Parallel GLC Circuits

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When a coil, capacitor, and resistor are connected in parallel (above figure), the resistance should be thought of as a conductance, whose value in siemens (symbolized S) is equal to the reciprocal of the value in ohms. Think of the conductance as all belonging to the inductor. Then you have two vectors to add, when finding the admittance of a parallel GLC (conductance-inductance-capacitance) circuit:
Y = (G + jBL ) + (0 + jBC)
= G + j (BL + BC)

Again, remember that BL is never positive! So, although the formulas here have addition symbols in them, you’re adding a negative number when you add in an inductive susceptance.

Problem 1

Suppose a resistor, a coil, and a capacitor are connected in parallel. Suppose the resistor has a conductance G = 0.10 S, and the susceptances are jBL = −j0.010 and jBC = j 0.020. What is the complex admittance of this combination?
Consider the resistor to be part of the coil. Then there are two complex admittances in parallel:
0.10 − j0.010 and 0.00 + j0.020. Adding these gives a conductance component of 0.10 + 0.00 = 0.10 and a susceptance component of −j0.010 + j0.020 = j 0.010. Therefore, the complex admittance is 0.10 + j0.010.

Problem 2

Suppose a resistor, a coil, and a capacitor are connected in parallel. Suppose the resistor has a conductance G = 0.0010 S, and the susceptances are jBL = −j0.0022 and jBC = j0.0022. What is the complex admittance of this combination?
Again, consider the resistor to be part of the coil. Then the complex admittances are 0.0010 − j0.0022 and 0.0000 + j0.0022. Adding these, the conductance component is 0.0010 + 0.0000 = 0.0010, and the susceptance component is −j0.0022 + j0.0022 = j 0. Thus, the admittance is 0.0010 + j 0. This is a purely conductive admittance.

Problem 3

Suppose a resistor, a coil, and a capacitor are connected in parallel. The resistor has a value of 100 Ω, the capacitance is 200 pF, and the inductance is 100 μH. The frequency is 1.00 MHz. What is the net complex admittance?
First, you need to calculate the inductive susceptance. Recall the formula, and plug in the numbers as follows:
jBL = −j[1/(6.28fL)]
= −j[1/(6.28 × 1.00 × 100)]
= −j0.00159

Megahertz and microhenrys go together in the formula. Next, you must calculate the capacitive susceptance. Convert 200 pF to microfarads to go with megahertz in the formula; thus C = 0.000200 μF. Then:
jBC = j(6.28fC )
= j(6.28 × 1.00 × 0.000200)
= j0.00126

Finally, consider the conductance, which is 1⁄ 100 = 0.0100 S, and the inductive susceptance as existing together in a single component. That means that one of the parallel-connected admittances is 0.0100 − j0.00159. The other is 0.0000 + j0.00126. Adding these gives 0.0100 − j0.00033.

Problem 4

Suppose a resistor, a coil, and a capacitor are in parallel. The resistance is 10.0 Ω, the inductance is 10.0 μH, and the capacitance is 1000 pF. The frequency is 1592 kHz. What is the complex admittance of this circuit at this frequency?
First, calculate the inductive susceptance. Convert the frequency to megahertz; 1592 kHz = 1.592 MHz. Plug in the numbers as follows:
jBL = −j[1/(6.28fL)]
= −j[1/(6.28 × 1.592 × 10.0)]
= −j0.0100

Next, calculate the capacitive susceptance. Convert 1000 pF to microfarads to go with megahertz in the formula; thus C = 0.001000 μF. Then:
jBC = j(6.28fC )
= j(6.28 × 1.592 × 0.001000)
= j0.0100

Finally, consider the conductance, which is 1/10.0 = 0.100 S, and the inductive susceptance as existing together in a single component. That means that one of the parallel-connected admittances is 0.100 − j0.0100. The other is 0.0000 + j0.0100. Adding these gives 0.100 + j 0.

Converting Complex Admittance to Complex Impedance

The GB plane is, as you have seen, similar in appearance to the RX plane, although mathematically they are different. Once you’ve found a complex admittance for a parallel RLC circuit, you will usually want to transform this back to a complex impedance. The transformation from a complex admittance G + jB to a complex impedance R + jX can be carried out using the following two formulas, one for R and the other for X:
R = G/(G2 + B2)
X = −B/(G2 + B2)

If you know the complex admittance, first find the resistance and reactance components individually using the preceding formulas. Then assemble the two components into the complex impedance, R + jX.

Problem 5

Suppose the complex admittance of a certain parallel circuit is 0.010 − j0.0050. What is the complex impedance of this same circuit, assuming the frequency does not change?
In this case, G = 0.010 S and B = −0.0050 S. First find G2 + B2, as follows:
G2 + B2 = 0.0102 + (−0.0050)2
= 0.000100 + 0.000025
= 0.000125

Now it is easy to calculate R and X, like this:
R = G/0.000125
= 0.010/0.000125
= 80 Ω
X = −B/0.000125
= 0.0050/0.000125
= 40 Ω

The complex impedance is therefore 80 + j40.