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Question 2ab: 2015 AP Physics 1 free response

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• Is it necessary to add another resistor in the circuit in addition to the bulb, as drawn in the diagram, or can we have a circuit with just the bulb? If I've thought it out correctly, we should be able to end up with the same result regardless. Or is there a mistake in my understanding?
• Hello AThont,

Lets remove the resistor and see what happens. We now have a series circuit consisting of a battery, ammeter #1, bulb, and ammeter #2. Let's assume our power supply is adjusted correctly so as NOT to damage the light bulb. WRT the current, nothing changes. Since this is a series circuit the current is the same in both ammeters. Actually we can generalize this statement and state that "current is the same at every point in a series circuit."

For voltage you could:

1) disconnect the light bulb

2) measure the voltage of the power supply (will be whatever you set it to)

3) measure the voltage on the light bulb (will be zero)

4) connect the light bulb to the power supply and measure the voltage (will be whatever you set it to). The power supply and the bulb now have the same voltage.

We can now make a statement about the power supply and bulb forming a loop. The power supply is the “source” (like a pump) that increases the electrical potential energy and the bulb is a “sink.” It reduces the electric potential energy. For good measure describe how the law of conservation of energy fits into this picture.

And yes, you could continue for the remainder of the lab to determine if the light bulb was nonohmic. Just measure the voltage across and the current through the light bulb as the power supply is varied. Plot the results and make your determination...

Please take a few minutes to look up Kirchoff’s Current Law (KCL) and Kirchoff’s Voltage Law (KVL) as these are really what we are talking about.

Keep up the good work!

Regards,

APD