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16-6 When conductance
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is present along with susceptance, admittance vectors point at angles; they are neither vertical nor horizontal.
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You ve seen how vectors add in the RX plane. In the GB plane, the principle is the same. The net admittance vector is the sum of the component admittance vectors.
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Formula for Complex Admittances in Parallel Given two admittances, Y1 = G1 + jB1 and Y2 = G2 + jB2, the net admittance Y of these in parallel is their vector sum, as follows:
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Y = (G1 + jB1 ) + (G2 + jB 2 ) = (G1 + G2 ) + j(B1 + B 2 ) The conductance and susceptance components add separately. Just remember that if a susceptance is inductive, then it is negative imaginary in this formula.
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When a coil, capacitor, and resistor are connected in parallel (Fig. 16-7), 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.
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Problem 16-13 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 = j0.020. 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 = j0.010. Therefore, the complex admittance is 0.10 + j0.010.
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conductanceinductance-capacitance (GLC) circuit.
254 RLC and GLC Circuit Analysis
Problem 16-14 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 = j0. Thus, the admittance is 0.0010 + j0. This is a purely conductive admittance. Problem 16-15 Suppose a resistor, a coil, and a capacitor are connected in parallel. The resistor has a value of 100 , the capacitance is 200 pF, and the inductance is 100 H. 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/(6.28f L)] = 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 F. Then: jBC = j(6.28f C ) = 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 16-16 Suppose a resistor, a coil, and a capacitor are in parallel. The resistance is 10.0 , the inductance is 10.0 H, 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/(6.28f L)] = 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 F. Then: