DC Circuits Lab

1. DC Circuits Lab ECE 002 Professor Ahmadi

2. Outline Basic Components of a Circuit Series Circuit Parallel Circuit Ohm’s Law Lab Overview

3. Basic Circuit Components 1.5 V 1.5V We represent real electrical components with symbols A Battery… …can be represented with this symbol …called a “DC voltage source” • A DC Voltage Source • Provides Power for our circuit • Battery or Lab ‘power supply’ is an example • DC voltage is supplied across the two terminals • Its voltage is VOLTS (V)

4. Basic Circuit Components A Light Bulb…or any ‘device’… RΩ • A Resistor • Represents any device that requires power to operate • Could be a light bulb, your computer, a toaster, etc. • Each device has a certain amount of ‘resistance’, R, in • the unit called: OHMS (Ω) We represent real electrical components with symbols …can be represented with this symbol …called a “resistor”

5. Basic Circuit Components • The Ground • Represents 0 volts • We use it as a ‘reference’ voltage…to measure other • voltages against it • The ‘Earth’ is at 0 volts, so we call this ground We represent real electrical components with symbols The Earth… …can be represented with this symbol …called the “ground” symbol

6. Building a Circuit… We wish to ‘power’ our flashlight’s light bulb… 1.5 V • We need a battery… • We need to attach the light bulb to the battery… • We use wires to connect the light bulb to the battery… Instead…let's represent the real components with their symbols

7. Building a Circuit… Replace the battery with a ‘DC Voltage Source’ symbol 1.5 V 1.5V creating a schematic • Replace the light bulb with a ‘Resistor’ symbol .5Ω • Mark the symbol’s values (V=, R=, etc.) Since this “node” is at GND (OV) this node must be 1.5Volts higher • Add the Ground reference 0V Instead…let's represent the real components with their symbols

8. Analyzing the Circuit…using Ohm’s Law Ohm’s Law (V=IR) ->Describes the relationship between the voltage (V), current (I), and resistance (R) in a circuit 1.5V • When we attach the resistor to the DC voltage source, current begins to flow • How much current will flow? .5Ω • Using Ohm’s Law, we can determine how much current is flowing through our circuit 0V

9. Analyzing the Circuit…using Ohm’s Law Use Ohm’s Law: V = I x R 1.5V = Ix .5 Ω Solve for I: I = 1.5V / .5 Ω = 3 Amps I = 3 Amps 1.5V • How much current will flow? .5Ω 0V So, 3 Amps will flow through the .5 Ohm resistor, when 1.5 Volts are across it

10. Resistors in Series 1.5V • Resistors connected by only 1 terminal, back-to-back, are considered to be in ‘series’ R1 = .5Ω • We can replace the two series resistors with 1 single resistor, we call Req • The value of Req is the SUM of R1 & R2: Req=R1+R2=.5 Ω + .5 Ω = 1Ω Req = 1Ω R2 = .5Ω 0V

11. Resistors in Series Use Ohm’s Law: V = I x Req 1.5V = Ix 1 Ω Solve for I: I = 1.5V / 1 Ω = 1.5 Amps I = 1.5 Amps 1.5V • Now we can find the current through the circuit using Ohm’s Law Req = 1Ω 0V The bigger the resistance in the circuit, the harder it is for current to flow

12. Resistors in Series I = 1.5 Amps 1.5V • Back to our original series circuit, with R1 and R2 • The current is the SAME through each resistor R1 = .5Ω • Current flows like water through the circuit, notice how the 1.5 Amp ‘stream of current’ flows through both resistors equally R2 = .5Ω • Ohm’s Law shows us voltage across each resistor: V(R1) = 1.5Amps x .5 Ω = .75V V(R2) = 1.5Amps x .5 Ω = .75V 0V

13. Resistors in Parallel 1.5V Req = .25 Ω • Resistors connected at 2 terminals, sharing the same node on each side, are considered to be in ‘parallel’ • Unlike before, we cannot just add them. We must add their inverses to find Req: R1 = .5Ω R2 = .5Ω 0V

14. Resistors in Parallel Use Ohm’s Law, we find the current through Req: V = I x Req 1.5V = Ix .25 Ω Solve for I: I = 1.5V / .25 Ω = 6 Amps I = 6 Amps 1.5V • This is the equivalent circuit Req = .25Ω 0V The small the resistance in the circuit, the easier it is for current to flow

15. Resistors in Parallel 1.5V • Back to our original series circuit, with R1 and R2 • The current is NOT the SAME through all parts of the circuit • Current flows like water through the circuit, notice how the 6 Amp ‘stream of current’ splits to flow into the two resistors R1 = .5Ω R2 = .5Ω • The Voltage across each resistor is equal when they are in parallel 0V

16. Resistors in Parallel I = 6 Amps I = 3 A I = 3 A 1.5V • The voltage is 1.5 V across each resistor • Ohm’s Law tells us the current through each: I(R1)= V / R = 1.5V / .5 Ω = 3A I(R2)= V / R = 1.5V / .5 Ω = 3A • The 6Amps of current has split down the two legs of our circuit • It split equally between the two legs, because the resistors have the same value R1 = .5Ω R2 = .5Ω 0V The current will split differently if the resistors are not equal…

17. In Summary… Ohm’s Law: V=IR Describes the relationship between the voltage (V), current (I), and resistance (R) in a circuit Current is equal through two resistors in series Voltage drops across each resistor Req = R1 + R2 + . . . Voltage is equal across two resistors in parallel Current splits through branches of parallel circuits 1/Req = 1/R1 + 1/R2

18. In Lab Today You will build series circuits Build parallel circuits Work with a breadboard Verify Ohm’s Law by measuring voltage using a multimeter And yes, there is HW!