Question

Lab 2: Ohm's Law and Battery and Resistor in a Circuit Date

Part A Simulation 1:

Play with Simulation 'Ohm's Law'

1. The equation in this simulation is shown a little differently. Write the equatio below and divide both sides by \(I\), the current. Do you recognize the equation now?

2. What happens to \(I\) when \(V\) increases?

3. What happens to I when R increases?

4. Let Voltage \(=6.0 \mathrm{~V}\). How many \(1.5 \mathrm{~V}\) batteries do you need? Let Resistanc 750 ohms. Calculate the current. Does your answer match with the simulati answer? (mA means milliamp.......1000 \(\mathrm{mA}=1 \mathrm{~A}\) )

Part B Simulation 2:

Play with Simulation 'Battery-Resistor Circuit'

Control Panel: Check "show inside battery"

1. Explain the following according to the circuit diagram:

Voltage

Resistance

2. What must happen to the voltage and resistance for the circuit to get hot?

3. What happens to the current in this scenario?

4. What must happen to the voltage and resistance for the circuit to get cold?

5. What happens to the current in this scenario?

6. Why do electrons (blue dots) move? Draw a diagram of the battery, label the flow of electrons. The flow of current ( \(t\) ) is opposite; draw this and note if toward or away from \(+\) terminal of the battery.

7. What does the Ammeter (on the left) measure? How is this shown in the sim?

8. What role do the "green dots" in the resistor play in the sim? What do you think they represent? What does this tell you about the effect of resistors in a circuit?

9. Increase the resistance (# green dots). What affect does this have on temperature? WHY?

10. To make the circuit "cold", what do you need to do? WHY?

Pre-Lab (Use the computer simulations here to answer these questions and submit your answers.)

Questions:

1. Predict what will happen to the current in a circuit if the voltage is kept constant but the wire diameter is increased.

Answer 1

Part A:

1. The Equation shown is V = IR, dividing both side by I we get V/I = R. This is the Ohm's law.

2. I increases when V increases.

3. When R increases I decreases.

4.We'll need 4 1.5V battries to create 6V. The Current is 8mA when R = 750 ohm.

Part B:

1. Voltage is how quickly the electrons (or charge carriers) made to move from positive to negative terminal inside of the battery.

Resistance is the slowing down of electrons in a conductor by the fixed metallic ion's collision with the electron.

2. Increasing the voltage and decreasing the resistance both need to increase in order to make the circuit hot.

3. The metallic ions starts to collide with the electrons and this turns the electromotive force in the battery to the vibration of the ions, this increases the thermal state of the metal and it becomes hot. The current increases.

4. Decreasing Voltage and increasing resistance will make the circuit cold.

5. As voltage is low in this state so electrons enter the metal with a lot less energy, the ions again collide with the electrons and further slow them down, so the current becomes near zero.

6. The chemical energy in the battery turns into electrical energy, we call this electromotive force, for this force the blue dots - electrons move inside the battery.

7. The ammeter measures the flow of current. It is shown if the blue dots are pushing its wings, when the wings move clockwise, it gets registered as positive current.

8.The green dots block the path of the electrons and vibrate when a collision occurs. The green dots are fixed metallic ions in the conductor. More metallic ions blocking the path of electron flow will increase the resistance of the conductor.

9. Increasing the green dots absorbs the electromotive force from the electrons increasingly and this slows down the flow in the circuit and the temperature decreases.

10. The resistance needs to be as high as possible and the voltage needs to be as low as possible (near zero) in order to decrease the temperature.

Pre lab:

1. As resistance is inversely proportional to the square of the radius of a wire, so when radius doubles , resistance decreases by a factor of 1/2² = 1/4.