https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html

Please find the following using the above link

Part 1: Single resistor Drag circuit elements from the left to create a circuit that looks like the one shown. It should include a battery, a switch, and a single light bulb. Notice the light bulb has two contact points. Each wire should connect to a different point on the light bulb. If the battery is on fire after closing the switch, you did something wrong! Drag the voltmeter and ammeter from the right side. The ammeter will show the current at any place you put the probe. The voltmeter measures the potential difference between the red and black probes. Close the switch at the top and enter the current and voltage measurements below. Use Ohm's Law (V = IxR) to calculate the resistance of the bulb and then calculate the power delivered to the resistor. 1. Current (1) = Amps 2. Voltage (V) = Volts 3. Resistance (R) = V/I = Ohms Voltage 0.000 V Current 0.000 A 4. Power = Watts You can click the bulb to check your value for the resistance. For a power, you can use P = IV, P = 12R, or P = V2R. You should get the same result all three ways.Part 2: Resistors in Series Add a second resistor to the circuit. To disconnect a wire, click the end and hit the delete key or the scissors on the screen. Both bulbs should have the same resistance that you found in part 1. Voltage Current 0.000 V 0.000 A Predict the total resistance. For a series circuit, the total resistance is R = R1 + R2 5. R = Now measure the current with the ammeter. You should find that the current is the same everywhere in the circuit. Use the voltmeter to find the potential difference across each individual resistor. Since they are the same resistance, the voltage should be the same across both. Then measure the voltage across both bulbs by placing the black probe to the left of the left bulb and the red probe to the right of the right bulb. Series Measurements: #6 - #8 Current, I AV across 1 bulb AV across both bulbs Check that your measurements match what we know about series circuits. . You should find that the voltage across both bulbs is the sum of voltages across each individual bulb: AV = V1 + V2. It should also be the same as the battery voltage. You can also find the total resistance of the circuit from R = V/I. If you use the voltage across both bulbs, you should get the total circuit resistance you predicted above. The brightness of the bulbs is shown by the length of the yellow lines. Two bulbs connected in series are dimmer than a single bulb connected to the same battery. The brightness depends on the power delivered to the light bulb. Find the power through a single resistor. Try it with the different equations and check that you get the same result. If you use an equation with voltage, make sure to use the voltage across that particular resistor, not the battery voltage.Part 3: Resistors in Parallel . . 1 2 Current 0.000 A 3 Voltage 0.000 V Now set up a circuit with two bulbs in parallel. They should both be the same resistance as all the other bulbs so far. Let's make some predictions. Calculate the total resistance of this circuit. For a parallel circuit, 1/R = 1/R1 + 1/Rz 10. Rtotal = Measure the voltage across each bulb and the current through each branch. The branches are numbered in the diagram. Parallel Measurements: #11-#14 11 12 13 V2 = V3 Check that your measurements match what we know about parallel circuits. From the junction rule, I1 = 12 + 13. . Since each resistor is the same, the current through each should also be the same. V2 and V3 are the same as the battery voltage. . Ohm's Law should work for each branch: V2 = Iz R2 Finally, you can find the total resistance of the circuit from Ohms Law: R = V/l3. This should match what you found above. For the bulbs in parallel, notice that they are the same brightness as the first circuit with one bulb. Find the power to one of the bulbs. Make sure you use the current and voltage for the bulb, not the whole circuit. 15. Power to one bulb =