Question #176810

1.a) Consider a series RLC circuit for which R= 1.50x10^2 ohms, L= 20mH, Vrms= 20V, and f= 796Hz. Determine the value of the capacitance for which the rms current is a maximum.

b) A series RLC ac circuit has resistance 2.50x10^2 ohms, inductance 0.60H, capacitance 3.50x10^-6F, frequency 60Hz and maximum voltage 1.50x10^2V.

Find the:

i) impedance

ii) maximum current in the circuit

iii) phase angle and its interpretation

iv) maximum voltage across the elements


1
Expert's answer
2021-03-31T09:39:11-0400

a) The current is maximum when the reactance of the capacitance and inductance are equal:


XC=XL, 1ωC=ωL, C=1ω2L=1(2πf)2L=1(2π796)20.02=8001012 F.X_C=X_L,\\\space\\ \frac{1}{\omega C}=\omega L,\\\space\\ C=\frac{1}{\omega^2L}=\frac{1}{(2\pi f)^2L}=\frac{1}{(2\pi·796)^20.02}=800·10^{-12}\text{ F}.

b) i) The impedance is


Z=R2+(XLXC)2= =R2+(2πfL12πfC)2=588 Ω.Z=\sqrt{R^2+(X_L-X_C)^2}=\\\space\\=\sqrt{R^2+\bigg(2\pi fL-\frac{1}{2\pi fC}\bigg)^2}=588\space\Omega.

ii) The maximum current is at resonance when the reactances are equal, simply


I=VR=0.6 A.I=\frac VR=0.6\text{ A}.

iii) Phase angle:


ϕ=atanXLXCR=65°,\phi=\text{atan}\frac{X_L-X_C}{R}=-65°,

which means that the circuit has a capacitive character.

iv) The maximum voltages are at the resonance, at the frequency of 690 Hz:


VR=IR=150 V,VL=VC=IXL=IXC=1561 V.V_R=IR=150\text{ V},\\ V_L=V_C=IX_L=IX_C=1561\text{ V}.

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