Page 79 - The Basics IA by ETron Circuit Labs

hole 16bhole 25c1K $ (Brown, Black, Red, Gold) hole 10ahole 16a1K $ (Brown, Black, Red, Gold) B1REDWIRESW1 = 10d & 20jBLACKred = hole 30e black = hole 20fSCAN HERE TO SEE MORE ONLINE0 MAHere’s What You’ll Need...RESISTORR1R2R3hole 25bhole 30b1K $ (Brown, Black, Red, Gold) 77LAB 39Step 3: ExperimentProcedure:Assemble the circuit and connect a 9-volt battery to the circuit and connect a voltmeter across each of the resistors. You should see approximately 3 volts across each resistor. If the resistors were exactly 1K Ohms each and the battery put out exactly 9 volts, 3 volts would be your exact measurement.Since the resistors are not exactly 1K Ohms and the battery is not exactly 9 volts, your readings will not be exactly 3 volts.The following case shows how Kirchhoff’s Voltage Law holds true when the values of the components are measured and known precisely. For instance, if your battery measures 8.7 volts and the resistors measure 990 Ohms, 1010 Ohms, and 1005 Ohms, the circuit’s total resistance is:990 + 1010 + 1005 = 3005 OhmsThis tells us that the 990 Ohm resistor is responsible for 990/3005 of the total resistance and 990/3005 of the total voltage drop. By Ohm’s Law, we know that:The voltage drop across the 990 Ohm resistor is:(8.7 Volts) x (990/3005) = 2.866 VoltsThe voltage drop across the 1010 Ohm resistor is:(8.7 Volts) x (1010/3005) = 2.924 VoltsThe voltage drop across the 1005 Ohm resistor is:(8.7 Volts) x (1005/3005) = 2.909 VoltsThe sum of these voltage drops is 8.699.This verifies Kirchhoff’s Voltage Law.“The sum of the voltage drops in a closed-loop is equal to the applied voltage.” Kirchhoff’s Voltage Law (KVL)Visit us online at www.ETronCircuit.com for more exciting projects