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Pure Susceptances Pure inductive susceptances (BL ) and capacitive susceptances (BC ) add together when coils and capacitors are in parallel Thus, B = BL + BC Remember that BL is never positive, and BC is never negative This is just the opposite situation from reactances In the GB plane, pure jBL and jBC vectors add Because such vectors always point in exactly opposite directions inductive susceptance down and capacitive susceptance up the sum, jB, inevitably points either straight down or straight up (Fig 16-5), unless the susceptances are equal and opposite, in which case they cancel and the result is the zero vector Problem 16-9 Suppose a coil and capacitor are connected in parallel, with jBL = j005 and jBC = j008 What is the net susceptance Just add the values as follows: jB = jBL + jBC = j005 + j008 = j003 This is a capacitive susceptance, because it is positive imaginary Problem 16-10 Suppose a coil and capacitor are connected in parallel, with jBL = j060 and jBC = j025 What is the net susceptance

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Again, add the values: jB = j060 + j025 = j035 This is an inductive susceptance, because it is negative imaginary

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Problem 16-11 Suppose a coil of L = 600 H and a capacitor of C = 150 pF are connected in parallel The frequency is f = 400 MHz What is the net susceptance First calculate the susceptance of the inductor at 400 MHz, as follows:

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jBL = j[1/(628f L)] = j[1/(628 400 600)] = j000663 Next, calculate the susceptance of the capacitor (converting its value to microfarads) at 400 MHz, as follows: jBC = j(628f C ) = j(628 400 0000150) = j000377 Finally, add the inductive and capacitive susceptances to obtain the net susceptance: jB = jBL + jBC = j000663 + j000377 = j000286 This is a pure inductive susceptance

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Problem 16-12 What is the net susceptance of the above parallel-connected inductor and capacitor at a frequency of f = 531 MHz

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First calculate the susceptance of the inductor at 531 MHz, as follows: jBL = j[1/(628f L)] = j[1/(628 531 600)] = j000500 Next calculate the susceptance of the capacitor (converting its value to microfarads) at 531 MHz, as follows: jBC = j(628f C ) = j(628 531 0000150) = j000500 Finally, add the inductive and capacitive susceptances to obtain the net susceptance: jB = jBL + jBC = j000500 + j000500 = j0 This means that the circuit has no susceptance at 531 MHz The situation in which there is no susceptance in an LC circuit is known as parallel resonance It is discussed in the next chapter

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