RLC CIRCUIT, Impedances and Phasors.
Cuthbert Nyack
The series RLC circuit is shown schematically above.
The Impedance of the circuit is
Z = R + jwL + 1/jwC
= R + j(wL - 1/wC)
The magnitude of the impedance of the circuit is given by:-
|Z| = Ö{R2 + (wL - 1/wC)
2 }
The phase of the impedance of the circuit is given by:-
Ðq = atan
{(wL - 1/wC)
/R}
The applet below shows several properties of series RLC circuits in the
frequency domain.
Fn = 0, Frequency variation of the component voltages and phases on a Bode plot. The inductor is assumed to have zero resistance.
eg parameters (1.0, 0.05, na, na, na, 375, na, 0, na, 2.5) show
resonance when Q ~10. Both Vc and Vl are ~ 10 * Vr. The voltages
across the components reach their peak value at 1rad/s.
eg parameters (1.0, 0.4, na, na, na, 375, na, 0, na, 7.0) show
resonance when Q ~1.25. Both Vc and Vl are ~ 1.25 * Vr at
resonance. Vc
reaches its peak value at ~0.822rad/s, Vr at 1rad/s and Vl at
~1.218rad/s. This is equivalent to the displacement, velocity and
acceleration of mechanical oscillators reaching their peak
values at slightly different frequencies.
Fn = 1, Real, imaginary part and magnitude of the circuit impedance.
eg parameters (1.0, 0.05, na, na, na, 375, na, 1, na, 1.0) show
that away from resonance, the magnitude of the impedance is very
close to the magnitude of the imaginary part of the impedance.
At resonance the impedance is a minimum and equal to the value of the resistance.
Fn = 2, Real, imaginary part and magnitude of the current.
eg parameters (1.0, 0.3, na, na, na, 375, na, 2, na, 8.0) show
the real part and magnitude of the current is a maximum at resonance
and equal to Vs/R. The imaginary part of the current peaks at
~0.7453rad/s and ~1.343rad/s respectively and is zero at resonance. The
power dissipated by the circuit is a maximum at resonance.
Fn = 3, the component voltages at w as
a phasor diagram.
eg parameters (1.0, 0.05, 1.0, na, na, na, 1.2, 3, na, na) show
the phasors at resonance.
eg parameters (1.0, 0.05, (0.951:1.051), na, na, na, 4.0, 3, na, na)
show the Q being determined from the half power points, giving a
value of 10.0.
Fn = 4, Frequency variation of the component voltages and phases on a Bode plot when the resistance of the inductor is included.
eg parameters (1.0, 0.2, na, na, 1.0, 375, na, 4, na, 5.0) show
an asymmetry has developed between Vc and Vl and Vr has decreased.
The phase of the inductor voltage now goes to that of the resistor
at low frequencies.
Fn = 5, Frequency variation of the component voltages and phases on a Bode plot with the resistance of the inductor included.
eg parameters (1.0, 0.1, 1.0, na, 1.0, na, 2.0, 5, na, na) show
the phasors at resonance for the case where Q ~ 5. The voltage
across the capacitor Vcp = 5V at -90°, across the inductive part of the
inductor Vlp = 5V at +90°, across the resistive part of the inductor is
0.5V at 0. Total voltage across the inductor Vlt = 5.02493V at 84.2894°, voltage across the resistor Vrp = 0.5V, net voltage across the capacitor and the inductive part of the inductor Vlcp = 0V and the total
voltage across the circuit is Vtp = Vs = 1V.
When applet is enabled its appearance is illustrated by the gif image below.
Return to main page
Return to page index
COPYRIGHT © 2007 Cuthbert A. Nyack.