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Homework Help: Solving for AC Voltmeters with a Pure Resistance and Pure Inductance

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    a) A pure resistance and a pure inductance are connected in series across a 100 V(rms) AC line. An AC voltmeter gives the same reading when connected across either the R or the L. What does it read? (Note: Meters read rms values.)

    b) The magnitudes of R and L are altered so that a voltmeter across L reads 50 V. What will the voltmeter read across R?

    2. Relevant equations

    3. The attempt at a solution
    Solution should be: a) 70.7V ; b) 86.7V
  2. jcsd
  3. Mar 30, 2013 #2
    The voltage across the line depends on a number of things:
    1) the position of the input waveform in terms of phase
    2) the total impedance of the line

    The total impedance of the line is the sum of the resistance and the inductance.

    You know that the average voltage on the line should be 100V under no load conditions.

    The voltage dropped across the resistor is a constant. It depends on the resistance of the resistor, as the current through the circuit is a fixed RMS current (same current flows through resistor, inductor and to/from source).

    the voltage dropped across the inductor depends on the rate of change of the current passing through it:
    v(t) = L (di/dt)

    So... we can say that the rate of change is pretty constant... as the frequency is constant... and because the question tells you that the two voltages are the same you can relate the inductance of the inductor and the resistance of the resistor.

    V1 = IR V2 = L(Di/dt)

    For the two things to have the same voltage... in terms of voltage divider theory.. they must have the same overall impedance...

    so IR = L(di/dt)

    You need to have values of R and L which provide the same numbers for all values of I.

    Lets consider I as a function of time I = sin(2*Pi*f*t)

    R*cos(2*Pi*f*t) = L d(sin(2*Pi*f*t))/dt

    we know that d/dx (sin (t) ) = cos(t)


    R*cos(2*Pi*f*t) = L*(2*Pi*f) cos(2*Pi*f*t))

    now we can divide....

    R = L (2*Pi*f)

    I guess f must be 1/2*Pi

    for R = L

    Not sure how to calculate an exact R or L without knowing the frequency...
    but if you knew the frequency and the R and the L you can find the voltages across them (which we know to be the same) without a problem.

    Hope this helps.
    Are you sure there are not more variables or given/known data??

    Good Luck :)
  4. Mar 30, 2013 #3


    User Avatar

    Staff: Mentor

    You should provide an attempt at a solution, if only your own thoughts about how to approach the problem.

    As a suggestion, investigate the voltage triangle for the circuit (similar to a power triangle, but places the real and reactive voltages on the "legs" of the triangle, with the hypotenuse representing the applied, or "apparent" voltage). This is analogous to a phasor analysis of the circuit.


    Attached Files:

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