Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

Create An Account

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #71 : Circuits

A scientist is testing resistors made of the same material. Resistor has a cross sectional area of 3mm^2 and a length of 3mm^2 . Resistor has a cross sectional area of 6mm^2 and a length of 3mm^2 . How will the resistance of compare to that of ?

Possible Answers:

The resistance will quadruple

None of these

The resistance will double

The resistance will be cut in half

The resistance will stay the same

Correct answer:

The resistance will be cut in half

Explanation:

Use the equation for resistivity to visualize the proportions:

$R=\rho*\frac{l}{A}$

$\rho$ will stay constant as the material is the same. doubles and stays the same. Thus, the resistance, , will get cut in half.

Report an Error

Example Question #71 : Circuit Properties

A resistor is made out of a material. The resistor has a cross-sectional area of 4mm^2 and a length of 8mm . It is found to have a resistance of $7\Omega$ . Determine the resistivity of the material.

Possible Answers:

$4.7\Omega\cdot mm$

$6.5\Omega\cdot mm$

$7.0\Omega\cdot mm$

$10.3\Omega\cdot mm$

$3.5\Omega\cdot mm$

Correct answer:

$3.5\Omega\cdot mm$

Explanation:

Using

$\rho=R\frac{A}{l}$

Plugging in values

$\rho=7*\frac{4}{8}$

$\rho=3.5\Omega\cdot mm$

Report an Error

Example Question #71 : Circuits

A resistor is made out of a material. The resistor has a cross-sectional area of 4mm^2 and a length of 8mm . It is found to have a resistance of $7\Omega$ . A new resistor is built that has a length of 6mm and a cross sectional area of 3.3mm^2 . The new resistor is placed in series with a battery. Determine the current through the resistor.

Possible Answers:

.94A

None of these

3.25A

1.94A

1.25A

Correct answer:

.94A

Explanation:

Using

$\rho=R\frac{A}{l}$

Plugging in values

$\rho=7*\frac{4}{8}$

$\rho=3.5\Omega*mm$

Using

$R=\rho*\frac{l}{A}$

Plugging in values

$R=3.5*\frac{6}{3.3}$

$R=6.36\Omega$

Using

V=IR

Solving for

$I=\frac{V}{R}$

$I=\frac{6}{6.36}$

I=.94A

Report an Error

Example Question #74 : Circuit Properties

A scientist is testing resistors made of the same material. Resistor has a cross sectional area of 4mm^2 and a length of 8mm . Resistor has a cross sectional area of 6mm^2 and a length of 3mm . How will the resistance of compare to that of ?

Possible Answers:

It is impossible to determine

None of these

The resistance of will be a quarter of that of

They will be the same

The resistance of will be a quarter of that of

Correct answer:

The resistance of will be a quarter of that of

Explanation:

Using

$R=\rho*\frac{l}{A}$

$\rho$ will stay constant as the material is the same. increases by a ratio of $\frac{3}{2}$ , and decreases by a ratio of $\frac{3}{8}$ . Thus, the resistance, , will decrease to a quarter of it's original value. $\left (\frac{\frac{3}{8}}{\frac{3}{2}}\right)$

Report an Error

Example Question #881 : Ap Physics 2

A cubic resistor is placed in a circuit. If each edge measures 1cm long, the voltage of the battery is and the current through the resistor is , determine the resistivity.

Possible Answers:

$1.67\Omega\cdot cm$

$18\Omega\cdot cm$

$6\Omega\cdot m$

None of these

$6\Omega\cdot cm$

Correct answer:

$6\Omega\cdot cm$

Explanation:

Determining the resistance

V=IR

6=1*R

$R=6\Omega$

Using

$\rho=R\frac{A}{l}$

$\rho=6*\frac{1}{1}$

$\rho=6\Omega\cdot cm$

Report an Error

Example Question #71 : Circuits

A scientist carves out a cylinder of a new material she developed. It is 5.0 cm tall, with a radius of .80 cm . She puts electrodes on each face of the cylinder. She determines the resistance to be $.75\Omega$ . What is the resistivity of this new material?

Possible Answers:

$1.5 *10^{ -3} \Omega \cdot m$

$4.5*10^{-3} \Omega \cdot m$

$1.75 *10^{ -3} \Omega \cdot m$

None of these

$3.0*10^{-3} \Omega\cdot m$

Correct answer:

$3.0*10^{-3} \Omega\cdot m$

Explanation:

Using the relationship:

$\rho =R\frac{A}{l}$

Here, $\rho$ is the resistivity, is the resistance, is cross sectional area, and is the length.

The area of a circle is:

$A=\pi r^2$

Substitute:

$\rho =R\frac{\pi r^2}{l}$

Plug in given values and solve.

$\rho =.75\Omega \frac{\pi (.008m)^2}{.05m}$

$\rho=3.0 *10^{ -3} \Omega \cdot m$

Report an Error

Example Question #1 : Circuit Power

Consider the circuit:

Circuit_1

What is the rate of power consumption in the circuit if every resistor has a resistance of $3\Omega$ ?

Possible Answers:

18W

21 W

15 W

24 W

12 W

Correct answer:

18W

Explanation:

To calculate the power consumption of the circuit, we need to first reduce it to an equivalent circuit with a single resistor. Since each resistor has the same resistance, this solution will keep resistance calculations as multiples of until the circuit is fully reduced.

Start with the two branches in parallel. We can condense R3 and R4, then solve for the total resistance of R2, R3, and R4.

$R_{34}=R+R=2R$

$\frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{R_2}+ \frac{1}{R_{34}}=\frac{1}{R}+\frac{1}{2R} = \frac{3}{2R}$

$R_{eq} = \frac{2}{3}R$

The equivalent circuit now has three resistors in series (R1, Req, and R5), so we can simply add them all up:

$R_{tot} = R+\frac{2}{3}R+R=\frac{8}{3}R$

Plug in the value for R:

$R_{tot}=\frac{8}{3}R=\frac{8}{3}(3\Omega)=8\Omega$

Now we can use the equation for power:

P = IV

Substituting in Ohm's law for current, we get:
$P = \frac{V^2}{R} = \frac{(12V)^2}{8\Omega} = 18W$

Report an Error

Example Question #2 : Circuit Power

You have 4 resistors, , , , and , set up like this:

4resistorcircuit

Their resistance are as follows:

$\begin{align*} A&=1\Omega \\ B&=3\Omega \\ C&=4\Omega \\ D&= 2\Omega \end{align*}$

If the battery has 8V, what is the total power dissipated through the resistors?

Possible Answers:

12W

16W

20W

Correct answer:

16W

Explanation:

The equation for power is

P=VI

In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law ( V = IR ).

To find the total resistance, remember the equations for adding resistors is this:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}$

Resistors and are in series, resistors and are in parallel, and resistors ABC and are in series.

$R_{AB}&=1+3$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}= 2$

$R_{total}&=R_{ABC}+R_{D}$

$R_{total}= 2+2$

$R_{total}= 4$

Now, we can find the current.

$I&=\frac{V}{R}$

$I=\frac{8}{4}$

I= 2

Finally, we can find the power.

P&=VI

P=(8)(2)

P= 16

Therefore, the power is 16W (watts).

Report an Error

Example Question #2 : Circuit Power

Physics2set1q6

Calculate the power consumed across resistor .

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

Possible Answers:

10W

3 W

Correct answer:

3 W

Explanation:

Physics2set1q6

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

To calculate power, we need two of the following three quantities: voltage, current, and resistance.

In this case, since we are lacking the voltage, let's try to find the current.

We can use Kirchoff's junction law to calculate current .

The current coming into the junction = the current coming out of the junction.

Let's take a look at the central junction to the right of resistor .

$I_{1}=I_{2}+I_{3}$

$2=I_{3}+1$

$I_{3}=1$

Now that we know and , we can calculate power across the resistor.

$P={I^2}R=3W$

Report an Error

Example Question #1 : Circuit Power

Physics2set1q4a Physics2set1q4b

Elements A-D represent light bulbs.

Which of the following is true about these two circuits? Assume voltage sources have the same value and all the light bulbs are all identical.

Possible Answers:

Bulbs C and D have different brightnesses.

All bulbs will have the same brightness.

Bulbs A and B will be brighter than bulbs C and D.

Bulbs A and B have different brightnesses.

Bulb A will be as bright as bulb C, but bulbs B and D will have a different brightness.

Correct answer:

Bulbs A and B will be brighter than bulbs C and D.

Explanation:

Since bulbs A and B are in parallel, they will have the same voltage, and since the bulbs are identical in resistance, they will have the same current running through them and will be just as bright.

Let's say the voltage source as a value of and each bulb has a resistance of .

The current going through bulbs A and B is $\frac{V}{R}$ .

However, the current going through bulbs C and D is $\frac{V}{2R}$ .

The current going through bulbs C and D is half as much as the other two, so their brightness will be less.

So, bulbs A and B will be brigher than bulbs C and D.

Report an Error

View Tutors

Samantha
Certified Tutor

Middlesex County College, Associate in Science, Sustainability Studies.

View Tutors

Richard
Certified Tutor

Northeastern University, Bachelor in Arts, Business, General.

View Tutors

Ebony
Certified Tutor

Johns Hopkins University, Masters in Education, Education.

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Atlanta, Math Tutors in Los Angeles, Spanish Tutors in New York City, Spanish Tutors in San Francisco-Bay Area, ACT Tutors in San Diego, Spanish Tutors in Houston, Calculus Tutors in Atlanta, Reading Tutors in Miami, Spanish Tutors in Philadelphia, ACT Tutors in Seattle

Popular Courses & Classes

GRE Courses & Classes in Seattle, MCAT Courses & Classes in San Diego, GMAT Courses & Classes in Houston, ACT Courses & Classes in Seattle, MCAT Courses & Classes in Los Angeles, ISEE Courses & Classes in Phoenix, SSAT Courses & Classes in Houston, SAT Courses & Classes in Boston, ISEE Courses & Classes in Chicago, ACT Courses & Classes in San Diego

Popular Test Prep

GRE Test Prep in Phoenix, ISEE Test Prep in New York City, SSAT Test Prep in Atlanta, ISEE Test Prep in Dallas Fort Worth, ACT Test Prep in San Diego, ACT Test Prep in Washington DC, LSAT Test Prep in Denver, SAT Test Prep in Denver, MCAT Test Prep in Seattle, ACT Test Prep in Seattle

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #71 : Circuits

Example Question #71 : Circuit Properties

Example Question #71 : Circuits

Example Question #74 : Circuit Properties

Example Question #881 : Ap Physics 2

Example Question #71 : Circuits

Example Question #1 : Circuit Power

Example Question #2 : Circuit Power

Example Question #2 : Circuit Power

Example Question #1 : Circuit Power

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: