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Capacitor - Resistor circuit.

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    This is an experiment Ill have to write up later, unfortunately I was out for 2 weeks of lectures due to illness. This is the experiment;

    Find the values of capacitance C and resistance R in a series combination, given he following apparatus;

    1. the unknown RC combination (both in the same enclosed box),
    2. a known, reference resistor, Rref,
    3. a variable frequency signal generator (0-1000Hz),
    4. a digital multimeter

    2. Relevant equations

    I = (Emax . Sin(wt + &)/ |Z|

    3. The attempt at a solution
    I really am completely stumped guys... I just need to know how to do the procedure and then the calculations should be ok... Any input would be GREATLY appreciated.
  2. jcsd
  3. Mar 2, 2008 #2
    We are completely stumped about how we can help you unless you can at least provide a start. It looks like you are supposed to use impedances to find the values, so just use the equations for impedances, and then map out what certain voltages would look like with different frequencies. If you have an op amp then the gain will roll off around the cutoff frequency, and all you have to do if find that cutoff frequency (the -3dB point). If it is just a voltage source you could make say a low pass filter, and find that cutoff frequency.

    [tex] Z_c = \frac{1}{j \omega C}[/tex]

    [tex]Z_R = R[/tex]
  4. Mar 2, 2008 #3
    Ok, so I guess Im making progress anyway. The resistor will not affect phase at all, so the impedance of that is just R... The capacitor shifts phase by +pi/2... The impedance being 1/wC, and w= 2pi.f.... I can use the multimeter to find the ac frequencies and voltages in the circuit (im not allowed to use it for current). Im still a little stumped at how to approach this though, how does the reference resistor come into play?
  5. Mar 2, 2008 #4
    Good so far, what kind of circuit do you want to set up? The reference resistor must come into play because you theoretically don't know its impedance, right?
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