1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electricity (probably easy

  1. Jan 14, 2008 #1
    [SOLVED] Electricity (probably easy

    1. The problem statement, all variables and given/known data

    In a circuit shown, the p.d. across the 4Ω resistor is 8 V.
    a)What is the current through the 4Ω resistor?
    b)What is the current through the 3Ω resistor?
    c)What is the p.d. across the 3Ω resistor?
    d)What is the p.d. applied by the battery?



    2. Relevant equations
    Ohms Law V=IR


    3. The attempt at a solution

    a)2A?
     

    Attached Files:

  2. jcsd
  3. Jan 14, 2008 #2
    Please show how you arrived at the solution. Since the attachment is pending approval from moderators, I can't see your circuit. A quicker way would be to upload the attachment to an online image hosting site like Photobucket and use the corresponding link in IMG tags while posting. Can you do that?
     
  4. Jan 14, 2008 #3
    Can you see it?
    <a href="http://s262.photobucket.com/albums/ii120/reika_xx/?action=view&current=Electricity.jpg" target="_blank"><img src="http://i262.photobucket.com/albums/ii120/reika_xx/Electricity.jpg" border="0" alt="Electricity Circuit"></a>


    a)V=IR
    8=I*4
    I=2A
     
  5. Jan 14, 2008 #4
  6. Jan 14, 2008 #5
    1
    i. For the 4[itex]\Omega[/itex] resistor,
    [tex]
    V = 8~NmC^{-1}
    [/tex]
    [tex]
    R = 4~NmC^{-2}s^{-1}
    [/tex]

    Hence,

    [tex]
    I = \frac{V}{R} = \frac{8}{4} \frac{NmC^{-1}}{NmC^{-2}s^{-1}}
    [/tex]
    [tex]
    I = 2~Cs^{-1} = 2~A
    [/tex]

    ii. Current through the 3[itex]\Omega[/itex] resistor is the same as through the 4[itex]\Omega[/itex] resistor as both of them are in series.

    iii. For the 3[itex]\Omega[/itex] resistor:

    [tex]
    I = 2~Cs^{-1}
    [/tex]
    [tex]
    R = 3~NmC^{-2}s^{-1}
    [/tex]

    Hence,

    [tex]
    V~=~IR = 2 \times 3~Cs^{-1}.NmC^{-2}s^{-1}
    [/tex]
    [tex]
    V = 6~NmC^{-1} = 6~V
    [/tex]

    iv. Do you mean the potential difference across the terminals of the battery? If that is the case, you can solve it using Kirchhoff's law. Refer to the figure in the attachment.

    Hence,

    [tex]
    \epsilon + 2(4)~NmC^{-1} + 2(3)~NmC^{-1} = 0
    [/tex]
    [tex]
    |\epsilon| = 14~NmC^{-1} = 14~V
    [/tex]

    So, the battery provides and e.m.f of 14V. [In case you can't see the attachment, go here: http://image.bayimg.com/iajbbaaba.jpg]
     

    Attached Files:

    Last edited: Jan 14, 2008
  7. Jan 14, 2008 #6
    thank you so much!
     
  8. Jan 14, 2008 #7
    I=U/R=8V/4om=2A
    I=8V/3om=8/3 A

    This questions I don't understand
    "c)What is the p.d. across the 3Ω resistor?
    d)What is the p.d. applied by the battery?"

    "sorry
    http://s262.photobucket.com/albums/i...lectricity.jpg"
    I=8V/(4om+3om)=8/7 A.
    If electricity is ~, then (8/7)A/2=8/14 A, becouse of diod.
     
    Last edited: Jan 14, 2008
  9. Jan 14, 2008 #8
    @fermio: I am not able what you are trying to say due to poor formatting. Please repost with proper formatting and punctuation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electricity (probably easy
  1. Probably really easy (Replies: 1)

  2. Entropy, probably easy (Replies: 2)

Loading...