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LCR circuit calculations

Discussion in 'Electronics Homework Help' started by evol_w10lv, Apr 17, 2013.

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  1. evol_w10lv

    evol_w10lv

    73
    0
    Feb 19, 2013
    I have to calculate result of voltmeter and wattmeter. And also have to find effective current in the branches


    [​IMG]

    Given:
    e = 180 V
    R1 = 8 Ohms
    C1 = 0.0001 F
    L1 = 0.0318 H
    R2 = 4 Ohms
    L2 = 0.0159 H
    L3 = 0.0094 H
    f = 50 Hz

    Here I started calculations.

    XL1 = 2*pi*f*L1 = 2*pi*50*0.0318 = ~10 Ohms
    XC1 = 1/(2*pi*f*C1) = 1/(2*pi*50*0.0001) = ~31.8 Ohms
    XL2 = 2*pi*f*L2 = 2*pi*50*0.0159 = ~5 Ohms
    XL3 = 2*pi*f*L3 = 2*pi*50*0.0094 = ~3 Ohms

    Z1 = R1 + j(XL1 - XC1) = 8 + j(10-31.8) = sqrt(8^2 + 21.8^2) * e^(-jarctg(21.8/8)) = ~ 23e^(-j69.85°) Ohms
    Z2 = R2 + jXL2 = 4 + j5 = sqrt(4^2 + 5^2) * e^(jarctg(5/4)) = ~ 6.4e^(j51.34°) Ohms
    Z3 = jXL3 = j3 = 3e^(j90°)
    Z2 + Z3 = (Z3*Z2)/(Z3+Z2) = ((3e^(j90°))*(6.4e^(j51.34°)))/((3e^(j90°))+(6.4e^(j51.34°))) = 0.45 + j2.1 = ~2.16*e^(j77.91°) Ohms
    Ztotal = (Z2+Z3) + Z1 = (0.45 + j2.1)+(8 + j(-21.8)) = 8.45-19.7j = 21.4*e^(-j66.78°)

    Can someone can check weather imdependances are calculated wright? What is the next step to calculate current?
     
  2. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Attachment shows your Z values are close but beginning to accumulate round-off error. Next step is to calculate V from the voltage divider formed by Z1 & Z23, then with V find the current through Z2. Not sure what to assume about the wattmeter and power factor.
     

    Attached Files:

  3. evol_w10lv

    evol_w10lv

    73
    0
    Feb 19, 2013
    Round-off error is not so important, the main thing is understanding, that's why I will use my calculation results to continue the task.

    I1 through R1, L1, C1:
    I1 = E/Ztoatl = 180/(21.4*e^(-j66.78°)) = ~8.4*e^(j66.78°) A
    V23 = I1*Z23 = (8.4*e^(j66.78°)) * (2.16*e^(j77.91°)) = ~18.2*e^(j144.69°) V
    I2 through R2, L2:
    I2 = V23/Z2 = (18.2*e^(j144.69°))/(6.4e^(j51.34°)) = ~2.8*e^(j63.53°) A
    I3 through L3:
    I3 = V23/Z3 = (18.2*e^(j144.69°))/(3e^(j90°)) = ~6*e^(24.69°) A

    And then result of voltmeter: I3 * Z3 [without complex part] = 3 * 6 = 18V (18.2V from V23), or am I wrong?

    Speaking about wattemer, it shows active power P as we know. Maybe we can use complex full power
    S = U*I = 180* (8.4*e^(j66.78°)) = 1512*e^(j66.78°) = 1512*cos(66.78°) + j1512*sin(66.78°)
    then S = P + jQ = ~596 + j1389.5 VA
    And it means that active power P = 596 W
    Reactive power Q = 1389

    But also not sure about it. What do you think, how to get result of wattmeter?
     
  4. john monks

    john monks

    693
    2
    Mar 9, 2012
    Without the charactoristics of the wattmeter the currents cannot be figured out.
    But if this is for understanding only then maybe we can assume the wattmeter is a short. If this is the case then you can figure out the reactances of the capacitor and inductors using your 50 hz and figuring out the currents exactly as you would if there were only resistors in the circuit. The formulas are exactly the same except that you must include the phase angles, or reactances. When I took circuit in engineering school I simply used my TI-83 calculator to do all the formulas because it handles imaginary numbers beautifully.
    Sorry to say but if you still have problems solving this circuit then you need to get some capacitors, inductors, and resistors and play with them using 50 cycles to get a full understanding of the circuit.
    From spending many years in college I discovered that most the students did not have a problem with the math or the formulas, it was with understanding the electronics.
     
  5. The Electrician

    The Electrician

    116
    11
    Jul 6, 2012
    Most of your values have some rounding error, but the angles of I2 and I3 seem to have greater error than can be blamed on rounding. See the correct values in red.

    The voltmeter will read the magnitude of V23 (unless it's a vector voltmeter) which is 18.2 volts.

    I assume the wattmeter is to be considered ideal, meaning no current taken by the voltage coil and a zero resistance current coil.

    Then the wattmeter is measuring the power loss in the Z2 branch. All you have to do is use the standard I^2*R formula, where I is the magnitude of I2 (ignore the angle of I2).

    The magnitude of I2 is 2.78 amps, and the loss occurs in the resistor R2 which is 4 ohms, so I would expect the wattmeter to read 30.91 watts.
     
    Last edited: Apr 20, 2013
  6. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Again your results are close but accumulating round-off error. Also, need to check arithmetic for I2 & I3 phase angle. It seems setup is right but the result is different than what MathCAD gives.

    Attached is an extract from an instrumentation reference book explaining wattmeters. Note that the dynamometer gives a reading that is the product of the voltage and current and the cosine of the phase angle between them, so the wattmeter measures real power or I^2R loss.

    Does your answer for measured power equal the I^2R loss in the resistor R2? The answer should be the same no matter how it is calculated.
     

    Attached Files:

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