Physical Meaning of Leading/Lagging Voltage/Current

Discussion in 'Homework Help' started by bobby19, Jun 20, 2007.

  1. bobby19

    Thread Starter Member

    Jun 3, 2007
    13
    0
    Hey, I'm trying to grasp a more physical meaning of what exactly is meant by a leading/lagging voltage/current.

    In terms of capacitors and inductors, I understand mathematically that the differential i-v relationship causes one waveform to lead and the other to lag, but in a physical circuit sense, what does this mean? Would it correspond to some sort of time-delay?

    Thanks
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    If you pick a point in an AC circuit to measure instantaneous current and voltage you will notice that one of two situations will occur.
    1. The current and the voltage are in phase. Peaks and zero crossings happen at the same time. Physically this means that the circuit contains pure resistances only. There are no reactive components.
    2. The current and voltage have their zero crossings and peaks at different times. Physically this indicates the presence of reactive components in the circuit.
    The terms lead and lag are relative. If voltage is leading current then current is lagging voltage. This is also called a phase difference.

    This is the relatively simple case where voltage and current appear to be functions of time alone. In tranasmission lines and RF circuits, voltage and current can be functions of time and space.
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    This is an important point. If I leads V by 90 degrees, then conversely one could also say V leads I by 270 degrees - it is relative to which variable leads/lags the other. As a general rule you would not quote a lead or lag in phase as being more than 180 degrees, therefore in the above example it would be more accurate to state that I leads V by 90 degrees.

    Dave
     
  4. bobby19

    Thread Starter Member

    Jun 3, 2007
    13
    0
    Thanks for the explanation! So just to confirm my thoughts...If I did have a capacitor connected to a sinusoidal voltage source, the instant it is turned on, the charge moving from one plate to the other (current) is at a maximum (corresponds with the lead). As the capacitor voltage in increasing in correspondance with the source, the amount of displacement charge is steadily decreasing. Then when the voltage reaches its peak, the capicitor is fully energized, and as a result, charge is no longer being displaced, so current is zero.

    In other words, this situation would correspond to current having a cosine waveform and the voltage (lagging) having a sine waveform. Is my intuition serving me right?

    Thanks again
     
  5. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    This can also be seen from the following expression:
    Code ( (Unknown Language)):
    1.  
    2. i(t) = C*d(v(t))/dt
    3.  
    So if the voltage looks like:
    Code ( (Unknown Language)):
    1.  
    2. v(t) = A*sin(wt)
    3.  
    then, the current looks like
    Code ( (Unknown Language)):
    1.  
    2. i(t) = C*A*w*cos(wt)
    3.  
    n'est pas?
     
  6. bobby19

    Thread Starter Member

    Jun 3, 2007
    13
    0

    Oui, Oui! Merci! :)
     
Loading...