Breadboards, Multimeters, and ResistorsYour MultimeterSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Breadboards, Multimeters, and ResistorsEGR1301Your Multimeterleadsprobespincer clips – good for workingwith Boe-Bot wiringYou will use the multimeter to understand and troubleshoot circuits, mostlymeasuring DC voltage, resistance and DC current. turn knob to what youwould like to measure(push these onto probes)Measure VinVin will be the same as your power supply voltage. For the case below, 4 AAbatteries are used resulting in approximately 6 V (5.79 V to be more exact).Vin = power supply voltage Vss = ground (negative side of battery)Switch to DC VoltsMeasure VddVdd will always be around 5V (it is 4.94 V here). A voltage regulator on the Board of Education reduces this voltage from Vin down to ~ 5V. The Boe-Bot operates on 5V DC. Vdd ~ 5V Vss = ground (negative side of battery)Build the Series Circuit Below470220the breadboard connectsthese resistors5VSelect Resistors Find the 220 and the 470 resistors from your Boe-Bot kit.Example: 470resistor:4 = yellow7 = violetAdd 1 zero to 47 to make 470, so 1 = brownNow, find the 220 resistor.So, 470 = yellow, violet, brownset multimeterto measure R ~ 470Check Resistance of ResistorsCompute the Voltage Drops Across the Two ResistorsOhms Law: V = I RR1 470  R2 220 Vdd 5 VReq R1 R2 Req 690Find the equivalent ResistanceIVddReqI 0.0072 AFind the current leaving VddVR1I R1 VR13.41 VFind the voltage across the 470  resistorVR2I R2VR21.59 VFind the voltage across the 220  resistorV = 3.41VV = 1.59V4702205V(Hint: compute the total current provided by the power source and then apply Ohm’s Law to each resistor)5V – 3.41V – 1.59V = 0Now, add the voltage rise of the power source (+5V) to the voltage drops across the resistors(negative numbers).SOLUTION:Your Multimeterleadsprobespincer clips – good for workingwith Boe-Bot wiringYou will use the multimeter to understand and troubleshoot circuits, mostlymeasuring DC voltage, resistance and DC current. turn knob to what youwould like to measure(push these onto probes)Use Multimeter to Measure Voltages Around Loop(1) From Vdd to Vss(2) Across the 470 resistor(3) Across the 220 resistorRemember . . . a RESISTOR is a voltage DROP and a POWER SOURCE is a voltage RISEV1 = _____V2 = _____V3 = _____V1V2V3Rises must balance drops!!!!V = 3.41VV = 1.59VCompare Measurements to Theory5V+-470220++-Pretty close!V = 3.41VV = 1.59V5V – 3.41V – 1.59V = 05V+-470220++-Kirchoff’s Voltage Law (KVL)Kirchoff’s Voltage Law says that the algebraicsum of voltages around any closed loop in a circuit is zero – we see that this is true for ourcircuit. It is also true for very complex circuits.Notice that the 5V is DIVIDED between the two resistors, with the larger voltage drop occurring across the larger resistor.Gustav Kirchoff (1824 – 1887) was a German physicist who made fundamental contributions to the understanding of electrical circuits and to the science of emission spectroscopy. He showed that when elements were heated to incandescence, they produce a characteristic signature allowing them to be identified. He wrote the laws for closed electric circuits in 1845 when he was a 21 year-old student.Photo: Library of CongressConnecting an LEDElectricity can only flow one way through an LED (or any diode). LED = Light Emitting DiodeDiagram from the Parallax Robotics bookBuilding an LED CircuitVdd = 5VLED5V+-470These circuit diagramsare equivalentthese holes are “connected”Vdd =LEDholes in this direction are not connectedDiagram from the Parallax Robotics bookReplace the 470 Resistor with the 10k ResistorWhat happens and Why??ANSWER: The smaller resistor (470) provides less resistance to current thanthe larger resistor (10k). Since more current passes through the smallerresistor, more current also passes through the LED making it brighter.What would happen if you forgot to put in a resistor? You would probably burn up your