Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask question.(5)

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

ITtutoria

ITtutoria Logo ITtutoria Logo

ITtutoria Navigation

  • Python
  • Java
  • Reactjs
  • JavaScript
  • R
  • PySpark
  • MYSQL
  • Pandas
  • QA
  • C++
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Python
  • Science
  • Java
  • JavaScript
  • Reactjs
  • Nodejs
  • Tools
  • QA
Home/ Questions/Find the emf E1 in the circuit of the figure (Figure 1) .
Next
Answered
lyytutoria
  • 6
lyytutoriaExpert
Asked: April 10, 20222022-04-10T06:56:03+00:00 2022-04-10T06:56:03+00:00In: Physics

Find the emf E1 in the circuit of the figure (Figure 1) .

  • 6

. Advertisement .

..3..

. Advertisement .

..4..

......... ADVERTISEMENT .........

..8..

Find the emf E1 in the circuit of the figure (Figure 1) . Find the emf...

 

 

 

 

 

 

 

Find the emf E1 in the circuit of the figure (Figure 1) .

Find the emf E2 in the circuit of the figure.

Find the potential difference of point b relative to point a


♦ Relevant knowledge
It is also known as the electromotive force or EMF is an electric force that is produced through transformation of every type of energy into electrical energy. It is the EMF is a term which is mentioned in explanations of the behaviour of electrochemical cells as well as electromagnetic induction. In the instance of a simple electrical circuit made up of the conductors of a wire, a generator and resistors, the EMF determines the energy generates (source of electromotive force) supplies to the load in order to allow it to move from one place to the next. In this way the term electromotive force could be described as the electric potential generated by an electromotive source force.

circuitemf e1emf e2
  • 2 2 Answers
  • 336 Views
  • 0 Followers
  • 0
Answer
Share
  • Facebook
  • Report

2 Answers

  • Voted
  • Oldest
  • Recent
  • Random
  1. Best Answer
    lyytutoria Expert
    2022-04-11T17:11:22+00:00Added an answer on April 11, 2022 at 5:11 pm

    • 20
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report
  2. tonytutoria
    2022-04-10T06:56:50+00:00Added an answer on April 10, 2022 at 6:56 am
    Concepts and Reason

    This problem can be solved using the concepts of Ohm’s law and Kirchhoff’s voltage laws, Electromotive force, Ohms law and Kirchhoff’s power law.

    Refer to the circuit diagram in the question. Find the [katex]{\varepsilon _1}[/katex] values later.

    Fundamentals

    The following is the potential difference between point and point , a:

    [katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]

    Here, [katex]{V_{\rm{b}}}[/katex]

    The Kirchhoff’s voltage law states that the sum of all voltages at point A is zero.

    To write the voltage inside the loop, you need to multiply the current and resistor by the sum of their resistance.

    [katex]V = IR[/katex]

    I here is the current, and R the resistance.

    (1)

    Below is an illustration of Kirchhoff’s voltage loop 1 & 2:


    The above illustration shows that there are loops. Loop 1 is the top loop, while loop 2 is the larger loop.

    The KVL loop 1 law states that the following expressions are allowed:

    [katex]20.{\rm{0 V}} – {\rm{1}}{\rm{.00 A}}\left( {1.00{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.00{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.00{\rm{ }}\Omega } \right) – {\varepsilon _1} – 6.00{\rm{ A}}\left( {1.00{\rm{ }}\Omega } \right) = 0[/katex]

    Solve for [katex]{\varepsilon _1}[/katex]

    [katex]\begin{array}{c}\\20{\rm{ V}} – {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.0{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) – {\varepsilon _1} – 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0\\\\20.{\rm{0 V}} – 1.00{\rm{ V}} + {\rm{4}}{\rm{.00 V}} + 1.{\rm{00 V}} – {\varepsilon _1} – 6.00{\rm{ V = 0}}\\\\{\varepsilon _1} = 18{\rm{ V}}\\\end{array}[/katex]

    [(1)]


    (1)

    [(1)]


    (2)

    The KVL loop law states that the following expressions are allowed:

    [katex]20{\rm{ V}} – {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {2.0{\rm{ }}\Omega } \right) – {\varepsilon _2} – 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0[/katex]

    Solve for [katex]{\varepsilon _2}[/katex]

    [katex]\begin{array}{c}\\20{\rm{ V}} – {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {2.0{\rm{ }}\Omega } \right) – {\varepsilon _2} – 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0\\\\20.{\rm{0 V}} – 1.00{\rm{ V}} – 2.{\rm{00 V}} – 4.{\rm{00 V}} – {\varepsilon _2} – 6.00{\rm{ V = 0}}\\\\{\varepsilon _2} = 7.0{\rm{ V}}\\\end{array}[/katex]

    [(2)]


    (2)

    [(2)]

    (3)

    Use the KVL laws to complete the loop and then write the following expression:

    [katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]0

    Solve for [katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]1

    [katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]2

    Substitute 18.0 V to [katex]{\varepsilon _1}[/katex]

    [katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]4

    [(3)]


    (3)

    [(3)]

    Ans:

    The [katex]{\varepsilon _1}[/katex] magnitude

    The [katex]{\varepsilon _2}[/katex] magnitude

    The potential difference between point b and point a is equal or -13.0 V.

    • 15
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

Sidebar

Ask A Question
  • How to Split String by space in C++
  • How To Convert A Pandas DataFrame Column To A List
  • How to Replace Multiple Characters in A String in Python?
  • How To Remove Special Characters From String Python

Explore

  • Home
  • Tutorial

Footer

ITtutoria

ITtutoria

This website is user friendly and will facilitate transferring knowledge. It would be useful for a self-initiated learning process.

@ ITTutoria Co Ltd.

Tutorial

  • Home
  • Python
  • Science
  • Java
  • JavaScript
  • Reactjs
  • Nodejs
  • Tools
  • QA

Legal Stuff

  • About Us
  • Terms of Use
  • Privacy Policy
  • Contact Us

DMCA.com Protection Status

Help

  • Knowledge Base
  • Support

Follow

© 2022 Ittutoria. All Rights Reserved.

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.