Download A Series Circuit

Electronics
Series
Resistive
Circuits
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Copyright © Texas Education Agency, 2014. All rights reserved.
What is a Series Circuit?
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A series circuit is one of the simplest
electrical circuits.
Because of this, it is the type of circuit used
for an introduction to problem solving and
circuit analysis.
Problem solving is a technique developed
through practice.
It uses math to calculate electrical values.

This is called applied math (based on Ohm’s
Law).
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Pre-Requisites
You have to know some basics:
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What an electrical circuit is
Common electrical components and their schematic
symbols
 Resistors, batteries, ground
 Current, voltage, resistance
 Switches, fuses, wires
You should gain this knowledge by completing the
prior electronics lessons in this sequence
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An Example Circuit
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An Example Circuit
Wires
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An Example Circuit
Battery
(DC voltage
source)
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An Example Circuit
Fuse
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An Example Circuit
Switch
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An Example Circuit
Load
(uses a
resistor
symbol)
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Typical Circuit Labels
F1
S1
VS
R1
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Circuit Operation
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Open switch, no current
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Resistance is infinite
Voltage is dropped across the switch
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Circuit Operation
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Closed switch, current flows
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Current flows from negative to positive
Amount of current determined by Ohm’s Law
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Back to Circuit Analysis
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Circuit analysis requires use of some
fundamental electrical laws
Ohm’s Law
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You should know this law by now
There are three forms of Ohm’s Law
Kirchhoff’s Law
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There are two parts to Kirchhoff’s Law
You need to know both
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Kirchhoff’s Laws
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Voltage law: the sum of all voltages in a
closed loop is equal to zero
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The sum of the voltage “drops” equals the sum of
the voltage “sources”
All of the voltage is always used in a loop
Current law: the sum of the currents “into” a
node is equal to the sum of the currents
“leaving” the node
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The current into a conductor is the same as the
current out of the conductor
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Where do the laws apply?
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These rules always apply to every DC circuit
with a resistive load
For AC circuits and active loads, the rules
generally apply, but not always
AC circuits have changing voltage and current
Active loads do not have constant resistance
An active load is like a variable resistor
A load is any device that the circuit is designed
to deliver power to
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The Simplest Circuit
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This is a trivial example of a series circuit
VS
R1
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A Series Circuit
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A series circuit has only one path for current
flow
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For every point in the circuit, the current “in”
equals the current “out”
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There are no branches going to another circuit
Kirchhoff’s current law
This means, in a series circuit, the current
has the same value everywhere
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Current is constant everywhere
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Using Kirchhoff’s Voltage Law
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To do Kirchhoff’s Law right, you must have
correct polarities
Make current loops going from the negative
side of the battery to the positive side
Current goes from negative to positive
everywhere outside the battery
Use arrows to indicate the direction of current
flow
The arrows point from negative to positive
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Using Arrows
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Use arrows to indicate the direction of current
flow and the polarity of voltage
VS
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+
R1
The red line represents a closed loop and
shows the path for current flow
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Using Kirchhoff’s Voltage Law
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Use the polarity that the arrow points to
The top arrow points to the positive side of
the battery
VS
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+
R1
The bottom arrow points to the negative side
of the resistor
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Using Kirchhoff’s Voltage Law
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Kirchhoff’s voltage law states that the sum of
the voltages in a closed loop equals zero
+ 𝑉𝑠 - 𝑉𝑅1 = 0
or
𝑉𝑆 = 𝑉𝑅1
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The voltage used equals the supply voltage
This is actually a lot easier to use than it looks
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Calculate Current
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The only voltage in this circuit is the supply
voltage, and it is dropped across R1
R1 is the only resistor so its value is the value
of total resistance
Use Ohm’s Law to calculate current:
𝐼𝑇 = 𝐼𝑅1 =
𝑉𝑇
𝑅𝑇
=
𝑉𝑆
𝑅1
VS
R1
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Adding Circuit Components
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Here is a slightly more complicated circuit:
R1
VS
R2
R3
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Apply Kirchhoff’s Law by drawing a current
loop with arrows
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Draw the Current Loop
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Place arrows according to the direction of
current flow
R1
VS
R2
R3
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Current flows from the negative terminal of
the battery to the positive terminal
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Place Polarity on Components
+
R1
VS
R2
R3

+
Current flows from negative to positive
outside the battery
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Write Kirchhoff’s Voltage Law
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Start with the arrow above the battery
+
R1
VS
R2
R3 +
+ VS – VR3 – VR2 – VR1 = 0
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Rearrange the Formula
VS = VR1 + VR2 + VR3
+
R1
VS
R2
R3

+
Each resistor drops some of the source
voltage; the three resistors together drop all
of the source voltage.
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Partial Summary
A partial summary of what we have learned so far:
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Current is the same everywhere in a series
circuit
𝐼𝑇 = I1 = I2 = I3
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Voltage drops add to equal the source voltage
V𝑇 = V1 + V2 + V3
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We have formulas for voltage and current—
now we need a formula for resistance
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Solve for Resistance
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From Ohm’s Law, V = I R
Substitute into the series voltage formula:
I𝑇RT = I1R1 + I2 R2+ I3R3
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Current is the same everywhere so it divides
out:
RT = R1 + R2+ R3
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Resistance adds in a series circuit
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Series Circuit Tool Kit
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Here are the three equations for a series
circuit:
𝐼𝑇 = I1 = I2 = I3
V𝑇 = V1 + V2 + V3
RT = R1 + R2+ R3
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These three equations, plus Ohm’s Law, form
a “tool kit” to analyze series circuits
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Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
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Solve for each of the voltage drops in this
circuit
In other words, solve for V1, V2, and V3
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Problem One Solution
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First, write the equation(s) that solve the
problem:
V1 = I1R1, V2 = I2R2, and V3 = I3R3
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Problem One Solution
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First, write the equation(s) that solve the
problem:
V1 = I1R1, V2 = I2R2, and V3 = I3R3
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Second, look for what is needed to solve
those equations
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Sometimes the information needed is given
Sometimes, like in this case, it is not
To solve for voltage, we need to know current
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Follow the Logic
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Now we are looking to solve this equation:
𝐼=

𝑅
Write down what we know in this formula
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
𝑉
We know VT = 12 V
Pay attention to the subscript, to use VT we need
RT
Solving this will give us IT (which is also I1,
etc.)
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Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
RT = R1 + R2 + R3
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Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
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Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
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Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
𝐼𝑇 =
𝑉𝑇
𝑅𝑇
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Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
𝐼𝑇 =
𝑉𝑇
𝑅𝑇
= 12 𝑉
600 Ω
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Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem One
R1 = 200 Ω
VS =
12 V
R2= 200 Ω
R3= 200 Ω
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
𝐼𝑇 =
𝑉𝑇
𝑅𝑇
= 12 𝑉
600 Ω
= 0.02 A = 20 mA
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Problem One Solution
IT = I1 = I2 = I3 = 20 mA
Plug these into the first set of equations:
V1 = I1R1 = 20 mA • 200 Ω = 4 V
V2 = I2R2 = 20 mA • 200 Ω = 4 V
V3 = I3R3 = 20 mA • 200 Ω = 4 V
In a series circuit, equal resistances drop
equal amounts of voltage
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How to Develop a Skill
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This may seem hard, but the point is to have
a systematic, step-by-step method
Problem solving is not magic; it is a skill
developed by following a procedure
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A valuable skill
Problem solving and troubleshooting requires
a logical, systematic, and consistent step-bystep process
Any skill also requires practice
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Circuit Example Two
R1 = 200 Ω
VS =
12 V
R2= 600 Ω
Solve for the voltage drops across R1 and R2
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1. Write the equations for VR1 and VR2
VR1 = I1 • R1 VR2 = I2 • R2
2. To use these equations to solve for voltage
we need current. Write the equation for current
I=V R
3. Looking for known values in this equation,
we have VT, we need RT
RT = R1 + R2 = 200 Ω + 600 Ω = 800 Ω
4. Substitute this into the formula from step two
IT = V𝑇 RT = 12 𝑉 800 Ω = 0.015 A = 15 mA
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Problem Two Solution
IT = I1 = I2 = 15 mA
5. Now solve for voltage drops from step one
VR1 = I1 • R1 = 15 mA • 200 Ω = 3 V
VR2 = I2 • R2 = 15 mA • 600 Ω = 9 V
Note that R2 has three times the resistance of
R1 and that VR2 has three times the voltage of
VR1. There is a rule for that; it’s called the
voltage divider rule.
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The Voltage Divider Rule
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The ratio of the voltages equals the ratio
of the resistances in a series circuit
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This rule is true because the current is the same
everywhere in a series circuit
This rule is typically expressed as a formula:

𝑉1
𝑉2

=
𝑅1
𝑅2
This formula applies to any ratio of voltage and
resistance in a series circuit as long as the equivalent
values are used properly
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Example Three
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Let’s work an example from the very
beginning
R1
R2
VS
R3
R4
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Example Three
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Draw the current loop
R1
R2
VS
R3
R4
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Place arrows negative to positive
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Example Three
+ VS – V4 – V3 – V2 – V1 = 0
+
R1
R2
VS
+R
+
3
R4
+
VS = V 1 + V 2 + V 3 + V 4
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Example Three
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VS = 9 V, R1 = 180 Ω, R2 = 330 Ω, R3 = 470 Ω, R4 = 150 Ω
Solve for each voltage drop
R1
R2
VS
R3
R4
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Go through the step-by-step process
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1. Write the equations we need:
VR1 = I1 • R1 ; VR2 = I2 • R2
VR3 = I3 • R3 ; VR4 = I4 • R4
2. Write the equation for current:
I=V R
3. We have VT, we need RT
RT = R1 + R2 + R3 + R4
= 180 Ω + 330 Ω + 470 Ω + 150 Ω = 1130 Ω
4. Substitute this into the formula from step two
IT = V𝑇 RT = 9 𝑉 1130 Ω = 0.008 A = 8 mA
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Problem Three Solution
IT = I1 = I2 = I3 = I4 = 8 mA
5. Now solve for voltage drops from step one
VR1 = I1 • R1 = 8 mA • 180 Ω = 1.4 V
VR2 = I2 • R2 = 8 mA • 330 Ω = 2.64 V
VR3 = I3 • R3 = 8 mA • 470 Ω = 3.76 V
VR4 = I4 • R4 = 8 mA • 150 Ω = 1.2 V
Note how the voltage drops are proportional to
the resistance values.
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Example Four
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Let’s try something a little different
R1 = 2 kΩ
R2= 3.5 kΩ
I2 = 2.5 mA
VS
R3= 2.5 kΩ
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Solve for VS
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1. Write the equation we need:
VS = VT = IT • RT
2. Look for what is needed to solve the equation:
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We need IT
We need RT
3. Solve for IT
IT = I1 = I2 = I3 = 2.5 mA
4. Solve for RT
RT = R1 + R2 + R3
= 2 kΩ + 3.5 kΩ + 2.5 kΩ = 8 kΩ
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Problem Four Solution
5. Plug these into the equation from step one:
VT = IT • RT = 2.5 mA • 8 kΩ = 20 V
Note: 1 mA = 0.001 A, 1 kΩ = 1000 Ω
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Example Five
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Now let’s try something a little harder
R1 = 1.2 kΩ
R2= 1.5 kΩ
I2 = 3 mA
VS
V3 = 6.9 V
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Solve for VS
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1. Write the equation we need:
VS = VT = IT • RT
2. Look for what is needed to solve the equation:
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We need IT
We need RT
3. Solve for IT
IT = I1 = I2 = I3 = 3 mA
4. Solve for RT
RT = R1 + R2 + R3
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We don’t know R3 so we need to solve for it
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Problem Five Solution
5. Write the formula for R3:
R3 = V3 I3 = 6.9 v 3 mA = 2.3 kΩ
6. Plug this into the equation from step four:
RT = R1 + R2 + R3 = 1.2 kΩ + 1.5 kΩ + 2.3 kΩ
= 5 kΩ
7. Plug this into the equation from step one:
VT = IT • RT = 3 mA • 5 kΩ = 15 V
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Summary
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There is a logical, step-by-step process to
solve a problem
Know the equations that form the tool kit for
series circuit analysis
I=V R
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What’s Next?
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Practice
Practice
Practice
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