Electronics Series Resistive Circuits 1 Copyright © Texas Education Agency, 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 Copyright © Texas Education Agency, All rights reserved.

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 3 Copyright © Texas Education Agency, All rights reserved.

An Example Circuit 4 Copyright © Texas Education Agency, All rights reserved.

An Example Circuit 5 Wires Copyright © Texas Education Agency, All rights reserved.

An Example Circuit 6 Battery (DC voltage source) Copyright © Texas Education Agency, All rights reserved.

An Example Circuit 7 Fuse Copyright © Texas Education Agency, All rights reserved.

An Example Circuit 8 Switch Copyright © Texas Education Agency, All rights reserved.

An Example Circuit 9 Load (uses a resistor symbol) Copyright © Texas Education Agency, All rights reserved.

Typical Circuit Labels 10 VSVS F1F1 S1S1 R1R1 Copyright © Texas Education Agency, All rights reserved.

Open switch, no current Resistance is infinite Voltage is dropped across the switch Circuit Operation Copyright © Texas Education Agency, All rights reserved.

Circuit Operation Closed switch, current flows Current flows from negative to positive Amount of current determined by Ohm’s Law Copyright © Texas Education Agency, All rights reserved.

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 13 Copyright © Texas Education Agency, All rights reserved.

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 conductor Copyright © Texas Education Agency, 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 Copyright © Texas Education Agency, All rights reserved.

The Simplest Circuit This is a trivial example of a series circuit 16 VSVS R1R1 Copyright © Texas Education Agency, All rights reserved.

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 Copyright © Texas Education Agency, All rights reserved.

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 Copyright © Texas Education Agency, All rights reserved.

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 19 + VSVS R1R1 Copyright © Texas Education Agency, All rights reserved.

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 20 + VSVS R1R1 Copyright © Texas Education Agency, All rights reserved.

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 21 Using Kirchhoff’s Voltage Law or Copyright © Texas Education Agency, All rights reserved.

Calculate Current The only voltage in this circuit is the supply voltage, and it is dropped across R 1 R 1 is the only resistor so its value is the value of total resistance Use Ohm’s Law to calculate current: 22 VSVS R1R1 Copyright © Texas Education Agency, All rights reserved.

Here is a slightly more complicated circuit: Apply Kirchhoff’s Law by drawing a current loop with arrows Adding Circuit Components 23 VSVS R3R3 R2R2 R1R1 Copyright © Texas Education Agency, All rights reserved.

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 24 VSVS R3R3 R2R2 R1R1 Copyright © Texas Education Agency, All rights reserved.

Place Polarity on Components Current flows from negative to positive outside the battery 25 VSVS R3R3 R2R2 R1R1 + + Copyright © Texas Education Agency, All rights reserved.

Start with the arrow above the battery 26 Write Kirchhoff’s Voltage Law VSVS R3R3 R2R2 R1R V S – V R3 – V R2 – V R1 = 0 Copyright © Texas Education Agency, All rights reserved.

Rearrange the Formula Each resistor drops some of the source voltage; the three resistors together drop all of the source voltage. 27 VSVS R3R3 R2R2 R1R1 + + V S = V R1 + V R2 + V R3 Copyright © Texas Education Agency, All rights reserved.

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 28 Copyright © Texas Education Agency, All rights reserved.

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 29 R T = R 1 + R 2 + R 3 Copyright © Texas Education Agency, All rights reserved.

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 30 R T = R 1 + R 2 + R 3 Copyright © Texas Education Agency, All rights reserved.

Example Problem One Solve for each of the voltage drops in this circuit In other words, solve for V 1, V 2, and V 3 31 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Problem One Solution First, write the equation(s) that solve the problem: V 1 = I 1 R 1, V 2 = I 2 R 2, and V 3 = I 3 R 3 32 Copyright © Texas Education Agency, All rights reserved.

Problem One Solution First, write the equation(s) that solve the problem: V 1 = I 1 R 1, V 2 = I 2 R 2, and V 3 = I 3 R 3 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 33 Copyright © Texas Education Agency, All rights reserved.

Follow the Logic 34 Copyright © Texas Education Agency, All rights reserved.

Example Problem One R T = R 1 + R 2 + R 3 35 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Example Problem One R T = R 1 + R 2 + R 3 = 200 Ω Ω Ω 36 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Example Problem One R T = R 1 + R 2 + R 3 = 200 Ω Ω Ω R T = 600 Ω 37 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Example Problem One 38 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Example Problem One 39 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Example Problem One 40 V S = 12 V R 3 = 200 Ω R 2 = 200 Ω R 1 = 200 Ω Copyright © Texas Education Agency, All rights reserved.

Problem One Solution 41 Plug these into the first set of equations: V 1 = I 1 R 1 = 20 mA 200 Ω = 4 V V 2 = I 2 R 2 = 20 mA 200 Ω = 4 V V 3 = I 3 R 3 = 20 mA 200 Ω = 4 V In a series circuit, equal resistances drop equal amounts of voltage Copyright © Texas Education Agency, All rights reserved.

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 practice 42 Copyright © Texas Education Agency, All rights reserved.

Circuit Example Two 43 V S = 12 V R 1 = 200 Ω R 2 = 600 Ω Solve for the voltage drops across R 1 and R 2 Copyright © Texas Education Agency, All rights reserved.

44 Copyright © Texas Education Agency, All rights reserved.

Problem Two Solution I T = I 1 = I 2 = 15 mA 5. Now solve for voltage drops from step one V R1 = I 1 R 1 = 15 mA 200 Ω = 3 V V R2 = I 2 R 2 = 15 mA 600 Ω = 9 V Note that R 2 has three times the resistance of R 1 and that V R2 has three times the voltage of V R1. There is a rule for that; it’s called the voltage divider rule. 45 Copyright © Texas Education Agency, All rights reserved.

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 46 Copyright © Texas Education Agency, All rights reserved.

Example Three Let’s work an example from the very beginning 47 VSVS R1R1 R2R2 R3R3 R4R4 Copyright © Texas Education Agency, All rights reserved.

48 Draw the current loop Place arrows negative to positive VSVS R1R1 R2R2 R3R3 R4R4 Example Three Copyright © Texas Education Agency, All rights reserved.

49 VSVS R1R1 R2R2 R3R3 R4R V S – V 4 – V 3 – V 2 – V 1 = 0 V S = V 1 + V 2 + V 3 + V 4 Example Three Copyright © Texas Education Agency, All rights reserved.

Example Three V S = 9 V, R 1 = 180 Ω, R 2 = 330 Ω, R 3 = 470 Ω, R 4 = 150 Ω Solve for each voltage drop Go through the step-by-step process 50 VSVS R1R1 R2R2 R3R3 R4R4 Copyright © Texas Education Agency, All rights reserved.

51 Copyright © Texas Education Agency, All rights reserved.

I T = I 1 = I 2 = I 3 = I 4 = 8 mA 5. Now solve for voltage drops from step one V R1 = I 1 R 1 = 8 mA 180 Ω = 1.4 V V R2 = I 2 R 2 = 8 mA 330 Ω = 2.64 V V R3 = I 3 R 3 = 8 mA 470 Ω = 3.76 V V R4 = I 4 R 4 = 8 mA 150 Ω = 1.2 V Note how the voltage drops are proportional to the resistance values. 52 Problem Three Solution Copyright © Texas Education Agency, All rights reserved.

Example Four Let’s try something a little different Solve for V S 53 VSVS R 3 = 2.5 kΩ R 2 = 3.5 kΩ I 2 = 2.5 mA R 1 = 2 kΩ Copyright © Texas Education Agency, All rights reserved.

1. Write the equation we need: V S = V T = I T R T 2. Look for what is needed to solve the equation: We need I T We need R T 3. Solve for I T I T = I 1 = I 2 = I 3 = 2.5 mA 4. Solve for R T R T = R 1 + R 2 + R 3 = 2 kΩ kΩ kΩ = 8 kΩ 54 Copyright © Texas Education Agency, All rights reserved.

Problem Four Solution 5. Plug these into the equation from step one: V T = I T R T = 2.5 mA 8 kΩ = 20 V Note: 1 mA = A, 1 kΩ = 1000 Ω 55 Copyright © Texas Education Agency, All rights reserved.

Example Five Now let’s try something a little harder Solve for V S 56 VSVS R 2 = 1.5 kΩ I 2 = 3 mA R 1 = 1.2 kΩ Copyright © Texas Education Agency, All rights reserved. V 3 = 6.9 V

1. Write the equation we need: V S = V T = I T R T 2. Look for what is needed to solve the equation: We need I T We need R T 3. Solve for I T I T = I 1 = I 2 = I 3 = 3 mA 4. Solve for R T R T = R 1 + R 2 + R 3 We don’t know R 3 so we need to solve for it 57 Copyright © Texas Education Agency, All rights reserved.

Problem Five Solution 58 Copyright © Texas Education Agency, All rights reserved.

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 59 Copyright © Texas Education Agency, All rights reserved.

What’s Next? Practice 60 Copyright © Texas Education Agency, All rights reserved.