qr code generator vb.net Use the results to sketch approximate curves of A versus d, for Q 10 and Q 20. in Visual Studio .NET

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Parallel Resonance
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" In this section we propose to analyze the parallel RLC network of Fig. 167, where Zp is the input impedance to the network, that is, the impedance a generator would see if connected to terminals a, b. The circuit of Fig. 167 is important because it nds wide use as a tuned load at higher frequencies, in both receiver and transmitter work. Note that resistance is assumed to exist only in the inductive branch of the circuit. This is because inductors (coils), being wound of wire, inherently have much more power loss* than capacitors (it being easy to obtain practically lossless capacitors).
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* The resistance R in Fig. 167 is the sum of the actual resistance of the coil itself and any resistance coupled into L due to transformer action (when L is the primary of a transformer).
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CHAPTER 8 Reactance and Impedance
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Fig. 167
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Now, applying the standard procedure for two impedances in parallel (product of the two, over the sum), we have that, in Fig. 167, jXC R jXL " Zp R j XL XC 249
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Now rationalize the above fraction; that is, multiply the numerator and denominator by R j XL XC . Doing this, then setting XL !L and XC 1=!C, you can verify that eq. (249) becomes " !# 1 R !2 L2 !L R2 " 2 2 j Zp 250   !C !C ! C 1 2 !2 C 2 2 R !L !C We now obviously have a considerably more complicated condition than we had for the series case of section 8.6. Let us therefore try to put eq. (250) in a somewhat better form, especially in regard to the quantity !L 1=!C . In an e ort to do this, let us rst write the imaginary component, inside the brackets, in the equivalent form     !2 L2 !C !LC R2 !C !C L 2 2 2 2 j 2 2 !2 L2 R2 j C !2 C 2 ! C C ! C ! C Thus, 1=!2 C 2 now factors out of the entire quantity inside the brackets in eq. (250); doing this, and remembering the algebraic fact that, A2 B2 AB 2 , you can verify that eq. (250) becomes   L 2 2 2 !C ! L R R C " j 251 Zp 2 2 2 2 2 2 2 2 2 R ! C ! LC 1 2 R ! C ! LC 1 To continue, let us now DEFINE that the resonant frequency, for the parallel case of " Fig. 167, is the frequency at which Zp becomes a PURE RESISTANCE. Let us denote the resonant frequency by !0 ; then, by de nition, !0 is the frequency at which the imaginary or reactive component of eq. (251) becomes equal to zero. Note that this requirement will be satis ed if the numerator of the imaginary part of eq. (251) is equal to zero; that is, if !C 0 or if  L ! L R2 C
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CHAPTER 8 Reactance and Impedance
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The rst possibility, !0 C 0, is true only if ! 0 or if C 0, and hence is of no practical importance. However, setting ! !0 in the second possibility gives, as you can verify, the meaningful, correct answer s 1 R2 2 252 !0 LC L in which only the positive value of the square root is to be taken (because negative frequency does not exist in the real world). Now let ! 0 ( omega prime ) be the frequency at which the reactances of the coil and capacitor are equal; that is, let ! 0 be the frequency at which !L 1=!C. Thus, ! 0 L 1=! 0 C; hence ! 02 1=LC and, putting this value in place of 1=LC, eq. (252) becomes s R2 !0 ! 02 2 L
253
" Thus, as eq. (253) shows, in the parallel circuit of Fig. 167 the frequency at which Zp is 0 a pure resistance is not the same as the frequency at which XL XC .* Thus !0 ! only for the theoretically ideal case of R 0. We have de ned that the resonant frequency for the parallel case is the frequency, " !0 , for which Zp in Fig. 167 becomes a pure resistance, which let us now denote by R0 . " Hence, if we set ! !0 in eq. (251), the imaginary part vanishes and Zp becomes equal to R0 ; thus R0 R R2 !2 C2 0 !2 LC 1 2 0
Now, in the above equation, replace !0 with the right-hand side of eq. (252); doing this, you can verify the important fact that R0 L ohms RC 254
where R0 is the pure resistance a generator sees looking into terminals a, b in Fig. 167, if the generator is operating at the frequency de ned by eq. (252). From inspection of eq. (254), note that the smaller the value of R, the larger is the value of R0 ; this is one reason why Fig. 167 is especially useful in certain practical applications (as will be commented on in the solution to the following problem). Problem 144 In Fig. 167, suppose L 100 microhenrys, C 100 picofarads (100 pF), and R 50 " ohms. What value of Zp would a generator, operating at a frequency given by eq. (252), see if connected to terminals a, b To continue our study of Fig. 167, it will be instructive to deal with the dimensionless " ratio of the general value of Zp (given by eq. (251)) to its value at resonance, R0 (given by
* This is di erent from the series case of Fig. 159, in which the generator sees a pure resistance at the same frequency at which XL XC . (See discussion just prior to eq. (238) in section 8.6.)
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