Series RLC circuits

Chapter Series RLC circuits

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

  • Calculating a vector sum using the formula is easier than doing it geometrically witha parallelogram. The arithmetic method is also more nearly exact. The resistance andreactance components add separately. That’s all there is to it.Series RLC circuitsWhen a coil, capacitor, and resistor are connected in series (Fig. 16-4), the resistance Rcan be thought of as all belonging to the coil, when you use the above formulas. (Think-ing of it all as belonging to the capacitor will also work.) Then you have two vectors toadd, when finding the impedance of a series RLC circuit:Z(RjXL)(0jXC)Rj(XLXC)288 RLC circuit analysis16-4A series RLC circuit.Problem 16-5A resistor, coil, and capacitor are connected in series with R50Ω, XL22Ω, and XC–33Ω. What is the net impedance, Z?Consider the resistor to be part of the coil, obtaining two complex vectors, 50 j22and 0 j33. Adding these gives the resistance component of 50 050, and the re-active component of j22j33j11. Therefore, Z50j11.Problem 16-6A resistor, coil, and capacitor are connected in series with R600Ω, XL444Ω, andXC444Ω. What is the net impedance, Z?Again, consider the resistor to be part of the inductor. Then the vectors are 600 j444 and 0 j444. Adding these, the resistance component is 600 0600, and thereactive component is j444j444j0. Thus, Z600j0. This is a purely resistiveimpedance, and you can rightly call it “600 Ω.”Problem 16-7A resistor, coil, and capacitor are connected in series. The resistor has a value of 330 Ω,the capacitance is 220 pF, and the inductance is 100 µH. The frequency is 7.15 MHz.What is the net complex impedance?First, you need to calculate the inductive and capacitive reactances. Rememberingthe formula XL6.28fL, multiply to obtainjXLj(6.28× 7.15× 100)j4490