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Electric Circuits Problem: Please Help

  1. Dec 7, 2004 #1
    My Physics teacher sent this out as a difficult problem, to us at least, to impress her if we could find the answers.

    Thanks for just guiding me on it; i dont want someone to do it for me, but rather, id just like someone to start me out

    basic Physics stuff i know:

    Ohm's Law, Parallel Series: amps different, voltage same, resistance added through (1/x) + (1/y) [+ (1/z)] = (1/total resistance). Series Circuit: voltage different, amps same, resistance added to get total

    Images take a little while to load. Sorry.


    image here: http://jon8rfc.homeip.net/uploader/uploaded/pic1.bmp

    Three identical resistors, each with resistance R, and a capacitor of 1.0 x 10-9 F are connected to a 30 V battery with negligible internal resistance, as shown in the circuit diagram above. Switches S1 and S2 are initially closed, and switch S3 is initially open. A voltmeter is connected as shown.
    (a) Determine the reading on the voltmeter.
    (b) Switches S1 and S2 are now opened, and then switch S3 is closed. Determine the charge Q on the capacitor after S3 has been closed for a very long time.

    After the capacitor is fully charged, switches S1 and S2 remain open, switch S3 remains closed, the plates are held fixed, and a conducting copper block is inserted midway between the plates, as shown below. The plates of the capacitor are separated by a distance of 1.0 mm, and the copper block has a thickness of 0.5 mm.

    2nd image here: http://jon8rfc.homeip.net/uploader/uploaded/pic2.bmp

    (c) i. What is the potential difference between the plates?
    ii. What is the electric field inside the copper block?
    iii. On the diagram above, draw arrows to clearly indicate the direction of the electric field between the plates.
    iv. Determine the magnitude of the electric field in each of the spaces between the plates and the copper block.

    Thanks for any help!!!
  2. jcsd
  3. Dec 7, 2004 #2
    Hi Jay,
    When uploading line drawings or other non-photorealistic illustrations to the internet, the optimal compression format is PNG (Portable Network Graphics). Bitmaps are uncompressed and are thus generally not used over the internet. If you don't have a program that saves in PNG, you can use the free http://www.numbera.com/software/pnggauntlet.aspx [Broken] converter.
    Following is your post with the PNG images added for future readers who may have a slower connection. :smile:

    In regards to 1(a), note that you can ignore the loop with the capacitor as the switch is open. This is a simple case of treating the 2 parallel resistors as one virtual series resistor (by your derived reciprocal relationship) and then calculating the voltage drops for the series circuit. Work symbolically at first.
    Regarding 1(b), you should know why they add "after it has been closed for a very long time". If you have not already learned about RC circuits, this is a test of how well you understand Kirchoff's rules (the way you derived the rules for resistors and capacitors using only voltage and current). The relevant rule is that the sum of voltage drops around a closed circuit must be zero. Ie., a closed circuit with a battery supplying voltage V and capacitor with capacitance C satisfies the equation V - Q/C = 0 (V is positive because electrons gain energy there).
    I've got to go, but others may help soon. :smile:
    Last edited by a moderator: May 1, 2017
  4. Dec 7, 2004 #3
    Okay, so for the first problem.
    a. Looking at the picture, the reading on the voltmeter across resistor R would be pretty simple. Since S3 is open, that part of the circuit doesnt matter, because no current can pass through it. This means that the only resistors you have to worry about are R, and the resistors next to S1 and S2 (call them R1 and R2). You can add the ones next to the switches in parallel, and add the result with the R in the series. Determine the current, and since R and (R1 + R2) are in series, the current times R will give you the voltage.

    b. Now that S1 and S2 are open, that part of the circuit doesnt matter. Since S3 is closed, the Capacitor and the Resistor that are in series will come into play, since there will be current passing through them. Now, the key is that she is asking you to find out the reading after S3 has been closed for a long time. This is important because after a long time, Q has reached its maximum, and you can find it by finding the voltage across it. Hope that helps.

    - harsh
  5. Dec 7, 2004 #4
    i dont exactly understand what u mean..i know u said symbolically..but what how am i supposed to get a number from a bunch of letters.. For instance:

    R + (1/R + 1/R)^-1 = total resistance

    and so it would be 30V / [R + {(1/R) + (1/R)}^-1] would = the current

    ok...? what did i do wrong? Should I substitute numbers or do i leave the answer as a bunch of letters?
  6. Dec 7, 2004 #5
    Hi Scott,
    Simplify your expression for the current (I) and remember that the voltage drop for the resistor in question is just IR. :smile:
  7. Dec 7, 2004 #6
    ok, but still i cant just substitute all of that for I, can I?

    I know the voltage drop for the resistor is IR, so what are u saying though, that the answer is just IR

    doesnt make sense
  8. Dec 7, 2004 #7
    Why not ? :smile: That's what Kirchoff's laws say I is equivalent to. Note that your expression for I can be simplified to 20/R.
  9. Dec 7, 2004 #8
    ok..so on with the problem..

    i think u meant 30/R represents I

    But how exactly do i find the amount of volts that that particular resistor uses. I know that the volts from the two parallel added with the original series resistor equal 30 V.
  10. Dec 7, 2004 #9
    No, I meant 20/R. I'll complete this first one just to clarify what's going on, but not the other 3.
    Your equation for I in the equivalent simple series circuit with 2 resistors (1 resistor with resistance R and the other with resistance 1/(1/R + 1/R) ) states that I = 30/(R + 1/(1/R + 1/R)) = 30/(R + R/2) = 30/(3R/2) = 20/R.
    Now use your value for I to find the voltage drop in the resistor with resistance R.
    Vd = IR = (20/R)*R = 20 [Volts]
    Did you see the difference between the two equations used ? The first equation used Kirchoff's laws to find the amperage for the entire circuit, while the second equation is just Ohm's law for the single resistor that has the same current I running through it.
  11. Dec 7, 2004 #10
    ok i see..thank u very much

    that jumpstart really helped me understand..

    so for part b

    u said earlier that V - Q/C = 0, so V must equal Q/C

    in other words, i can find the charge by VC, or (30V)(1 x 10^-9 F)

    is this a valid approach?
  12. Dec 7, 2004 #11
    While it will yield the correct answer, it is the wrong approach. My equation was just an example of Kirchoff's loop rule (the potential drops in a closed circuit sum to 0) with a simple circuit consisting of just a voltage source and a capacitor. Your circuit also has a resistor. Write the circuit's equation, but note that this circuit has an interesting property. When you first connect the switch, current flows through the resistor according to Ohm's law. However the circuit is not closed (because of the capacitor), so the current is not maintained over time; at some future time, the current in the circuit will be zero and the capacitor will be fully charged. The actual expressions for current and charge over time are easily derived using calculus, but I doubt your teacher meant for you to use calculus at this level. It is just enough to note that in your circuit equation, the capacitor is fully charged when the current goes to zero.
  13. Dec 8, 2004 #12

    when u multiply F and C, do u just get Coulombs? no prefixes?

    second of all, finding the potential difference between the two plates of a capicitor would be the same as charge, as CV is the potential difference...what is the unit here?

    Next, in Dielectrics, the electric fields are proportional (electric field without/electric field with)

    Since the electric field is calculated E = V/d, the electric field INSIDE THE COPPER BLOCK would be (30/.5) or 60 V/mm???

    Lastly, the magnitude of electric field in each of the spaces would be 120 V/mm in the spaces between the plates and copper surface and 60 V/mm between the surfaces of the copper??

    A lot of calculations here...please help me figure out which ones are right

    Last edited: Dec 8, 2004
  14. Dec 9, 2004 #13
    anyone? i need an answer by friday please

    i know this cant be that advanced...this is just basic physics
  15. Dec 9, 2004 #14
    nvm...i guessed and got it right

    thanks guys
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