Converting from admittance to impedance

Chapter Converting from admittance to impedance

Teach Yourself Electricity and Electronics Third Edition Book
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Teach Yourself Electricity and Electronics Third Edition Book

  • 1/R1/10.00.100.) Let the capacitance alone be the other vector, 0 j0.0100. Thenthe sum is 0.100 – j0.0100j0.01000. 100 j0. This is a pure conductance of 0.100siemens.Converting from admittance to impedanceThe GB plane is, as you have seen, similar in appearance to the RX plane, although math-ematically the two are worlds apart. Once you’ve found a complex admittance for a par-allel RLC circuit, how do you transform this back to a complex impedance? Generally, itis the impedance, not the admittance, that technicians and engineers work with.The transformation from complex admittance, or a vector GjB, to a complex im-pedance, or a vector R jX, requires the use of the following formulas:RG/(G2B2)XB/(G2B 2)If you know the complex admittance, first find the resistance and reactance com-ponents individually. Then assemble them into the impedance vector, RjX.Problem 16-17The admittance vector for a certain parallel circuit is 0.010 –j0.0050. What is the im-pedance vector?In this case, G0.010 and B0.0050. Find G2 B2 first, because you’ll need touse it twice as a denominator; it is 0.0102( 0.0050)20.0001000.0000250.000125. ThenRG/0.0001250.010/0.00012580XB/0.0001250.0050/0.00012540The impedance vector is therefore RjX80j40.Putting it all togetherWhen you’re confronted with a parallel RLC circuit, and you want to know the compleximpedance RjX, take these steps:1. Find the conductance G = 1/R for the resistor. (It will be positive or zero.)2. Find the susceptance BL of the inductor using the appropriate formula. (Itwill be negative or zero.)3. Find the susceptance BC of the capacitor using the appropriate formula. (Itwill be positive or zero.)4. Find the net susceptance B = BL + BC. (It might be positive, negative, or zero.)5. Compute R and X in terms of G and B using the appropriate formulas.6. Assemble the vector R + jX.Problem 16-18A resistor of 10.0 Ω, a capacitor of 820 pF, and a coil of 10.0 µH are in parallel. The fre-quency is 1.00 MHz. What is the impedance RjX ?294 RLC circuit analysis