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Current and e.m.f

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data
    The cell supplies 8.1×10^3 J of energy when 5.8×10^3 C of charge moves completely round the circuit. The current in the circuit is constant.
    A. Deduce that the e.m.f of the cell E is about 1.4 V
    B. The resistor R has resistance 6.0 Ω. The potential difference between its terminals is 1.2 V. Determine the internal resistance r of the cell.
    C. Calculate the power dissipated by the resistor R.

    2. Relevant equations



    3. The attempt at a solution
    For A, I noticed that if you divide 8.1 x 10^3J by 5.8 x 10^3C you get 1.39 which is very close to what I am supposed to deduce (1.4). Is this a correct way of finding what the e.m.f of the cell is?

    For B, I think I need the equation E = DeltaV + (I)(r) . In this equation what does E stand for? Can someone help me solve for I?

    For C, I'm not even sure how to go about this other than for this equation: P=I^2 * R
     
    Last edited: Apr 26, 2009
  2. jcsd
  3. Apr 26, 2009 #2
    Hi there.

    a) You're quite right. The emf is the energy transfered into the circuit per unit of charge that passes through the cell. So e = W/Q (1 Volt is 1 Joule per Coulomb)
    b)If the emf of the cell is 1.4V but the terminal pd is only 1.2V then the missing voltage must be dropped across the internal resistance. As the cell and the 6ohm resistor are in series then they both have the same current flowing through them. Find the current through the 6ohm resistor and you'll have enough information to find the internal resistance.
    c)You're quite right. Since you found the current through the 6ohm resistor in (b) then you can use your formula to find the power.
     
    Last edited: Apr 26, 2009
  4. Apr 26, 2009 #3
    so I found V=IR for the 6 ohm resistor
    1.4= I (6)
    so I equal .2333

    so in E = DeltaV + Ir
    E = .2 + (.2333)(r)

    Is this correct? What does E stand for again?
     
  5. Apr 26, 2009 #4
    Oops, sorry. That should be 6ohms not 8! Anyway...

    When calculating the current through the 6ohm resistor you need to use the terminal pd. So the current you're looking for is I = 1.2/6. This is the current through the 6ohm resistor and the cell (because they're in series).

    Now that you know the current through the cell and the fact that the internal resistance of the cell has 1.4 - 1.2 = 0.2V across it you can find the internal resistance.
     
  6. Apr 26, 2009 #5
    alright so using that, 1.2/6= .2 so I is .2.

    In the equation its then E = .2 + .2r

    In order to solve for r though I need to know what E is. What does E stand for?
     
  7. Apr 26, 2009 #6
    Your equation is something like

    emf of the cell = potential difference across the load + potential difference across the internal resistance

    so....

    e = IR + Ir = I(R+r)

    I assume delta V in your equation represents the pd across the load resistor.
     
  8. Apr 26, 2009 #7
    So your equation is

    1.4 = 1.2 + Ir
     
  9. Apr 26, 2009 #8
    I was thinking though, normally to find (I) don't you need to have the total resistance in the circuit? Since r and R are in series it should be
    V = I (R + r)
    right?

    1.4 = .2(6 + r)
    1.4/.2 = 6 + r
    7 = 6 + r
    1 = r

    is that correct?
     
  10. Apr 26, 2009 #9
    Yep. Although I'd write it as e = I(R + r) just to be clear that the emf is equal to the TOTAL pd around the circuit whereas V is usually just the pd across the load.

    You might find it helpful to draw out a simple circuit with just two series resistors of 1ohm and 6ohms and a cell of emf 1.4V and zero internal resistance. If you apply ohms law to this circuit you'll see that it works out exactly the same as this internal resistance problem.
     
  11. Apr 26, 2009 #10
    Would you mind helping me with the follow up question?
    It is:
    The resistor R is replaced with an electromagnet and a switch. When the current is switched on, small pieces of iron, initially on the ground below the electromagnet, are attracted to, and stick to, the electromagnet.
    (A) State briefly the energy changes occurring from the time that the current is switched on.

    I'm not sure what they're asking for here. I think that the electromagnet will disperse heat right? So therefore the iron will heat up according to its specific heat (Q = mc * deltaT). The important thing is that the electromagnet turns the current that runs through it into something else. Is it in this case the electromagnetic field it is producing?
     
  12. Apr 26, 2009 #11
    Hmmm, well it seems to me that since the question asks for a brief statement you could just think about the energy changes taking place. You're right about the heat, and you've got the iron being pulled up against gravity. I think you've got the right idea.

    chemical energy stored in the cell ---> electrical energy in the circuit ---> heat due to the resistence of the wires and the coil ---> increase in kinetic energy of the iron & increase in the potential energy of the iron as it's pulled up against gravity (via the magnetic field that the current produces around the coil.)
     
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