Travis Lick
Homework #4: Electric circuits
1A: When the switch is open, bulbs A and C create a series circuit. In class, during our
experimental setups, bulbs connected in a series like this produced the same brightness.
Because the switch is open, bulb B will bypassed and therefore won’t light at all.
1B: When the switch is closed, bulbs A and B are in parallel so they will be the same
brightness as each other. Bulb C will be the brightest because it receives the full output
of the battery while A and B share the output.
2: There are four possible scenarios for this problem.
a.
b.
c.
d.
Both switches open
Both switches closed
First switch open, second closed
First switch closed, second open
a- Only bulb A will light because it is the only bulb that lies on a closed circuit.
b- If both switches are closed, all bulbs will light but with varying degrees of
brightness. A > B = C > D = E = F
The more bulbs that are placed in series the dimmer they light due to the fact that
the increase in bulbs in a series, also increases resistance. As a group, we
discussed this during Experiment 3.5 of our inquiry.
c- Only bulb A will light because it is on the path of least resistance in the circuit so
the current will not approach bulbs D, E and F.
d- In this case, bulb A, B and C will all light. However, A > B = C, because B and
C are in a series circuit so they will be less bright than A which is solo.
3A: Because this is a parallel circuit setup, the other bulbs would be unaffected by the
lack of current going through one of the bulbs. There would however, be a change in the
output from the battery. The less bulbs in a series circuit, the less resistance and therefore
a greater output would be produced. So, in this case, the resistance would be lessened by
a bulb burning out, so the output from the battery would increase.
3B: In this setup, the bulbs would all go out if one were to burn out. The circuit would
be incomplete and none would light.
3C: A series circuit would be better for the ornamental light company, but not the
homeowner! A string of lights arranged in parallel would be better because it would
allow some of the bulbs to burn out and still light the remaining lights. In a series circuit,
if one goes out, they will all go out.
4A: A = D = E = F > B = C = G = H:
We observed in experiment 2.6 that a single bulb circuit and a two bulb parallel circuit
create the same bulb brightness when connected. Given this information, bulbs A, D, E,
and F will all have the same brightness. All of the previously mentioned bulbs will be
brighter than B, C, G and H because they are all included in series circuits that were
observed in experiment 2.6 to be less bright than both individual bulbs and bulbs attached
in parallel.
4B: F = A = D = E > G = H = B = C: This rationalization comes from part A. The
brightness of the bulbs is a result of the current that is flowing through it. So, according
to A, this equation should stand up.
4C: 3 > 4 > 1 > 2: Circuit 3 would have the greatest current going through the battery
because it has less resistance than 4 which is also a parallel circuit, but contains a series
circuit that would increase resistance, Circuit 1 is a single bulb circuit and as seen in
many of our experiments, would have less resistance than a two bulb series circuit and
would therefore have more current through it.
4D: 2 > 1 > 4 > 3
Circuit 2 is a series circuit that would have the most resistance. We observed that series
circuits have more resistance than parallel circuits and the more bulbs in a series circuit,
the greater the resistance, so that is why it is greater than circuit 1. Circuits 3 and 4 are
both parallel but because 4 has more bulbs, we saw that this decreases resistance and
increases current flow.
5A: C > A > B = D: Because bulb C is going to accept all of the current coming from
the battery before it splits towards the other bulbs, it will be the brightest. Once the
current leaves bulb C, it will split towards bulbs A, B and D. These bulbs will have the
same brightness as each other but because bulbs B and D are in series, they will receive
less current flow than bulb A.
5B: B > A > C = D: When the current leaves the battery it can choose to go in the
direction of either A or B. Because A and C are in series with each other