Series RLC Filter: Phase Requirement versus Impedance Magnitude
A technician is tuning a series RLC filter connected to a 50 Hz AC supply. The resistance is R=20 Ω and the inductance is L=0.15 H. The supply angular frequency is ω=2π×50=100π rad/s. The impedance is Z=R+i(ωL−1/(ωC)), where C is the capacitance in farads. The filter is required to make the current lead the voltage by 30°, which means the impedance argument must be −30°. The technician first chooses C to make the impedance magnitude 25 Ω, because that value appears on a component rating, but then notices that two different positive capacitances can produce the same magnitude while giving opposite signs for the reactance. A second requirement is that the current must lead, not lag. You are asked to resolve the ambiguity, compute the required capacitance, and explain why the magnitude condition alone is insufficient. Use exact surds where useful and give C in microfarads to three significant figures.
1. Compute the inductive reactance and find the capacitance that makes the impedance argument exactly −30°.
Read the worked answer
The inductive reactance is X_L=ωL=100π(0.15)=15π≈47.124 Ω. For impedance Z=R+iX with X=X_L−1/(ωC), the argument is arctan(X/R). To have argument −30°, we need X=R tan(−30°)=20(−1/sqrt3)=−20/sqrt3≈−11.547 Ω. Hence X_L−1/(ωC)=−20/sqrt3, so 1/(ωC)=15π+20/sqrt3≈47.124+11.547=58.671 Ω. Therefore C=1/[100π(15π+20/sqrt3)]≈5.425×10^-5 F≈54.3 μF. This gives a negative reactance, so current leads voltage as required.
2. Solve |Z|=25 Ω for all positive capacitances. Identify which solution makes the current lead and which makes it lag.
Read the worked answer
For |Z|=25, R^2+X^2=25^2, so 20^2+X^2=625, hence X^2=225 and X=±15. Case 1: X=+15, so X_L−1/(ωC)=15. Then 1/(ωC)=15π−15≈32.124 Ω, C=1/[100π(15π−15)]≈9.91×10^-5 F≈99.1 μF. Here X>0, so the current lags the voltage. Case 2: X=−15, so 1/(ωC)=15π+15≈62.124 Ω, C=1/[100π(15π+15)]≈5.12×10^-5 F≈51.2 μF. Here X<0, so the current leads the voltage. Thus magnitude alone gives two capacitances with opposite phase behaviour; the 51.2 μF value is the lead solution, but it is not the same as the 54.3 μF value required for a −30° phase.
3. If the technician mistakenly uses the larger capacitance from the magnitude condition, what is the phase angle? Explain the consequence for the filter requirement.
Read the worked answer
The larger capacitance is about 99.1 μF, corresponding to X=+15 Ω. The phase angle φ satisfies tanφ=X/R=15/20=0.75. Thus φ=arctan(0.75)≈36.9°. Because X is positive, the impedance angle is positive and the current lags the voltage by about 36.9°. This does not satisfy the requirement that current lead by 30°: the sign is wrong and the magnitude of the phase is also different. The technician must choose capacitance by imposing the phase condition, not merely the impedance magnitude.