Electronics Series Resistiv e Circuits 1 Copyright © Texas Education Agency, 2014. All rights reserved.
Jan 01, 2016
Electronics
Series Resistive
Circuits
1
Copyright © Texas Education Agency, 2014. All rights reserved.
What is a Series Circuit?
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). 2
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Pre-Requisites
You have to know some basics: 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
5
Wires
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An Example Circuit
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Battery(DC voltagesource)
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An Example Circuit
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Fuse
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An Example Circuit
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Switch
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An Example Circuit
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Load(uses a resistorsymbol)
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Typical Circuit Labels
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VS
F1 S1
R1
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Open switch, no current Resistance is infinite Voltage is dropped across the switch
Circuit Operation
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Circuit Operation
Closed switch, current flows Current flows from negative to positive Amount of current determined by Ohm’s Law
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Back to Circuit Analysis
Circuit analysis requires use of some fundamental electrical laws
Ohm’s Law You should know this law by now There are three forms of Ohm’s Law
Kirchhoff’s Law There are two parts to Kirchhoff’s Law You need to know both
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Kirchhoff’s Laws
Voltage law: the sum of all voltages in a closed loop is equal to zero 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 The current into a conductor is the same as the
current out of the conductorCopyright © Texas Education Agency, 2014. All rights reserved.
Where do the laws apply?
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
This is a trivial example of a series circuit
VS R1
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A Series Circuit
A series circuit has only one path for current flow There are no branches going to another circuit
For every point in the circuit, the current “in” equals the current “out” Kirchhoff’s current law
This means, in a series circuit, the current has the same value everywhere Current is constant everywhere
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Using Kirchhoff’s Voltage Law
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
Use arrows to indicate the direction of current flow and the polarity of voltage
The red line represents a closed loop and shows the path for current flow
+VS R1
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Using Kirchhoff’s Voltage Law
Use the polarity that the arrow points to The top arrow points to the positive side of
the battery
The bottom arrow points to the negative side of the resistor
+VS R1
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Kirchhoff’s voltage law states that the sum of the voltages in a closed loop equals zero
The voltage used equals the supply voltage This is actually a lot easier to use than it looks
Using Kirchhoff’s Voltage Law
- = 0or
=
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Calculate Current
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:
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= = = VS
R1
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Here is a slightly more complicated circuit:
Apply Kirchhoff’s Law by drawing a current loop with arrows
Adding Circuit Components
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VS
R3
R2
R1
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Draw the Current Loop
Place arrows according to the direction of current flow
Current flows from the negative terminal of the battery to the positive terminal
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VS
R3
R2
R1
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Place Polarity on Components
Current flows from negative to positive outside the battery
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VS
R3
R2
R1+
+
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Start with the arrow above the battery
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Write Kirchhoff’s Voltage Law
VS
R3
R2
R1+
+
+ VS – VR3 – VR2 – VR1 = 0
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Rearrange the Formula
Each resistor drops some of the source voltage; the three resistors together drop all of the source voltage.
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VS
R3
R2
R1+
+
VS = VR1 + VR2 + VR3
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Partial Summary
A partial summary of what we have learned so far: Current is the same everywhere in a series
circuit
Voltage drops add to equal the source voltage
We have formulas for voltage and current— now we need a formula for resistance
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= I1 = I2 = I3V = V1 + V2 + V3
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Solve for Resistance
From Ohm’s Law, V = I R Substitute into the series voltage formula:
Current is the same everywhere so it divides out:
Resistance adds in a series circuit
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IRT = I1R1 + I2 R2+ I3R3RT = R1 + R2+ R3
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Series Circuit Tool Kit
Here are the three equations for a series circuit:
These three equations, plus Ohm’s Law, form a “tool kit” to analyze series circuits
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= I1 = I2 = I3V = V1 + V2 + V3RT = R1 + R2+ R3
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Example Problem One
Solve for each of the voltage drops in this circuit
In other words, solve for V1, V2, and V3
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VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Problem One Solution
First, write the equation(s) that solve the problem:
V1 = I1R1, V2 = I2R2, and V3 = I3R3
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Problem One Solution
First, write the equation(s) that solve the problem:
V1 = I1R1, V2 = I2R2, and V3 = I3R3
Second, look for what is needed to solve those equations 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
Now we are looking to solve this equation:
Write down what we know in this formula 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
RT = R1 + R2 + R3
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VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Example Problem One
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
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VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Example Problem One
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
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VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Example Problem One
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
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VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Example Problem One
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
= 39
VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Example Problem One
RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω
RT = 600 Ω
= = 0.02 A = 20 mA40
VS =12 V
R3= 200 Ω
R2= 200 Ω
R1 = 200 Ω
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Problem One Solution
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= I1 = I2 = I3 = 20 mAPlug these into the first set of equations:V1 = I1R1 = 20 mA • 200 Ω = 4 VV2 = I2R2 = 20 mA • 200 Ω = 4 VV3 = I3R3 = 20 mA • 200 Ω = 4 VIn a series circuit, equal resistances drop
equal amounts of voltage
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How to Develop a Skill
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 A valuable skill
Problem solving and troubleshooting requires a logical, systematic, and consistent step-by- step process
Any skill also requires practice42
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Circuit Example Two
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VS =12 V
R1 = 200 Ω
R2= 600 Ω
Solve for the voltage drops across R1 and R2
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1. Write the equations for VR1 and VR2VR1 = I1 • R1 VR2 = I2 • R2
2. To use these equations to solve for voltage we need current. Write the equation for currentI = 3. Looking for known values in this equation, we have VT, we need RT RT = R1 + R2 = 200 Ω + 600 Ω = 800 Ω4. Substitute this into the formula from step two
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IT = = 0.015 A = 15 mACopyright © Texas Education Agency, 2014. All rights reserved.
Problem Two Solution
IT = I1 = I2 = 15 mA5. Now solve for voltage drops from step oneVR1 = I1 • R1 = 15 mA • 200 Ω = 3 V VR2 = I2 • R2 = 15 mA • 600 Ω = 9 VNote 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
The ratio of the voltages equals the ratio of the resistances in a series circuit
This rule is true because the current is the same everywhere in a series circuit
This rule is typically expressed as a formula:
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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=
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Example Three
Let’s work an example from the very beginning
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VS
R1
R2
R3
R4
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Draw the current loop
Place arrows negative to positive
VS
R1
R2
R3
R4
Example Three
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VS
R1
R2
R3
R4
+
++
+
+ VS – V4 – V3 – V2 – V1 = 0
VS = V1 + V2 + V3 + V4
Example Three
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Example Three
VS = 9 V, R1 = 180 Ω, R2 = 330 Ω, R3 = 470 Ω, R4 = 150 Ω
Solve for each voltage drop
Go through the step-by-step process
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VS
R1
R2
R3
R4
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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 = 3. We have VT, we need RT RT = R1 + R2 + R3 + R4 = 180 Ω + 330 Ω + 470 Ω + 150 Ω = 1130 Ω4. Substitute this into the formula from step two
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IT = = 0.008 A = 8 mACopyright © Texas Education Agency, 2014. All rights reserved.
IT = I1 = I2 = I3 = I4 = 8 mA5. Now solve for voltage drops from step oneVR1 = I1 • R1 = 8 mA • 180 Ω = 1.4 V VR2 = I2 • R2 = 8 mA • 330 Ω = 2.64 VVR3 = 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. 52
Problem Three Solution
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Example Four
Let’s try something a little different
Solve for VS
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VS
R3= 2.5 kΩ
R2= 3.5 kΩI2 = 2.5 mA
R1 = 2 kΩ
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1. Write the equation we need:VS = VT = IT • RT
2. Look for what is needed to solve the equation: We need IT
We need RT
3. Solve for IT
IT = I1 = I2 = I3 = 2.5 mA4. Solve for RTRT = 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
Now let’s try something a little harder
Solve for VS
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VS
R2= 1.5 kΩI2 = 3 mA
R1 = 1.2 kΩ
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V3 = 6.9 V
1. Write the equation we need:VS = VT = IT • RT
2. Look for what is needed to solve the equation: We need IT
We need RT
3. Solve for IT
IT = I1 = I2 = I3 = 3 mA4. Solve for RTRT = R1 + R2 + R3
We don’t know R3 so we need to solve for it57
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Problem Five Solution
5. Write the formula for R3:R3 = = = 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 58
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Summary
There is a logical, step-by-step process to solve a problem
Know the equations that form the tool kit for series circuit analysis
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I =
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What’s Next?
Practice Practice Practice
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