05 March, 2007

Vector Diagrams - RLC Series and Parallel circuit

RLC Series circuits
When we add a resistance to a series LC circuit, as shown in the schematic diagram to the right, the behavior of the circuit is similar to the behavior of the LC circuit with no resistance, but there are some variations.We wil see the affects of added resistance with the parameters given below
  • f = 1 MHz
  • e = 10 vrms
  • L = 150 µh
  • C = 220 pf
  • R = 100 Ω
With these measured voltages across R L and C we get are
  • vL = 39.1v
  • vC = 30.0v
  • vR = 4.15v
We must take into account the different phase angles between voltage and current for each of the three components in the circuit. The vector diagram to the right, while not to scale, illustrates this concept.
Since this is a series circuit, the current is the same through all components and is therefore our reference at a phase angle of 0°. This is shown in red in the diagram. The resistor's voltage, vR, is in phase with the current and is shown in green. The blue vector shows vL at +90°, while the gold vector represents vC, at -90°. Since they oppose each other diametrically, the total reactive voltage is vL - vC. It is this difference vector that is combined with vR to find vT (shown in cyan in the diagram).
We already know that vT = 10 vrms. Now we can see that vT is also the vector sum of (vL - vC) and vR. In addition, because of the presence of R, the phase angle between vT and i will be arctan((vL-vC)/vR), and can vary from -90° to +90°.

For calculations and more information visit the following site :

RLC Parallel circuits
The schematic diagram to the right shows three components connected in parallel, and to an ac voltage source: an ideal inductance, and ideal capacitance, and an ideal resistance. We will use the following values for our components
  • VAC = 10 vrms.
  • f = 1 MHz. ( = 6283185.3 rad/sec)
  • L = 150 µh. (XL = 942.4778 )
  • C = 220 pf. (XC = 723.43156 )
  • R = 1000
According to Ohm's Law:
iL =vL/XL = 10/942.4778 = 0.01061033 = 10.61033 mA.
iC =vC/XC = 10/723.43156 = 0.013823008 = 13.823008 mA.
iR = vR/R = 10/1000 = 0.01 = 10 mA.

If we measure the current from the voltage source, we find that it supplies a total of 10.503395 mA to the combined load — only about half a milliamp more than iR alone.
So we now have 10 mA of resistive current and just over 3.2 mA of reactive current, and yet the measured total current is just over 10.5 mA.

As usual, the vectors, shown to the right, tell the story. Since this is a parallel circuit, the voltage, v, is the same across all components. It is the current that has different phases and amplitudes within the different components.

Since voltage is the same throughout the circuit, we use it as the reference, at 0°. Current through the resistor is in phase with the voltage dropped across that resistor, so iR also appears at 0°.

Current through an inductor lags the applied voltage, so iL appears at -90°. Current through a capacitor leads the applied voltage, so iC appears at +90°. Since iC is greater than iL, the net reactive current is capacitive, so its phase angle is +90°.
Now the total current, iT, is the vector sum of reactive current and resistive current. Since iR is significantly greater than the difference, iC - iL, the total impedance of this circuit is mostly resistive, and the combined vector for iT is at only a small phase angle, as shown in the diagram.
For calculations and more information visit the following site :http://www.play-hookey.com/ac_theory/ac_rlc_parallel.html
The above are the cases of ideal inductor and capacitor.If those are not ideal the vector diagram wil be as shown below

(a)Series RLC circuit

(b)In (a) XL > XC, and the driving voltage (V) leads the current by a phase angle of f.
(c)In (b) XC > XL, and the driving voltage (V) lags the current by a phase angle of f.

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