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Phase difference in CR and LR circuits

  1. Oct 2, 2006 #1
    I have just done an experiment on CR and LR circuits.

    http://img9.picsplace.to/img9/21/RC.jpg [Broken]
    Connect channel 1 of the CRO across both the capacitor and resistor and channel 2 across the resistor.The trace on channel 1 is taken as the p.d. across the capacitor and that on channel 2 as ths current throught the capacitor.

    http://img9.picsplace.to/img9/21/RL.jpg [Broken]
    Replace the capacitor waith a high inductance coil fitted on a double C-core.

    I have some questions about this.

    Why is channel 1 is connected across both the capacitor and resistor and not only the capacitor?

    Why the trace on channel 2 gives the current throught the capacitor?

    When we increased the resistance value, the phase difference decreased.Why?

    I have so many question to ask:confused: ..
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 3, 2006 #2
    I have thought about this today.

    (1) The channel 2 will be earthed if channel 1 is only connected the capacitor. Thus, we cannot observe the trace 2 and cannot compare the phase difference.

    (2) There is no phase difference between trace 2 and the current since this is the voltage across the resistor only. Therefore, trace 2 gives the current through the capacitor.

    (3)I have no idea about the third question...:cry:
  4. Oct 3, 2006 #3


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    Homework Helper

    One of the leads of the oscilloscope is earthed. This means that one observe the potential differences at certain points in a circuit relative to earth. Channel 1 will thus in effect show the ouput voltage of the signal generator and channel 2 that of resistor, which in effect is the current in the circuit since the current through the resistor and the potential over a resistor is in phase - so your reasoning is correct since it is a series circuit.

    Remember the what the RC charge - discharge curves look like? With a larger resistance less current flows in the circuit so the capacitor need to store less charge and will be on the parts of the curve where it "responds quicker" than if more charge would have been stored.
    Last edited: Oct 3, 2006
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