Question: (12 points, suggested time 25 minutes)
Some students want to know what gets used up in an incandescent lightbulb when it is in series with a resistor:
current, energy, or both. They come up with the following two questions.
(1) In one second, do fewer electrons leave the bulb than enter the bulb?
(2) Does the electric potential energy of electrons change while inside the bulb?
The students have an adjustable power source, insulated wire, lightbulbs, resistors, switches, voltmeters, ammeters, and other standard lab equipment. Assume that the power supply and voltmeters are marked in 0.1 V increments and the ammeters are marked in 0.01 A increments.
(a) Describe an experimental procedure that could be used to answer questions (1) and (2) above. In your description, state the measurements you would make and how you would use the equipment to make them. Include a neat, labeled diagram of your setup.
(b)
i. Explain how data from the experiment you described can be used to answer question (1) above.
ii. Explain how data from the experiment you described can be used to answer question (2) above.
A lightbulb is nonohmic if its resistance changes as a function of current. Your setup from part (a) is to be used or modified to determine whether the lightbulb is nonohmic.
(c)
i. How, if at all, does the setup need to be modified?
ii. What additional data, if any, would need to be collected?
(d) How would you analyze the data to determine whether the bulb is nonohmic? Include a discussion of how the uncertainties in the voltmeters and ammeters would affect your argument for concluding whether the resistor is nonohmic.
▶️Answer/Explanation
Ans:
To answer questions 1 and 2, the students would use the voltameters and ammeters to measure the current and voltage of the circuit both before and after the lightbulb for 3 trials of 1 second. The student would record this information for calculations.
(b) i. The data from the experiment above can be used to answer question (i) by comparing the current in the circuit at the ammeter before and the ammeter after the light bulb. If a discrepancy between the two measurements in earn trial is found, fever electrons may have left the light bulb.
ii. The data from the experiment above can be used to answer question (2) by comparison the voltage measured by the voltmeters placed before and after the light bulb. If a discrepancy between the values from each voltmeter in a given trial is found, the electrical potential energy of the electrons may have changed which inside the bulb.
(c) i. The voltage of the adjustable power source must be varied for a given number of trials to create varying currents because current = \(\frac{voltage}{resistance}= I = \frac{V}{R}\)
ii. Additional data reavired would be a number of trials at given voltages from the adjustable power source. This would allow different data sets of voltage and current to be used to determinse the resistance of the bulb.
(d) To determine whether the bulb is nonohmic, I would determine the resistances of the lightbulb in each trial according to the formula \(I = \frac{\Delta V}{R}\) assuming that in series no current is lost in the lightbulb (or circuit for that matter). If the resistance of the lightbulb is found to vary, it may be assumed to be nonohmic assuming the uncertainties in the voltmeters and ammeters did not affect the resorts. Because the uncertainties exist, hoever, If the change in voltage is less than or eacal to. IV, The conclusion as to whether or not the lightbulb is nonohmic cannot be verified using the given ealipment.
Question
Three identical light bulbs are connected in the circuit shown above. The switch S is initially in the open position and then is closed at time t.
(a) Describe any changes that occur in the current through each bulb when the switch is closed. Justify your answers.
(b) Describe any changes that occur in the brightness of each bulb when the switch is closed. Justify your answers.
(c) When the switch is closed, does the power output of the battery increase, decrease, or remain the same? Justify your answer.
(d) When the switch is closed, the current in Bulb 1 changes. Explain why this change in current does not violate the law of conservation of charge.
▶️Answer/Explanation
Ans:
Part (a)
Initially, the circuit is just Bulbs 1 and 3 in series. When Bulb 2 is added, the voltage from the battery is unchanged. Yet the total resistance of the circuit decreases, because an additional parallel path is added. Therefore, by V = IR with constant V, the total current in the circuit increases. Bulb 1 takes the total current, so Bulb 1’s current increases. For Bulb 1 only, the resistance is a property of the bulb and thus doesn’t change. So by V = IR with constant R, Bulb 1 takes an increased voltage, too. Then by Kirchoff’s loop rule, an increase voltage across Bulb 1 means a decreased voltage across Bulb 3. And for Bulb 3 only, by V = IR with constant R, Bulb 3’s current also decreases. (Obviously Bulb 2’s current increases from nothing to something.)
Part (b)
All bulbs have an unchanging resistance. Brightness depends on power, which is \(I^{2}R\). With constant R, a bigger current means more brightness; a smaller current means less brightness. So Bulb 1 gets brighter and Bulb 3 gets dimmer.
Part (c)
For the whole circuit, use power = \(V^{2}/R\). The voltage of the battery is unchanged because it’s still the same battery. The resistance of the circuit decreases because of the extra parallel path. Decreasing the denominator increases the entire value of the equation, so power increases.
Part (d)
Conservation of charge in circuits is expressed in Kirchoff’s junction rule—the current entering a junction equals the current leaving the junction. At any given moment of time, the junction rule holds. Now, when the switch is closed, more current flows from the battery than before. That’s not a violation of charge conservation, because the materials in the battery contain way more charged particles than are ever flowing through the wires. After the switch is closed, more current flows into the junction right before the switch than before, but more current also flows out of that junction than before. Charge conservation doesn’t mean that the same current must always flow in a circuit, it just says that whatever charge does flow in a circuit must flow along the wires.