.0 Ω
ID = 6.0 A
Response
SKETCH AND ANALYZE THE PROBLEM
• Draw and label a circuit diagram.
• List the knowns and unknowns.
I
SOLVE FOR THE UNKNOWN
• The current through the battery is equal to
the potential difference across the battery
divided by the equivalent resistance.
I battery 
IA
IB
IC
ID
RA
RB
RC
RD
Req = 7.79 Ω
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.
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.
• An analogy of this rule is walking
around a side of a hill in a loop.
Kirchhoff’s Rules
• 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 bottom figure.
Kirchhoff’s Rules
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.
Kirchhoff’s Rules
• 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.
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.
Kirchhoff’s Rules
• 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, I1 = I2 + I3
at junction A, and
I2 + I3 = I1 at junction B.
Quiz
1. Which type of circuit is shown?
A
series
B
resistant
C
parallel
D
nonsequential
CORRECT
Quiz
2. What is the formula for equivalent resistance for
this circuit?
1 1
+
R1 R2
A
R=
B
R = R1
C
R = R1 + R2
D
R = R2
CORRECT
Quiz
3. Which term describes this
circuit?
A
parallel circuit
B
voltage divider
CORRECT
C
nonfunctioning
D
incomplete
Quiz
4. What type of circuit is shown?
A
parallel
B
series
C
Kirchhoff’s
D
incomplete
CORRECT
Quiz
5. What is the total current entering junction B?
A
I2
B
I3
C
I2 + I3
D
1/(I2+ I3)
CORRECT