6/24 Discussion notes prepared by Jonny Trampe
*Reminder that there will be discussion after the exam on Wednesday, July 2 nd .
- The exam material covers chapters 15-18.
Conceptual Problems
1. (Clarence borrowed it from 2005 Practice Exam, #17)
Ten coulombs of charge start from the negative terminal of a battery, flow through the battery and then leave the positive terminal through a wire, flow through a resistor and then return to the starting point on this closed circuit. In the complete process, the ten coulombs: a. have no net change in potential energy b. receive heat energy from the resistor c. have a net loss of potential energy d. do positive work on the battery e. are dissipated in the resistor, causing fewer than 10 coulombs to return to the battery.
Answer=a. because for any charge leaving and coming back to the same point has the same potential energy even if it goes through a resistor.
2. (Clarence borrowed it from 2004 Practice Exam, #16)
Which of the following is false?
a. the electrons in a wire carrying an electric current normally move very slowly (<1 m/s) b. A battery supplies the same current to any object to which it is connected c. The electric current leaving the positive terminal of a battery is the same as the current returning to the negative terminal d. A conductor in which an electric current is flowing has an electric field in its interior e. The electric current flows in a direction opposite to the motion of the electrons.
Answer=b. because of Ohm’s Law, I=V/R. If the resistance is different, then the current will change.
Answer a. is true due to drift speed. Answer c. is true due to the conservation of charge. Answer d. is true because F=qE and the battery sets up the electric field, which is why charges move.
Answer e. is true because electrons carry the charge in a circuit and move the opposite way as the electric field. However, the current flows from the positive to negative direction.
3. (Michelle Lalich borrowed it from Essentials of College Physics, pg. 461 #11)
Two wires A and B with circular cross sections are made of the same metal and have equal lengths, but the resistance of wire A is three times greater than that of wire B. What is the ratio of their cross-sectional areas? How do the radii compare?
R= ρ l/A. π r ² = cross sectional area
Ra=3Rb

Ra= ρ l/Ab Rb= ρ l/3Ab
4. (Michelle Lalich borrowed it from past exam 2005)
Two capacitors with CA greater than CB and are connected in a series with a battery. Which of the following is true?
a. There is more charge stored on CA.
b. There is more charge stored on CB.
c. There is the same charge stored on each capacitor.
d. There is the same potential difference across both capacitors.
e. There is a larger potential difference across the capacitor with CA.
Ctot.=(1/Ca) + (1/Cb)
Qtot.=Q1=Q2
Answer=c. because due to being connected to a battery and being in a series, there’s the same electric potential and the same charge in each capacitor.
Answer a. and b. are false because the charge is the same on each capacitor because it is in a series; therefore, Q1=Q2. Being connected in a series means that charges will redistribute themselves until there’s no electric difference between the two capacitors. Answers d. and e. are false because the same charge is stored on each capacitor. C=Q/V, so if C stays the same, then Q and V have to change accordingly.
Quantitative Problems
1. (Aaron Wessell, borrowed it from Essentials of College Physics p. 462, #25)
(a) A 34.5-m length of copper wire at 20.0 ºC has a radius of 0.25 mm. If a potential difference of
9.0-V is applied across the length of the wire, determine the current in the wire.
(b) If the wire is heated to 30.0 ºC while the 9.0-V potential difference is maintained, what is the resulting current in the wire?
(The resistivity of copper is 1.7x10-8 ohms x m; the temperature coefficient of resistivity is
3.9x10-3 ºC-1) l=34.5m T=20 ºC r=0.25 x 10-3m V=9.0 V ρ copper= 1.7x10-8 ohmsxm I=?
A= π r ² = π (.25x10-3) ² =1.96x10-7m ²
R= ρ (1/A)=(1.7x10-8)(34.5/1.96x10-7)=2.98 ohms
Δ V=IR, so I= Δ V/R
(a) I=(9.0 V)/(2.98 ohms)= 3.01 V/ohms or 3.01 A
(b) R=Ro + α Ro (T-To)
R= (2.98 ohms) + (3.9x10-3) (2.98)(30 ºC – 20 ºC)
R=3.096
*An increase in temperature=an increase in resistivity=less current flow*

2. (Alana borrowed it from physics 104 practice exam from 2005)
An 8.00 ohm resistor is dissipating 100 watts. What are the current through it, and the difference of potential across it?
a. 3.54 A, 28.3 V b. 3.54 A, 12.5 V c. 12.5 A, 28.3 V d. 28.3 A, 3.54 V e. 12.5 V, 28.3 A
Answer = a.
-Current through it -Difference of potential across it
Power=I ² R Power=VI
100=I ² (8) 100=V(3.54)
I= 3.54 A V=28.3 V