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A series RLC circuit , driven with E_(rms)=120 V at frequency f_d=60.0 Hz, contains a resistance R=200 Omega, and inductance with inductive reactance X_L=80.0 Omega, and a capacitance with capacitive reactance X_C=150Omega. (a) What are the power factor cos phi and phase constant phi of the circuit ? |
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Answer» Solution :The power factor cos `phi` can be found from the resistance R and impedance Z (cos `phi` =R/Z). Calculations : To calculate Z, we use `Z=sqrt(R^2+(X_L-X_C)^2)` `=sqrt((200OMEGA)^2+(80.0 Omega- 150Omega)^2)` =211.90 `Omega` `cos phi = R/Z = (200 Omega)/(211.90 Omega)=0.9438 APPROX 0.944` Taking the inverse cosine then yields `phi=cos^(-1) 0.944=pm 19.3^@` The inverse cosine on a calculator givens only the positive answerhere , but both `+19.3^@` and `-19.3^@` have a cosine of 0.944. To determine which sign is correct , we must consider whether the current leads or lags the driving emf . Because `X_C gt X_L` , this circuit is mainly capacitive, with the current leading the emf. Thus, `phi` must be negative . `phi=-19.3^@` We could , instead, have found `phi`. A calculator would then have given us the answer with the minus sign. (b) What is the average rate `P_(avg)` at which energy is dissipated in the resistance ? There are TWO ways and two ideas to use : (1) Because the circuit is assumed to be in steady -state operation, the rate at which energy is dissipated in the resistance is equal to the rate at which energy is supplied to the circuit . `(P_(avg)=E_(rms) I_(rms) cos phi)`. (2) The rate at which energy is dissipated in a resistance R DEPENDS on the square of the rms current `I_(rms)` through it, `(P_(avg)=I_(rms)^2 R)`. First way : We are given the rms driving emf `E_(rms)` and we already know cos `phi` from part (a). The rms current `I_(rms)` is determined by the rms value of the driving emf and the circuit impedance Z (which we know ). `I_(rms)=E_(rms)/Z` . `P_(avg)=E_(rms)I_(rms)cos phi=E_(rms)^2/Z cos phi` `=(120V)^2/(211.90Omega)(0.9438)`=64.1 W Second way : Instead, we can write `P_(avg)=I_(rms)^2 R=E_(rms)^2/Z^2 R` `=(120V)^2/(211.90Omega)^2 (200Omega)`=64.1W (c) What new capacitance `C_"new"` is needed to maximize `P_(avg)` if the other parameters of the circuit are not changed? (1) The average rate `P_(avg)` at which energy is supplied and dissipated is maximized if the circuit is brought into resonance with the driving emf. (2) Resonance occurs when `X_C=X_L`. Calculations : From the given data, we have `X_C gt X_L` . Thus , we must decrease `X_C` to reach resonance .From `(X_C=1//omega_d C)`, we see that this means we must increase C to the new value `C_"new"`. we can write the resonance CONDITION `X_C=X_L` as `1/(omega_d C_"new")=X_L`. Substituting `2pif_d` for `omega_d` (because we are given `f_d` and not `omega_d` ) and then solving for `C_"new"`, we find `C_"new" = 1/(2pi f_d X_L)=1/((2pi)(60 Hz)(80.0 Omega))` `=3.32xx10^(-5)F=33.2 muF` Following the procedure of part (b), you can show that with `C_"new"`, the average power of energy dissipation `P_(avg)` would then be at its maximum value of `P_(avg. max)`=72.0 W |
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