Physics quiz question: ammeter reading after switch is closed

Physics 205B Quiz 5, spring semester 2018
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to two light bulbs, a resistor, and an ideal ammeter, and an open switch. When the switch is closed, the ammeter reading will:
(A) decrease.
(B) remain constant.
(C) increase.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (A)

When the switch is open, the 1.0 Ω light bulb will be dark as no current will pass through it.  The current in this circuit will start at the 6.0 V emf, pass through the ammeter, go through the 7.5 Ω resistor, and then through the 2.0 Ω resistor, and back to the 6.0 V emf.  

The equivalent resistance R_eq of this circuit is 7.5 Ω + 2.0 Ω = 9.5 Ω.  

The current I through this circuit is (6.0 V)/(9.5 Ω) = 0.63 A, which is the ammeter reading.

When the switch is closed, then the 1.0 Ω light bulb is in parallel with the 7.5 Ω resistor.

The equivalent resistance R_eq of this circuit is 2.0 Ω + ((1/1.0 Ω) + (1/7.5 Ω))^–1 = 2.88 Ω.

The current I_emf through the emf is (6.0 V)/(2.88 Ω) = 2.08 A.

Now let's figure out how much current goes through the ammeter when the switch is closed, as the 2.08 A that passes through the 6.0 V emf will split with some either going through the switch path or going through the ammeter path, as given by Kirchhoff's junction rule:

I_emf = I_switch + I_ammeter.  

Let's apply Kirchhoff's loop rule for the clockwise emf-ammeter-7.5 Ω-2.0 Ω round trip path:

voltage rises = voltage drops,

6.0 V = ∆V_{7.5 Ω} + ∆V_{2.0 Ω},

and then apply Ohm's law to the right-hand side terms:

6.0 V = I_ammeter·(7.5 Ω) + I_emf·(2.0 Ω),

and since we already know I_emf= 2.08 A, then:

6.0 Ω = I_ammeter·(7.5 Ω) + (2.08 A)·(2.0 Ω),

0.25 A = I_ammeter,

which means the ammeter reading will decrease from its previous reading of 0.63 A when the switch was still open.

Sections 30882, 30883
Exam code: quiz05z0m6
(A) : 22 students
(B) : 8 students
(C) : 4 students
(D) : 0 students

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.70

Physics quiz question: power dissipated by resistor in parallel circuit

Physics 205B Quiz 5, spring semester 2018
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to a resistor and two light bulbs. The electrical power used by the 4.0 Ω resistor is:
(A) 1.5 W.
(B) 9.0 W.
(C) 24 W.
(D) 96 W.

Correct answer (highlight to unhide): (B)

The basic equation for the power dissipated by the 4.0 Ω resistor is:

P_resistor = I_resistor·ΔV_resistor,

where the current through the resistor I_resistor is not equal to the current passing through the 6.0 V emf source (I_circuit = ΔV_resistor/R_eq), due to the junction rule. However, we do not need to find I_resistor, as we can appeal to Ohm's law:

I_resistor = ΔV_resistor/R_resistor,

such that we can substitute this into the basic power equation, and result in a "specialized" form of the power equation for this resistor:

P_resistor = I_resistor·ΔV_resistor,

P_resistor = (ΔV_resistor/R_resistor)·ΔV_resistor,

P_resistor = (ΔV_resistor)²/R_resistor.

To find the ΔV_resistor voltage used by the resistor, we apply the loop rule in the clockwise direction, starting from lower right-hand corner, through the 6.0 V emf source, then through the 4.0 Ω resistor before returning to the starting point in the lower right-hand corner (the loop rule can be applied to any round-trip loop in a circuit, even if there are other parts of this circuit):

"voltage supplied = voltage used,"

∆V_rises = ∆V_drops,

(6.0 V) = ΔV_resistor.

Then we can evaluate the "specialized" form for the power used by the resistor:

P_resistor = (ΔV_resistor)²/R_resistor,

P_resistor = (6.0 V)²/(4.0 Ω) = 9.0 W.

(This "specialized" equation for power should only be used if the voltage used by the circuit element is known already either from the loop rule (as was done here) or from Ohm's law.)

(Response (A) is the numerical value for the current flowing through the resistor; response (C) is ΔV_resistor·R_resistor; response (D) is ΔV_resistor)²·(R_resistor)².)

Sections 30882, 30883
Exam code: quiz05z0m6
(A) : 5 students
(B) : 19 students
(C) : 9 students
(D) : 1 student

Success level: 56%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.83

Physics quiz question: equivalent resistance of serio-parallel circuit

Physics 205B Quiz 5, spring semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 18.112

An ideal 12 V emf source is connected to three resistors, as shown at right. The equivalent resistance of this circuit 
is:
(A) 3.4 Ω.
(B) 4.7 Ω.
(C) 9.3 Ω.
(D) 14.0 Ω.

Correct answer: (C)

The 2.0 Ω resistor and the 4.0 Ω resistor are directly connected in parallel to each other, such that their equivalent resistance is:

R_2,4 = [ (R₂)^–1 + (R₄)^–1 ]^–1,

R_2,4 = [ (2.0 Ω)^–1 + (4.0 Ω)^–1 ]^–1,

R_2,4 = 1.333... Ω.

Then the 8.0 Ω resistor is in series with the combined R_2,4 equivalent resistor, so the final equivalent resistance of the circuit is:

R_eq = R₈ + R_2,4,

R_eq = 8.0 Ω + 1.333... Ω = 9.3 Ω, to the significant tenths decimal place.

(Response (A) is where the 2.0 Ω resistor and the 4.0 Ω resistor are first combined in series, and their resulting equivalent resistance R_2,4 combined in parallel with the 8.0 Ω resistor; response (B) is the average of all three resistance values; response (D) is the sum of all three resistance values.)

Section 30882
Exam code: quiz04eQu7
(A) : 1 student
(B) : 4 students
(C) : 25 students
(D) : 2 students

Success level: 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.38

Physics midterm question: comparing voltmeter readings

Physics 205B Midterm 2, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 9.0 V emf source is connected to a light bulb and two resistors, with a voltmeter connected to the light bulb, and another voltmeter connected to one of the resistors. Discuss why the two voltmeters have the same reading (in volts). Show your work and explain your reasoning using Kirchhoff's laws, Ohm's law, and properties of voltmeters.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates how the two voltmeters have the same reading because:
  1. due to Kirchhoff's junction rule, the current flowing through each of the 4.0 Ω resistors is one-half of the current flowing through the 2.0 Ω resistor; and
  2. since each voltmeter will read the voltage drop (I⋅R) of their respective resistors, the smaller current (factor of one-half) flowing through the 4.0 Ω resistor will be compensated by its larger resistance (factor of two).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying properties of Kirchhoff's rules, Ohm's law, and properties of ammeters.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at systematically applying Kirchhoff's rules, Ohm's law, and properties of ammeters.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 28 students
r: 4 students
t: 2 students
v: 8 students
x: 0 students
y: 0 students
z: 1 student

A sample "p" response (from student 1950):

Physics midterm question: ammeter reading after switch is closed

Physics 205B Midterm, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to an ammeter, a resistor, a light bulb, and an open switch. When the switch is closed, determine whether the ammeter reading (in amps) will decrease, increase, or remain the same, and explain why. 
Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and properties of ammeters.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates how ammeter reading will increase when the switch is closed because:
  1. when the switch is open, the equivalent resistance is 3.0 Ω, and the ammeter will read the current of this circuit I = ε_eq/R_eq = (6.0 V)/(3.0 Ω) = 2.0 A;
  2. when the switch is closed, no current will flow through the 0.5 Ω resistor (flowing only through the zero resistance path of the closed switch), such that the equivalent resistance decreases to 2.5 Ω, such that the ammeter will read a higher amount of current in this circuit I = ε_eq/R_eq = (6.0 V)/(2.5 Ω) = 2.4 A.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying properties of Kirchhoff's rules, Ohm's law, and properties of ammeters.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at systematically applying Kirchhoff's rules, Ohm's law, and properties of ammeters.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 20 students
r: 4 students
t: 10 students
v: 6 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1001):
Another sample "p" response (from student 1810):

Physics midterm problem: brightness of light bulbs in circuit

Physics 205B Midterm 2, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 9.0 V emf source is connected to several light bulbs that all have the same resistance. Calculate the powers dissipated (in watts) for each of these light bulbs. Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and properties of electrical power.

Solution and grading rubric:

• p:
  Correct. Solves for the powers dissipated by each light bulb by:
  1. finding equivalent resistance of the circuit by recognizing that the top light bulb is in series to the lower three parallel light bulbs);
  2. applying Ohm's law to determine the current of the equivalent circuit, which is the current flowing through the top light bulb;
  3. determines the power dissipated by the top light bulb;
  4. applies Kirchhoff's loop and/or junction rules to solve for the voltage difference used by and/or the current flowing through each of the lower three parallel light bulbs; and
  5. determines the power dissipated by each of the lower three parallel light bulbs.
• r:
  Nearly correct, but includes minor math errors. Typically incorrect calculation in (1) or in (5), but otherwise everything else is consistent with this error.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Multiple issues in (1)-(5), but still attempts to systematically analyze most of (1)-(5) even with wrong numerical values.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at applying Kirchhoff's rules, Ohm's law, and properties of electrical power.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying Kirchhoff's rules, Ohm's law, and properties of electrical power.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 10 students
r: 6 students
t: 7 students
v: 18 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 1982):
Another sample "p" response (from student 8812):

Physics quiz question: power dissipated by a resistor in series to parallel light bulbs

Physics 205B Quiz 5, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 9.0 V emf source is connected to a resistor, and two light bulbs that are each connected to ideal ammeters. The electrical power used by the 3.0 Ω resistor is:
(A) 2.6 W.
(B) 3.4 W.
(C) 21 W.
(D) 27 W.

Correct answer (highlight to unhide): (C)

The 0.5 Ω and the 2.5 Ω light bulbs are in parallel with each other, and are together in series with the 3.0 Ω resistor, such that their equivalent resistance is given by:

R_eq = (3.0 Ω) + ((0.5 Ω)^–1 + (2.5 Ω)^–1)^–1,

R_eq = (3.0 Ω) + (2.4 Ω^–1)^–1 = 3.4166666667 Ω.

Then Ohm's law is applied to the entire circuit, to find the current flowing through the entire circuit:

I_circuit = ε/R_eq,

I_circuit = (9.0 V)/(3.4166666667 Ω) = 2.6341463415 A.

So the power used by the 3.0 Ω resistor is then given by:

P = I²·R,

P = (2.6341463415 A)²·(3.0 Ω) = 20.8161808453 W,

or to two significant figures, the power used by the 3.0 Ω resistor is 21 W.

(Response (A) is the current flowing through the 3.0 Ω resistor; response (B) is the equivalent resistance of the circuit; response (D) is (9.0 V)²/(3.0 Ω), which would be the power used by the 3.0 Ω resistor if it were somehow able to use up all of the 9.0 V from the emf source that is supplied to the circuit as a whole, leaving nothing for the light bulbs.)

Sections 30882, 30883
Exam code: quiz05eXpL
(A) : 1 student
(B) : 5 students
(C) : 4 students
(D) : 29 students

Success level: 11% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33

Physics quiz archive: circuits (2)

Physics 205B Quiz 5, spring semester 2019
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz05eXpL



Sections 30882, 30883 results

0- 6 :
7-12 :	** [low = 9]
13-18 :	****************
19-24 :	***************** [mean = 20.2 +/- 4.7]
25-30 :	**** [high = 30]

Physics quiz archive: capacitors, circuits

Physics 205B Quiz 4, spring semester 2019
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz04KhhF



Sections 30882, 30883 results

0- 6 :
7-12 :	**** [low = 9]
13-18 :	*****
19-24 :	*************** [mean = 23.4 +/- 6.0]
25-30 :	***************** [high = 30]

Online reading assignment: advanced electricity (review)

Physics 205B, spring semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on re-reading textbook chapters and reviewing presentations on advanced electricity concepts.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"How to calculate equivalent resistance for parallel circuits and series."

"In series resistors will add up to attain the resistor equivalent and that resistors in parallel will be the inverse of the resistance added together and then inversed again."

"Why an ammeter has to have a resistance of zero, because it measures current and if it had a resistance it would interfere in that."

"Since we had lab, I have a much better understanding of how voltmeters and ammeters work and why they need to have the resistance they have. Since ammeters are measuring the current through a circuit, they need to have a very very low resistance so they don't disturb the circuit they are measuring. The voltmeter measures the voltage difference for a resistor like a bulb. They have to have a very high resistance so they don't create a short in the circuit."

"When the amount of resistors in parallel goes up the resistance gets dangerously low and this causes current to get very high at which point it can become dangerous because its heat can melt the material used to insulate wires, thus exposing wires to other surrounding surfaces and possibly sparking a fire if they come in contact with each other."

"Grounding is an important aspect of safety when it comes to electrical appliances such as clothes dryers. A three-pronged plug connects the metal casing of a dryer to a copper pipe in the ground so that if wiring becomes lose inside the dryer and touches the casing, a person who is also touching the casing will not be shocked. This is because the copper pipe would have much less resistance than the person's body."

"I understand the relationship between resistance and current in series and parallel circuits a lot better after reviewing for this quiz. I understand the measurements that ammeters and voltmeters make better after reviewing the presentation."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I'm still a little confused on the junction and loop rules."

"I'm still a little confused on why ammeters need to have a low resistance wile voltmeters want to have a high resistance. I'm also confused on the measurement process on a voltmeter."

"I do not quite understand joule heating."

"I'm still confused about power and how the amount of power used varies for resistors with different magnitudes in series or parallel circuit."

"I couldn't quite grasp the idea of the power dissipation. I guess I don't know what to use it for or how to use it..."

State the unit of electrical power, and give an equivalent definition in terms of other SI units.

"Watts, amps times volts."

"Watts, which is joules per second."

What are the resistances of these (ideal) devices?
(Only correct responses shown.)

Ideal light bulb: some finite value between 0 and ∞ [65%]
Burnt-out light bulb: ∞ [35%]
Ideal wire: 0 [58%]
Ideal (non-dead) battery: 0 [42%]
Real (non-dead) battery: some finite value between 0 and ∞ [61%]
Ideal switch, when open: ∞ [36%]
Ideal switch, when closed: 0 [42%]

Two light bulbs with different resistances r and R, where r < R, are connected in series with each other to an ideal emf source. Select the light bulb with the greater quantity.
(Only correct responses shown.)

More current flowing through it: (there is a tie) [46%]
Larger potential potential difference: light bulb R [45%]
More power used: light bulb R [52%]

Two light bulbs with different resistances r and R, where r < R, are connected in parallel with each other to an ideal emf source. Select the light bulb with the greater quantity.
(Only correct responses shown.)

More current flowing through it: light bulb r [55%]
Larger potential potential difference: (there is a tie) [23%]
More power used: light bulb r [36%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"Starting a fire from bubble gum wrapper was very interesting. I love it when the concepts we are learning are applied to survival skills."

"No questions today just hope class will clarify my confusion."

"I think you didn't post the blog that pertains to this reading assignment?" (Actually the presentations for this assignment were for review; and these are more advanced questions that build on top of what you've already seen in class and from the textbook, as opposed to being directly off of the presentation slides.)

"I like to get more examples about resistances problems like this in class and get to do a hands on problems."

"How does the internal resistance differ for an ideal versus real battery?" (Ideal batteries have zero internal resistance, they purely provide an emf (potential difference, voltage) without any other complications. A real battery does have an internal resistance, as the chemical reactions that release energy to produce its emf (voltage) also produce inert waste products that still sit inside the battery to impede current. A fresh new battery will have very little internal resistance (very little waste products), and an old battery will have some internal resistance (more waste products), while a dead battery will have a very high internal resistance (nearly all of it inside is waste products). You'll get to investigate this in a later lab.)

"You need to 'make it rain' more in terms of extra credit. You must increase the extra credit limit! This is because, among other things, the actual quizzes do not accurately reflect what is taught."

Online reading assignment: advanced electricity

Physics 205B, spring semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and reviewing presentations on circuit analysis and previewing presentations on advanced electricity concepts.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"An ammeter must have a resistance that is ideally 0 while a voltmeter must have a resistance that is ideally infinity. Joule heating is when the power used by the circuit's resistance due to the about of current flowing through it."

"Power measures the rate of electric potential energy usage. It can be calculated as current squared times resistance or current times potential. When resistors are connected in parallel, resistance decreases and current increases. This increase in current can overload the circuit."

"Voltmeters measure the amount of electrical potential used by an element in the circuit and ammeters measure the amount of current passing through an element of a circuit."

"Ammeters measure the current of any circuit element that is "broken" open by measuring the current after it travels through the element. The ideal resistance of an ammeter is zero. Voltmeters measure the amount of electric potential used by any circuit potential, and measures the current of both before and after the element. The ideal resistance of a voltmeter is infinite."

"Series wiring is when devices are connected so that the same current runs through both devices. Parallel wiring is when devices are connected so that the same voltage is applied across each device. The junction and loop rules set the parameters through which current and voltage can move through a circuit."

"If there is too much current going through something, that something will heat up and it can become dangerous and lead to a runaway current which can cause fires and other damages. I also understand how series resistors and parallel resistors divide/use the current in a circuit. Whatever potential a battery adds to a circuit, the resistors will use up the same amount. Circuit breakers are designed to trip a current before it becomes dangerously high turns to a runaway current."

"That runaway currents can be prevented by circuit breaker that interrupts the current if it becomes to high due to excessive appliances being plugged into the same outlet. Each appliance plugged in is considered a resistor, so when many appliances are plugged into the same outlet the resistance becomes dangerously low and the current becomes dangerously high. Outlet overload if dangerous because it causes the wires to heat up thus melting the outer coating that insulates the current flowing through the wire. When the coating is melted off it can make contact with other surfaces and create a spark which could cause a fire."

"Honestly not a lot. Parallel circuits and resistors. Circuit breakers."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I'm going to need more help plugging in the numbers for the different equations."

"The equations may need more examples. Need extra practice on equations."

"I understood junction rule but could use another run through on it. Just in case."

"Can you describe more about the concept of parallel resistors?"

"The section on circuits wired in series and parallel. I am still trying to understand where one starts, the series or parallel."

"I do not really understand how a voltmeter makes measurements. I understand the equations involving power, but am unsure of how to apply the equations. I think it will make sense after practicing with lots of examples."

"I do not really understand why ammeters need to have a low resistance and voltmeters want to have a high resistance. Also (this might be from what we covered earlier) I do not understand why the resistance will decrease when multiple things are plugged into a circuit. This applies to my house because it happens fairly often and my circuit breaker stops the current."

"I am not sure why the resistance for an ideal ammeter is equal to zero while the resistance of and ideal voltmeter should be very high."

"I think I am kinda fuzzy on some of the terminology and how each of the aspects are used. I get the basic concepts of what is going on in each part of the presentation, but if I have to describe it using the right terminology, I am pretty lost."

"Not too confused after this chapter."

"It wasn't too confusing, but I need to read through it once more to get the concept a bit better."

"All pretty good I think! Maybe?"

What are the resistances of these (ideal) devices?
(Only correct responses shown.)

Ideal ammeter: 0 [75%]
Ideal voltmeter: ∞ [69%]

Determine what will happen to the following parameters when additional electrical appliances are plugged in and turned on in the same household circuit.
(Only correct responses shown.)

Equivalent resistance R_eq of circuit: decreases [78%]
Current I flowing through emf source: increases [56%]

A fuse or circuit breaker is designed to prevent too much __________ in household wiring.

current.	********************** [22]
voltage.	*** [3]
(Both of the above choices.)	****** [6]
(Neither of the above choices.)	[0]
(Unsure/guessing/lost/help!)	* [1]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"This subject can be shocking."

"I really liked the 'Twinkle, twinkle, little star, Power equals I-squared R' nursery rhyme. I'll definitely remember this."

"This was a super-interesting lecture because my house burnt down and there were suspicions that it was because of a faulty circuit breaker! So now I understand how that could be majorly bad. I wonder how likely that is?"

"I guess the electrons could just be looked at as the current (or the amount of electrons flowing through a circuit is equal to the current)?" (Yes, but remember that the direction of current is actually the reverse of the actual direction of electrons, due to the arbitrary assignment of negative charge to electrons.)

"Why should a voltmeter have an ideal resistance of infinity?" (When current flows through a resistor, the current doesn't get "used up" (the same amount that goes in comes back out), but the potential/voltage/energy per charge gets "used up." So a voltmeter that measures how much voltage gets "used up" by a resistor is attached on either side of the resistor to compare how much voltage there is in the current before entering the resistor, to how much voltage there is in the current after exiting the resistor. You don't want current to actually flow through the voltmeter when taking these measurements (as you don't want it to affect the current flowing through the resistor that it's measuring), so it should ideally have a resistance of infinity in order to "block" current flowing through itself.)

"What is more dangerous volts/voltage or current? Are they the same?" (As for which is more dangerous, current is the actual (backwards) flow of electrons, and too much of that going through your body is bad. Voltage by itself is just the measure of how much energy per charge there is that current could potentially use to flow through you. A very crude analogy is that a bowling ball dropped on your foot is bad. A lot of bowling balls continuously dropped on your foot is very bad. (This is analogous to the amount of current.) Whether a bowling ball starts from an inch above your foot, or from up on a very high shelf above you is a measure of how much potential energy there is should it drop. (This is analogous to the amount of voltage.) So lots of current with a high voltage is dangerous, but high voltage by itself (with no current) is not dangerous, as long as you don't touch anything that would conduct current through yourself.)

Online reading assignment: circuit analysis

Physics 205B, spring semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing presentations on circuit analysis.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"In a series the current must pass through the resistors in sequence while in a parallel, portions can pass through separately and independently."

"There are two types of configurations for equivalent resistances and they each have a different way of calculating them. First, if there is a series configuration--meaning that the resistors are connected in a chain pattern--to calculate the resistance all you do is add them. Secondly, if there is the parallel configuration, to calculate the resistance you take the inverse of each resistor and add them up, and invert the resulting sum."

"When the resistors are in series, more resistors would mean that the resistance increases; however, when the resistors are in parallel, more resistors would mean that the resistance decreases, which is good for an ideal circuit."

"Current conservation (what flows in must also come out). Current leaving a junction must equal current entered."

"I get the basic concept of what goes in must come out. Any potential increase has to equal the potential drop that occurs from the current flowing through the resistors and bulbs."

"Resistor drops downstream and rises upstream. Emf rises during 'power ups,' drops during 'penalties.'"

"If we follow a complete loop in an electric circuit such that we wind up back at our starting point all the electric rise this potential added together will equal of electric potential that dropped together. This is having the same location as the final and initial points travel in a complete Loop forming an electrical circuit."

"How resistors are connected in series and in parallel and the equivalent resistance calculations. I also understood Kirchhoff's rules."

"I am beginning to understand voltages and currents but I need more practice using Kirchoff's rules."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"Kirchoff's junction rule and loop rule are both confusing. Might need some examples in class to clarify."

"I am still wrapping my brain around why the series and parallel resistors are so different."

"I was a little confused how when resistors are connected in parallel the equivalent resistance is smaller than either resistor."

"I am confused about emf rises/drops with respect to the battery terminals and resistor rises/drops with respect to current."

"I am hazy about voltage potential difference drop and electric potential decrease when the circuit moves from positive to negative. Also voltage potential difference and electric potential increase when the circuit moves from negative to positive."

"I understand the basic concepts, but I think I could use some practice with the actual calculations and logistics of what happens when in regards to the potential and traveling in directions in a circuit."

"I found everything very interesting and understand just about everything in this presentation."

"I don't understand most of this terminology."

"Sorry, so much chemistry."

Determine what happens to the following parameters as current flows through an ideal wire.
(Only correct responses shown.)

Current: remains the same [59%]
Voltage: remains the same [45%]

Determine what happens to the following parameters if you go through a resistor along the direction of current.
(Only correct responses shown.)

Current: remains the same [34%]
Voltage: decreases [59%]

Determine what happens to the following parameters if you follow a path (regardless of current direction) into the (–) terminal and out of the (+) terminal of an ideal battery.
(Only correct responses shown.)

Current: remains the same [45%]
Voltage: increases [55%]

Briefly explain what quantity is conserved when applying Kirchhoff's junction rule.

"Current (amperage) is conserved."

"Charge flow per time is conserved."

"The quantity of current flowing into a junction is equal to the quantity of current flowing out of the junction."

"I think it is 'what goes in must come out' which apparently seems like a simple concept but is useful to enforce mathematically as well to analyze electrical circuits."

Briefly explain what quantity is conserved when applying Kirchhoff's junction rule.

"Electric potential is conserved."

"Energy per charge."

"Kirchhoff's loop rule: the conservation of electric potential (electric potential energy per charge). The sum of voltages around any closed loop in a circuit must equal zero (charge conservation and conservation of energy)."

"No idea."

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"I was following this up until about halfway then I started to get lost with these concepts."

"I am still a little hazy on some of the topics, but I really liked that there were some smaller pictures in the presentation that gave us examples of what was happening while we were reading the descriptions. That really helped and I liked the way it was set up :)" (Hopefully those pictures are what's in your head from now on when you visualize what's going on with the currents and potential rises/drops in circuits.)

"Hi, sorry I was studying for a chemistry test and pretty much just remembered about this assignment at the last minute. :/ "

"I thought I had a good understanding until I saw these examples. The amount of current in must eqaul the amount of current that out of a voltage source. Same goes for the potential difference, the sum of the electric potential rises must equal the sum of the electric potential drops." (That sounds pretty good, so far.)

"Is there anything covered early this semester that will not be on the upcoming midterm? (The study guide for the midterm next Wednesday is already up (five key topics, anything not listed will not be on the midterm), and for this weekend relevant practice problems have been assigned for you to work on, before our review session next Monday.)

Online reading assignment: circuit basics

Physics 205B, spring semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing presentations on circuit basics.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"Most basic circuits that we can build will have an electromotive source of voltage connected to the resistor. In this way the charges can flow continuously around and around. I also learned that an ideal battery uses the chemical reactions that occur change charges in an order to release electric potential energy. I also learned that different chemical reactions will release different amounts of electrical potential energy which creates different voltages."

"An ideal circuit is one where charges can flow continuously through it. Electrons flow in the opposite direction of the current, which is how much positive charge in coulumbs that circulates per time in seconds. Coulombs per second is referred to as amps. Ideal batteries are used to release electric potential energy by exchanges charges. A circuit also consists of a resistor in which all different materials have different resistance values. We can use Ohm's law to determine how much current will flow through a circuit."

"When you stack ideal batteries, you add the total value of voltages. A good conductor has a low resistance value and a poor conductor has a high resistance value. When you string together resistors, you have to add their individual values together for an equivalent resistance value. Ohm's law can help determine how much current will flow given the total amount of voltage and resistance."

"In a basic circuit, charges can flow continuously around. A current is the positive charge that circulates while the electrons will flow around the other way."

"A basic circuit involves an electromotive force connected to a resistor so that a charge may flow continuously. A current is defined by the amount of positive charge flowing (out from the (+) terminal of the battery), but it is the electrons that are flowing in the opposite direction (out from the (–) terminal of the battery)."

"Amps (A) is what a positive current charge that circulates in coulombs per second."

"There is a direct relationship between resistance, voltage and current. These three variables describe different characteristics of a circuit. The equation is able to be manipulated easily to solve for the desired variable."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I don't understand the idea of battery stacking in order to increase the voltage output. Does the electric current just flow between the batteries, how do the stacked batteries relate to each other in order to increase the voltage output?"

"What is amperage and how is it measured in contrast to voltage? Why are these two aspects inversely related and how does that apply to each circuit?"

"I think I'm understanding the basics, but it also feels like I'm really missing something."

"We hear AC/DC all our lives (not the band necessarily). I don't know that I've ever realized how simple that concept really is. Direct current is moving one way all the time and alternating current is alternating back-and-forth. what a concept!"

"For some reason this is really hard for my brain to wrap around. I get the basic concepts of how a circuit functions, but when it comes to amps and current and voltage and stuff like that."

"Ohm's law--I do not understand the difference between the voltage and the current. Also I do not understand why the ratio of ∆V/I remains constant."

"I am sorry, I will read this before class. Chemistry is killing me :("

"I need to be more in-depth with all the material in this reading."

"I will take great notes in class on Monday."

A wire is used to complete a circuit with a single 9.0 V battery. When a wire is used to complete a circuit with a system of 244 "stacked" 9.0 V batteries, there will be __________ voltage and __________ current, compared to the single 9.0 V battery circuit.

less; less.	* [1]
less; more.	** [2]
more; less.	****** [6]
more; more.	********************** [22]
(Unsure/guessing/lost/help!)	***** [5]

An emf source is connected to a container of water. When salt is dissolved in the water, there will be __________ resistance and __________ current, compared to the pure deionized water circuit.

less; less.	[0]
less; more.	**************************** [28]
more; less.	** [2]
more; more.	* [1]
(Unsure/guessing/lost/help!)	***** [5]

A metal screw completes a "short circuit" with a transformer emf source. This is dangerous due to the very __________ resistance of the metal screw, and the very _________ current flowing through it.

low; low.	* [1]
low; high.	*************************** [27]
high; low.	* [1]
high; high.	** [2]
(Unsure/guessing/lost/help!)	***** [5]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"I didn't quite understand what you meant by 'current flowing clockwise through this circuit, while the electrons actually circulate in the opposite counterclockwise direction through this circuit.' Does this mean that the negative charge (electrons) is going one direction and the positive charge (protons) is going the opposite direction?" (Almost; electrons are traveling in one direction around the circuit, and the lack of electrons (which makes a neutral object have a positive charge) travels around in the opposite direction. You can think of bumper-to-bumper traffic, where each car moves forward, but the gap between cars moves backwards down the freeway.)

"So the current moves in one direction?" (For the steady-state, direct current circuits (which we are considering here), the current keeps moving in one direction around the circuit, at a constant value. However, what is really going on is that electrons are going in the opposite direction of how we define currents, but that's a result of electrons arbitrarily being labeled as having a negative charge instead of positive charge.)

"This was very helpful. The circuits seem super-dangerous." (These were only dangerous not just because of the voltage differences, but because the resistances were low, making the resulting amounts of current very high.)

"I don't understand if short-circuiting is due to too much or too little resistance." (Too little. Even with a modest amount of voltage (∆V), completing a circuit with a low resistance object (metal, water, unprotected sweaty palms) will make the resistance (R) in the denominator in Ohm's law (I = ∆V/R) very small, making the current (I) very high, which is what can kill you.)

"Can you explain what happens when you are putting up your Christmas lights and one bulb is dead then the series of bulbs after it don't work." (If a bulb completely burns out ), then no current can pass through it (as its filament is broken), which prevents the rest of the bulbs from being lit, as the circuit is now "open." However, a newer "shunt" type of bulb has a backup path for current to pass through it even after the filament is broken, although you may notice the rest of the bulbs are a little dimmer afterwards.)

"I'm curious to how my portable charger works and what makes it so easy to charge as well as why it charges the phones so fast in addition to lasting so long to die off. I'm curious to know what materials they used, but of course I'm not going to dismantle it to see what's inside." (Most likely it contains a lithium-ion battery inside. And yes, good on deciding to not take it apart.)

"I am just a little confused on the stacking component of batteries (and even batteries in general). I understand there is a chemical reaction but how does that create a charge? Do batteries constantly have a charge and chemical reactions are happening inside the battery at all times?" (By "charge" let's make sure we're talking about actual electrons that flow, instead of "putting energy into" the battery. So chemical reactions occur in batteries by materials exchanging electrons (to fill or to empty their orbitals and bonds), releasing energy in the process. If the battery is part of a complete circuit, then electrons are free to flow through the rest of the circuit to return to the battery, and take part in further chemical reactions, releasing more energy, etc. Ideally, if the battery is disconnected from a circuit, then the chemical reactions will stop, because no more electrons are available to be exchanged.)

"It would be cool to make a battery for lab." (That sounds like chemistry to me. However, we'll be building a capacitor in lab this week, and taking a look at thermocouples for next week's lab, which can basically be thought of as temperature-dependent batteries.)

Physics quiz archive: circuits (2)

Physics 205B Quiz 5, spring semester 2018
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz05z0m6



Sections 30882, 30883 results

0- 6 :
7-12 :	***** [low = 9]
13-18 :	*************** [mean = 18.4 +/- 4.4]
19-24 :	******************* [high = 24]
25-30 :

Physics quiz archive: capacitors, circuits

Physics 205B Quiz 4, spring semester 2018
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz04Md1o



Sections 30882, 30883 results

0- 6 :
7-12 :	*** [low = 9]
13-18 :	************
19-24 :	************ [mean = 20.3 +/- 5.3]
25-30 :	***** [high = 27]

Online reading assignment: advanced electricity (review)

Physics 205B, spring semester 2018
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on re-reading textbook chapters and reviewing presentations on advanced electricity concepts.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"The review of Kirckhoff's laws and how to apply them to various levels of complicated circuits. I also sort of understand ammeters."

"The equation for series and parallel, and the addition rules for them are pretty simple, especially after the lab."

"That series resistors are chain like and parallel resistors are when the current can divide up in a fork kind of way and portions of it can pass through two resistors simultaneously. I also understand the equations that come with these concepts."

"An effective voltmeter 'feels' the current without taking anything from it. We are reported the difference between the high and low potentials."

"According to the loop rule, the total change in rises equals the total change of drops."

"When measuring the amount of electrical potential used by a light bulb (or any other circuit element), the digital multimeter must be connected to both before and the current flows through the light bulb. This means that the wiring in the circuit is not modified in order to connect the digital multimeter to measure electric potential (making it an voltmeter), as it 'feels' the amount of electric potential before and after the light bulb, and reports the difference (whether a drop or rise)."

"The difference between an anameter and a voltmeter and how you need to wire them into a circuit to get the readings you want. I also understand the power equation given in the presentation and how you can substitute in Ohm's law and get different variations."

"Ammeters measure the amount of resistance in an item on a circuit like bulb or resistor and has a resistance of zero while voltmeters need a before and after and have a resistance of infinity."

"The circuit analysis presentation, especially after the review in class. The equivalent resistances for series and parallel circuits especially make sense to me: for series, you can simply add the resistances together since the current will only be able to take one path whereas with parallel circuits you have to add the inverses (and take the inverse of that amount) since the circuits have different junctions to go through."

"Joule heating is the rate of energy used per time. Joules per second is equivalent of watts. Ammeters must have resistance ideally zero. Voltmeter resistance ideally should be infinity."

"The power of a circuit is the product of the current and the voltage of that circuit."

"When a circuit is grounded, charge can flow through back to the circuit itself, where it can be released into the ground instead of running through a person's body. Household circuits are wired to have outlets in parallel, meaning that it's important not to overload them as the resistance could become too low."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I was not sure where we were supposed to find the resistances of the list of devices below so I guessed."

"I would like an introduction on voltmeters. I think they are more confusing than ammeters."

"I don't understand the equations for the potential drops and rises."

"I am finding Kirchhoff's rules (especially the loop one) really confusing. In addition, I'm still not quite sure about voltmeters and ammeters. I don't know some examples using one or both of the ideas would I think help me better understand these ideas."

"I am still confused with the loop rule and would like more of an explanation on it in class."

"I found the section about runaway current and circuit breakers to be confusing. Also I would like to know how the resistance of a voltmeter is infinite."

"I have not read enough of the chapter to determine what I find confusing. At the moment almost all of it is confusing."

"Pretty chill lesson."

State the unit of electrical power, and give an equivalent definition in terms of other SI units.

"Watts, joules."

"Watt, one joule per second."

"Electric power is measured in watts, which equates to Joules per second when using other SI units to define watts."

"Watt (W), also amps times volts."

What are the resistances of these (ideal) devices?
(Only correct responses shown.)

Ideal light bulb: some finite value between 0 and ∞ [71%]
Burnt-out light bulb: ∞ [54%]
Ideal wire: 0 [58%]
Ideal (non-dead) battery: 0 [71%]
Real (non-dead) battery: some finite value between 0 and ∞ [63%]
Ideal switch, when open: ∞ [33%]
Ideal switch, when closed: 0 [29%]

Two light bulbs with different resistances r and R, where r < R, are connected in series with each other to an ideal emf source. Select the light bulb with the greater quantity.
(Only correct responses shown.)

More current flowing through it: (there is a tie) [58%]
Larger potential potential difference: light bulb R [46%]
More power used: light bulb R [75%]

Two light bulbs with different resistances r and R, where r < R, are connected in parallel with each other to an ideal emf source. Select the light bulb with the greater quantity.
(Only correct responses shown.)

More current flowing through it: light bulb r [46%]
Larger potential potential difference: (there is a tie) [38%]
More power used: light bulb r [46%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"It will be better after spring break...hopefully!"

"Can we talk about Kirchhoff's loop rule, and power? I feel kind of lost."

"The potential drop is still not clear to me, reading isn't helping me understand it either." ("Potential" is just another term for voltage, so "potential drop" is the change in voltage ∆V, which measures how much voltage a resistor uses up while current is passing through it. If a Christmas light bulb is connected to an ideal 1.5 V emf, the emf provides 1.5 V to the current passing through it, and then the current as it passes through the light bulb uses up 1.5 V.)

"I didn't understand the upstream/downstream example. Will we be given a path direction and use that to check our answer? Or do we use our answer to determine the direction the current is traveling?" (For very complicated circuits it will not be apparent what direction(s) the current has, but for the scope of this course the direction(s) of current will be given, or will be apparent if you inspect the polarity (the +/− ends) of the emfs on the circuit.)

"Please explain examples above."

"Could you explain the last two examples in class? I'm confused on them (and I hope I'm not the only one)."

"I still don't understand resistors in series and the electric potential difference. How can something with a higher resistance have a lower potential difference than something with a lower resistance?" (Actually, that's impossible (at least for either of the two examples above):

1. For the two light bulbs in series, the same current I passes through them (from the junction rule where there is no junction). Using Ohm's law for each light bulb, ∆V = I⋅R and ∆V = I⋅r, since I is the same for both bulbs, the higher resistance bulb must have the higher potential drop (since it has a higher resistance, it "costs more" in voltage to pass through it).
2. For the two light bulbs in parallel, both light bulbs have the same ∆V potential drop (from the loop rule, as whatever loop you take, each light bulb is supplied directly from the same emf). Using Ohm's law for each light bulb, ∆V = I⋅R and ∆V = I⋅r, since ∆V is the same for both bulbs, the higher resistance bulb will have less current flowing through it, and the lower resistance bulb will have more current flowing through it).)