AP and Honors: Circuits with Bulbs

I have uploaded a solution for a problem essentially identical to the one I gave honors as their circuits lab and also similar to the questions in the AP circuits lab. Honors students please note that here I have used bulbs A and B as 3 W and C as 6 W. These are all the rated power when there is 12 V across the bulb. If the voltage across bulb A, VA, is less than 12 the power used by bulb A will be reduced.

Many of you in AP and Honors still do not have a clear understanding of the fact that current is caused by a voltage difference (or voltage drop) across an element. If you remove that element its current no longer exists. Sometimes some current will increase elsewhere in the circuit, but this is because voltage across that part of the circuit has increased. Batteries produce nearly fixed voltages. The current flowing to and from the battery varies inversely with the circuit resistance. Removing part of a parallel branch removes a number of electrons that were subject to fields in that part of the circuit and thus reduces the current in the circuit. Another way of saying this is removing a parallel element increase the circuit resistance (decreases the circuit conductance).


Here is the solution set

BulbCircuitSolns.PDF