Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it.

Transcription

1 Learning Goal: To understand the relationships between voltage, current, resistance, and power for a simple circuit containing one resistor and one battery. For this tutorial, use the PhET simulation Circuit Construction Kit (DC Only). This simulation allows you to build circuits using wire, resistors, batteries, and other circuit components. The voltage across any two locations on the circuit can be measured using a voltmeter, and the current can be measured using an ammeter. In this tutorial, only direct current (DC) circuits are considered. Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it. You should see a variety of circuit components (named and pictured) near the right edge of the blue panel. You can click and drag any of these components into the blue panel and construct a circuit. The components can be connected to each other by overlapping the red circles (the junction becomes a brown circle). Each component can be rotated by dragging it by the red circle, and the wire can also be lengthened or shortened this way. To disconnect two components, right-click (or control-click) on the brown circle of the junction and select Split Junction in the small menu that appears. To change the resistance of a resistor or the emf of the battery, right-click (or control-click) the component, select the appropriate choice in the pop-up menu, and make the adjustment using the pop-up adjustment panel. The two tools you will use in this tutorial are the voltmeter and the noncontact ammeter. The voltmeter gives you the voltage (potential difference) between the two locations of the probes. Simply drag the red and black probes and place the tips at any two locations on the circuit. The output of the voltmeter is the potential of the red probe minus the potential of the black probe. The noncontact ammeter allows you to measure the current simply by dragging the transparent circle (with cross hairs) over a wire. Feel free to play around with the simulation. When you are done, click Reset All before beginning Part A. Part A Drag a battery into the construction panel, and use the voltmeter to determine which end of the battery is the positive terminal. The positive terminal has a higher potential than the negative terminal (recall that the voltmeter measures the potential difference between the red probe and the black probe). Which end of the battery is the positive terminal? a. the grey end b. the black end

2 Part B Construct a circuit containing one battery, one resistor, and wire to close the circuit. The order and orientation doesn t matter, but it should look something like the figure below. You can show the values of the components by right-clicking (control-clicking) on each component and selecting Show Value in the pop-up menus. Use the default values of the battery and resistor. You should see the blue electrons flowing through the circuit. In what direction is the current flowing through the circuit? Recall that current is the flow of positive charge. Part C a. The current flows from the positive terminal, through the wires and resistor, and into the negative terminal. b. The current flows from the negative terminal, through the wires and resistor, and into the positive terminal Use the noncontact ammeter to measure the current flowing through the circuit. What is the current? Part D a. 9 A b. 10 A c. 0.9 A d. 90 A For the circuit in the previous part, the current flowing in the wire between the positive terminal of the battery and the resistor is the current flowing between the resistor and the negative terminal of the battery. a. Equal to b. Greater than c. Less than Part E Double the resistance of the resistor by changing it from 10 Ω to 20 Ω. What happens to the current flowing through the circuit? a. The current does not change. b. The current decreases by a factor of four. c. The current decreases by a factor of two. d. The current increases by a factor of two.

3 Part F For the circuit containing one resistor and one battery, what happens to the current if the voltage is doubled? a. The current does not change. b. The current increases by a factor of two. c. The current increases by a factor of four. d. The current decreases by a factor of two. Part G For the circuit containing one resistor and one battery, what happens to the current if the voltage is tripled and the resistance is doubled? a. The current increases b. The current does not change c. The current decreases. Part H What is the voltage reading across the resistor, as shown in the figure below? Part I a. 9.0 V b V c. 4.5 V d V A light bulb is basically a resistor that gets so hot that it glows, emitting light. For this tutorial we will assume the resistor in the light bulb is ohmic (that means Ohm s law applies to the resistor). The rate of energy emitted by the light bulb is its output power, commonly referred to as luminosity (brighter means more luminous). Hook up a light bulb to a 5-V battery. Right-click (or control-click) on the light bulb, and change its resistance. How does the brightness of the light bulb depend on its resistance? a. The brightness of the light bulb is independent of its resistance. It only depends on the voltage of the battery. b. The light bulb gets brighter as the resistance is increased. c. The light bulb gets dimmer as the resistance is increased.

4 Part J In Part H, you discovered that the luminosity of a light bulb increases if the current increases. The rate at which electric potential energy is converted into heat depends on the current flowing through the bulb and the voltage across the bulb. This energy is supplied by the battery. Mathematically, the luminosity P of the light bulb is given by P=ΔVI, where ΔV is the voltage across the bulb and I is the current. What happens to the luminosity of the light bulb if the voltage of the battery is doubled? (Note that the PhET simulation does not display a numerical value for the luminosity, so you should use the relationship between the luminosity, the voltage across the bulb, and the current.). a. The luminosity doubles. b. The luminosity increases by a factor of eight. c. The luminosity increases by a factor of four. d. The luminosity does not change. Part A PART 2 Let s start with a simple circuit that you will have seen in the Ohm s Law and Power PhET tutorial, if you completed that one. What is the voltage reading across the resistor, as shown in the figure below? Part B a. 4.5 V b V c. 9.0 V d V Construct a circuit containing two 10-Ω resistors in series, as shown below.

5 Place the crosshairs of the noncontact ammeter just before each resistor to measure the current through the top and bottom resistors. How do these currents compare? Part C a. The currents are the same. b. There s more current through the top resistor than the bottom resistor. c. There s less current through the top resistor than the bottom resistor. The current through each resistor in the two-resistor circuit is the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is the voltage across the resistor in the one-resistor circuit. a. half / half. b. half / twice. c. twice / twice. d. twice / half. e. half / the same as. f. the same as / half. Part D For the series circuit in the previous part, change the resistance of the bottom resistor to 20 Ω. What is the voltage across this 20-Ω resistor? For the series circuit in the previous part, change the resistance of the bottom resistor to 20. What is the voltage across this 20- resistor? a. 6.0 V b. 3.0 V c. 4.5 V Part E What must the resistance be of a single resistor connected to a 9-V battery for the current coming out of the battery to be the same as that of the circuit in Part D? a. 7 Ω b. 5 Ω c. 15 Ω d. 20 Ω

6 Part F Construct a circuit having two resistors connected in parallel, as shown in the figure below. Change the value of the battery emf to 10.0 V, and make sure the middle resistor is set to 20 Ω. Use extra wire for each leg of the circuit so you can measure the current through all legs with the ammeter. How does the current coming out of the battery change when the switch is closed? Part G a. The current decreases. b. The current increases. c. The current does not change. With the switch closed, how does the voltage across the 20-Ω resistor compare to the voltage across the 10-Ω resistor? a. The voltage across the 20-Ω resistor is less than that across the 10-Ω resistor. b. The voltage across the 20-Ω resistor is equal to the voltage across the 10-Ω resistor. c. The voltage across the 20-Ω resistor is greater than that across the 10-Ω resistor. Part H For the parallel circuit in the previous part (with the switch closed), the current through the 20-Ω resistor is the current through 10-Ω resistor. a. greater than. b. less than. c. equal to. Part I For the circuit in the previous part, what happens to the current flowing through the resistor on the right when the switch is closed (allowing current to flow through the resistor on the left)? a. The current flowing through the resistor on the right does not change. b. The current flowing through the resistor on the right decreases. c. The current flowing through the resistor on the right increases.

7 Part J With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be the same as when the switch is closed (and the resistances of the two resistors are 20 Ω and 10 Ω)? a. 7 Ω b. 5 Ω c. 30 Ω d. 15 Ω Part K The brightness of a light bulb having a specific resistance increases if the current through it increases. For the circuit shown in the figure below, how does the brightness of the light bulb change when the resistance of the 10-Ω resistor is decreased? a. The brightness decreases. b. The brightness does not change. c. The brightness increases.