Reducing complicated RLC circuits

Chapter Reducing complicated RLC circuits

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

  • Proceed by the steps as numbered above.1. G1/R1/ BL1/(6.28fL)1/(6.28× 1.00× 10.0) = 0.0159.3. BC6.28fC6.28× 1.00× 0.0008200.00515. (Remember to convertthe capacitance to microfarads, to go with megahertz.)4. BBLBC0.01590.005150.0108.5. First find G 2 B20.1002(-0.0108)20.010117. (Go to a couple ofextra places to be on the safe side.) Then RG/0.010117 = 0.100/0.010117= 9.88, and XB/0.0101170.0108/0.0101171.07.6. The vector RjX is therefore 9.88 j1.07. This is the complex impedanceof this parallel RLC circuit.Problem 16-19A resistor of 47.0 Ω, a capacitor of 500 pF, and a coil of 10.0 µH are in parallel. What istheir complex impedance at a frequency of 2.252 MHz?Proceed by the steps as numbered above.1. G1/R1/47.00.0213.2. BL1/(6.28fL)1/(6.28× 2.252× 10.0)0.00707.3. BC6.28fC6.28× 2.252× 0.0005000.00707.4. BBLBC0.007070.007070.5. Find G2B 20.021320.00020.00045369. (Again, go to a couple ofextra places.) Then RG/0.000453690.0213/0.0004536946.9, and XB/0.000453690.6. The vector RjX is therefore 46.9 j0. This is a pure resistance, almostexactly the value of the resistor in the circuit.Reducing complicated RLC circuitsSometimes you’ll see circuits in which there are several resistors, capacitors, and/or coilsin series and parallel combinations. It is not the intent here to analyze all kinds of bizarrecircuit situations. That would fill up hundreds of pages with formulas, diagrams, and cal-culations, and no one would ever read it (assuming any author could stand to write it).A general rule applies to “complicated” RLC circuits: Such a circuit can usually bereduced to an equivalent circuit that contains one resistor, one capacitor, and one in-ductor.Series combinationsResistances in series simply add. Inductances in series also add. Capacitances in seriescombine in a somewhat more complicated way. If you don’t remember the formula, it is1/C1/C11/C2…1/Cnwhere C1, C2, …, Cn are the individual capacitances and C is the total capacitance. Onceyou’ve found 1/C, take its reciprocal to obtain C.Reducing complicated RLC circuits295