Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Electric Circuits

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Electricity And Magnetism

What is the voltage of a circuit with 10A $\displaystyle 10A$ of current and $20\Omega$ $\displaystyle 20\Omega$ of resistance?

Possible Answers:

10V $\displaystyle 10V$

30V $\displaystyle 30V$

0.5V $\displaystyle 0.5V$

200V $\displaystyle 200V$

$\displaystyle 2V$

Correct answer:

200V $\displaystyle 200V$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ . In this equation $\displaystyle V$ is the voltage, $\displaystyle I$ is the current, and $\displaystyle R$ is the resistance.

Plug in the given values and solve for the voltage.

$V=10A* 20\Omega$ $\displaystyle V=10A* 20\Omega$

V=200V $\displaystyle V=200V$

Report an Error

Example Question #2 : Electricity And Magnetism

What is the current of a circuit with a voltage of 12V $\displaystyle 12V$ and a total resistance of $5\Omega$ $\displaystyle 5\Omega$ ?

Possible Answers:

60A $\displaystyle 60A$

0.42A $\displaystyle 0.42A$

$\displaystyle 7A$

4.8A $\displaystyle 4.8A$

2.4A $\displaystyle 2.4A$

Correct answer:

2.4A $\displaystyle 2.4A$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ . In this equation $\displaystyle V$ is the voltage, $\displaystyle I$ is the current, and $\displaystyle R$ is the resistance.

We can re-arrange the equation to solve specifically for $\displaystyle I$ .

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

Plug in the given values for voltage and resistance to solve for the current.

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

$\frac{12V}{5\Omega}=I$ $\displaystyle \frac{12V}{5\Omega}=I$

2.4A=I $\displaystyle 2.4A=I$

Report an Error

Example Question #1 : Circuit Calculations And Concepts

What is the resistance in a circuit with a voltage of 50V $\displaystyle 50V$ and a current of $\displaystyle 8A$ ?

Possible Answers:

$6.25\Omega$ $\displaystyle 6.25\Omega$

$58\Omega$ $\displaystyle 58\Omega$

$0.16\Omega$ $\displaystyle 0.16\Omega$

$400\Omega$ $\displaystyle 400\Omega$

$42\Omega$ $\displaystyle 42\Omega$

Correct answer:

$6.25\Omega$ $\displaystyle 6.25\Omega$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ . In this equation $\displaystyle V$ is the voltage, $\displaystyle I$ is the current, and $\displaystyle R$ is the resistance.

We can re-arrange the equation to solve specifically for $\displaystyle R$ .

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

Plug in the given values for voltage and current and solve for resistance.

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

$\frac{50V}{8A}=R$ $\displaystyle \frac{50V}{8A}=R$

$6.25\Omega=R$ $\displaystyle 6.25\Omega=R$

Report an Error

Example Question #1 : Circuit Calculations And Concepts

A $\displaystyle 9V$ battery is connected to a circuit. The measured current is $\displaystyle 2A$ . What is the equivalent resistance?

Possible Answers:

$10\Omega$ $\displaystyle 10\Omega$

$4.5\Omega$ $\displaystyle 4.5\Omega$

$18\Omega$ $\displaystyle 18\Omega$

$9\Omega$ $\displaystyle 9\Omega$

$0.22\Omega$ $\displaystyle 0.22\Omega$

Correct answer:

$4.5\Omega$ $\displaystyle 4.5\Omega$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ .

We are given the voltage and current, allowing us to solve for the resistance.

$\displaystyle 9V=2A* R$

$\frac{9V}{2A}=R$ $\displaystyle \frac{9V}{2A}=R$

$4.5\Omega=R$ $\displaystyle 4.5\Omega=R$

Report an Error

Example Question #3 : Electricity And Magnetism

A closed electric circuit is set up so that there is a current of 9.9A $\displaystyle 9.9A$ and a voltage of 30.22V $\displaystyle 30.22V$ . What is the resistance in the circuit?

Possible Answers:

$20.32\Omega$ $\displaystyle 20.32\Omega$

$299.18\Omega$ $\displaystyle 299.18\Omega$

$0.33\Omega$ $\displaystyle 0.33\Omega$

$3.05\Omega$ $\displaystyle 3.05\Omega$

$6.1\Omega$ $\displaystyle 6.1\Omega$

Correct answer:

$3.05\Omega$ $\displaystyle 3.05\Omega$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ .

We are given the current and the voltage. Using these terms, we can solve for the resistance.

V=IR $\displaystyle V=IR$

$\displaystyle 30.22V=(9.9A) R$

$\frac{30.22V}{9.9A}=R$ $\displaystyle \frac{30.22V}{9.9A}=R$

$3.05\Omega=R$ $\displaystyle 3.05\Omega=R$

Report an Error

Example Question #1 : Electric Circuits

The current in a circuit is 20A $\displaystyle 20A$ . If the voltage is 35V $\displaystyle 35V$ , what is the total resistance?

Possible Answers:

$70\Omega$ $\displaystyle 70\Omega$

$3.5\Omega$ $\displaystyle 3.5\Omega$

$0.57\Omega$ $\displaystyle 0.57\Omega$

$700\Omega$ $\displaystyle 700\Omega$

$1.75\Omega$ $\displaystyle 1.75\Omega$

Correct answer:

$1.75\Omega$ $\displaystyle 1.75\Omega$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the current and the voltage, allowing us to solve for the resistance.

$\displaystyle 35V=(20A)*R$

$R=\frac{35V}{20A}$ $\displaystyle R=\frac{35V}{20A}$

$R=1.75\Omega$ $\displaystyle R=1.75\Omega$

Report an Error

Example Question #3 : Electric Circuits

An electrical circuit has a total of $3.22\Omega$ $\displaystyle 3.22\Omega$ resistance and $\displaystyle 8V$ . What is the current?

Possible Answers:

0.403A $\displaystyle 0.403A$

2.48A $\displaystyle 2.48A$

11.22A $\displaystyle 11.22A$

25.76A $\displaystyle 25.76A$

0.089A $\displaystyle 0.089A$

Correct answer:

2.48A $\displaystyle 2.48A$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the resistance and the voltage, allowing us to solve for the current.

$8V=I*(3.22\Omega)$ $\displaystyle 8V=I*(3.22\Omega)$

$I=\frac{8V}{3.22\Omega}$ $\displaystyle I=\frac{8V}{3.22\Omega}$

I=2.48A $\displaystyle I=2.48A$

Report an Error

Example Question #2 : Electric Circuits

An electrical circuit has a current of $\displaystyle 8A$ and $32\Omega$ $\displaystyle 32\Omega$ of resistance. What is the voltage?

Possible Answers:

$\displaystyle 4V$

128V $\displaystyle 128V$

256V $\displaystyle 256V$

0.25V $\displaystyle 0.25V$

40V $\displaystyle 40V$

Correct answer:

256V $\displaystyle 256V$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(8A)(32\Omega)$ $\displaystyle V=(8A)(32\Omega)$

V=256V $\displaystyle V=256V$

Report an Error

Example Question #3 : Electric Circuits

An electrical circuit has a current of 13A $\displaystyle 13A$ and $26\Omega$ $\displaystyle 26\Omega$ of resistance. What is the voltage?

Possible Answers:

338V $\displaystyle 338V$

$\displaystyle 2V$

13V $\displaystyle 13V$

39V $\displaystyle 39V$

0.5V $\displaystyle 0.5V$

Correct answer:

338V $\displaystyle 338V$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(13A)(26\Omega)$ $\displaystyle V=(13A)(26\Omega)$

V=338V $\displaystyle V=338V$

Report an Error

Example Question #2 : Circuit Calculations And Concepts

A circuit has a current, $\displaystyle I$ , a voltage, $\displaystyle V$ , and a resistance, $\displaystyle R$ . If the voltage remains constant, but the current is doubled ( $\displaystyle 2I$ ), what must the new resistance be?

Possible Answers:

$\frac{1}{4}R$ $\displaystyle \frac{1}{4}R$

$\displaystyle 4R$

$\displaystyle 2R$

There is insufficient information to solve

$\frac{1}{2}R$ $\displaystyle \frac{1}{2}R$

Correct answer:

$\frac{1}{2}R$ $\displaystyle \frac{1}{2}R$

Explanation:

To solve this problem, use Ohm's law:

V=IR $\displaystyle V=IR$

Since we are doubling the current, but the voltage is remaining the same, we can set our old and new equations equal to each other.

$\displaystyle I_1R_1=V=I_2R_2$

We know that the second current is equal to twice the first current.

$\displaystyle I_2=2I_1$

Use this equation to substitute current into the first equation.

$\displaystyle I_1R_1=(2I_1)R_2$

The initial current now cancels out from both sides.

$\displaystyle R_1=2R_2$

Divide both sides by two to isolate the final resistance variable.

$\frac{1}{2}R_1=R_2$ $\displaystyle \frac{1}{2}R_1=R_2$

Report an Error

View Tutors

Salman
Certified Tutor

Payam Noor University,, Bachelor of Science, Physics. University of Kashan, Doctor of Philosophy, Physics.

View Tutors

Varun
Certified Tutor

University of Arizona, Bachelor of Science, Aerospace Engineering.

View Tutors

Peter
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Computer Science.

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in Chicago, Calculus Tutors in Houston, Physics Tutors in Washington DC, GRE Tutors in San Diego, GRE Tutors in Philadelphia, ISEE Tutors in Phoenix, Statistics Tutors in New York City, LSAT Tutors in Miami, Computer Science Tutors in New York City, ACT Tutors in Boston

Popular Courses & Classes

Spanish Courses & Classes in Washington DC, SAT Courses & Classes in Washington DC, ACT Courses & Classes in Chicago, MCAT Courses & Classes in Houston, GRE Courses & Classes in Chicago, GMAT Courses & Classes in Boston, SAT Courses & Classes in Chicago, GRE Courses & Classes in Boston, ACT Courses & Classes in Houston, SAT Courses & Classes in San Diego

Popular Test Prep

ACT Test Prep in Houston, GRE Test Prep in Boston, SSAT Test Prep in Seattle, MCAT Test Prep in Philadelphia, LSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Atlanta, GRE Test Prep in Philadelphia, SAT Test Prep in Washington DC, MCAT Test Prep in Dallas Fort Worth

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

High School Physics : Electric Circuits

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Electricity And Magnetism

Example Question #2 : Electricity And Magnetism

Example Question #1 : Circuit Calculations And Concepts

Example Question #1 : Circuit Calculations And Concepts

Example Question #3 : Electricity And Magnetism

Example Question #1 : Electric Circuits

Example Question #3 : Electric Circuits

Example Question #2 : Electric Circuits

Example Question #3 : Electric Circuits

Example Question #2 : Circuit Calculations And Concepts

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: