Supplementary problems
EC
503
Problem 5.3
Consider the four-way node shown on the right.
The total current flowing into the node is i. The
current in branch A is three times the current in
branch B. The current in branch C is two times the
current in branch B. Write algebraic expressions in
terms of i for the current in each branch.
B
Problem 5.4
What is the current through each of bulbs in the
circuit shown on the right? Explain your
reasoning.
1.2 A
.4 A
Problem 5.5
A. In the network above, rank the bulbs
according to brightness. Explain your
reasoning. (Note that you may not be
able to compare all the bulbs with each
gather. Do the ranking only for the
bulbs you can compare.)
B . How will the brightness of bulb/ 1 , 1/
change if we add a lengtbrof
nichrome wire in series with bulb 6?
Explain your reasoning.
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(jyc <tu.se o c_<-j f < t«T^
C. How will the brightness of bulbs 1, 4,
and 7 change if we add a length of
nichrome wire by connecting it between
points D and H? Explain your reasoning.
D. How will the brightness of bulbs 1, 2, 6, and 7 change if we connect a conductor
between points B and H? Explain your reasoning. j-i ^i/f </#)4-tz> & <^^~
aft 0->rnapt4- <spes ^rou^? /ve.cuf^ec^i/i,
Problem 5.6
The currents flowing out of a four-way node are i and 3//4. The currents flowing into the
node along the two other paths are in the ratio of 2 to 5. Write expressions in terms of i
for the currents flowing into the node.
J/g- <.
^
Copyright © John Wiley & Sons, Inc.
McDermott & P.E.G., U.Wash./ Physics by Inquiry
Physics homework week 4
Ken O'Rourke
5.1
The section of nichrome wire must be identical in resistance to the resistance of the bulbs.
The brightness of the bulbs are dependent on how much current is flowing through the
bulbs. If bulb A & C are the same brightness the same amount of current must be
flowing through them. Since both circuits are series circuits, that means that they have
identical currents flowing in each circuit. If the current is the same in both circuits, the
total resistance is the same in both circuits. Since both circuits have 2 components in
them, and one component in each circuit is identical in resistance, that means that the
other components in each circuit must be identical in resistance. Since all the bulbs are
identical in resistance, then the nichrome section of wire must also be identical in
resistance to the bulbs.
i=3B+B+2B=6B
A=3B
C=2B
I=A+B+C
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0^°^
fa
d
5.4
The current leaves the battery and splits unevenly into 2 parallel branches. One branch
has bulb B and the other branch has bulbs A & C. The current comes back together and
all of it flows through bulb D, As a result bulb D will be brightest, bulb B is second
brightest, and bulbs A & C are of equal brightness and the are the dimmest.
f Ok.5.5
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A.
)/y
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i
Brightest to lowest- 1, 7, (2 & 3 $4), (4 & 5)
There are 2 branches parallel to the battery. 1 branch contains bulbs 1,2,&3. The 2nd
branch contains bulbs 4,5,6&7. The total resistance in branch one is less than the total
resistance in branch^" The branches are identical except for the added resistor in branch
2, which must increase the resistance in branch 2. As a result buhVA will have all of the
current flowing through it from the branch that has the least resistance, thus the most
current. The next brightest bulb will be bulb 7. Bulb 7 has all of the slightly less current
than branch 1 flowing in branch 2 go through it. Bulbs 2 & 3 should be of equal
brightness but dimmer than bulbs 1 &J^~as bulbs 2 & 3 splits the current evenly between
them--Bulb 6_should be of equal brightness to bulbs 2 &3. The current does not split
evenly in the 2 branches in the parallel branch in series with bulb 7. The branch with
bulb 6 will be the^ame as in 2&3, the difference in overall current in the 2 branches only
because of the branch with bulbs 4 & 5. That branch will not allow as much current to
pass as in branch 6 as there are 2 bulbs in series instead of 1. This will cut the current in