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# Alternating Currents: LRC circuits, resonance

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An L-R-C series circuit is connected to an AC source of constant voltage amplitude V and variable angular frequency w.

A) Show that the current amplitude, as a function of w

See Eq 1

B) Show the average power dissipated in the resistor

See Eq 2

C) Show that I and P are both maximum when the source frequency equals the resonance frequency of the circuit.

See Eq 3

D) Graph P as a function of for V=100V, R=200 ohms, L=2 H, and C= 0.5 micro F. Discuss the behavior of I and P in the limits w=0 w-->infinity

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LRC series circuit 5
An L-R-C series circuit is connected to an AC source of constant voltage amplitude V and variable angular frequency w.

A) Show that the current amplitude, as a function of w

See Eq 1
The Net oppose to the current in an A.C. circuit is not only due to the resistance but due to the inductance and capacitance as well. As there is no loss of energy against the inductance and capacitance their oppose to the current is called reactance. The reactance depends on the frequency of the current. If the angular frequency of the source is w then
The inductive reactance is given as and
The capacitive reactance is given by
The current at any instant in the circuit I remain same through the circuit. The voltage across the inductor is ahead in phase then current by 900 and lagging behind by 900 across the capacitor. Thus using phasor diagram the resultant voltage at that instant is given as

Or
Or
Or
Thus for V as voltage amplitude the current amplitude is given by

Here
Gives the total oppose to the current in the circuit due to resistance and reactance and is called impedance of the circuit denoted by Z.
Thus

B) Show the average power dissipated in the resistor

See Eq 2
The power is consumed only in the resistor. As the power dissipated in a resistor is given by i2R and the current is continuously varies with time, the power dissipated also varies with time as p = (I sin t)2*R, where I is the current amplitude. The average power for a complete cycle of period T is thus given by

Or
Or
Or
Or
As  = 2/T we get

Or
Substituting the current amplitude I from part (A) we get average power as

Or

C) Show that I and P are both maximum when the source frequency equals the resonance frequency of the circuit.

See Eq 3

Both the current and power in the circuit varies with the angular frequency w. As w appears in the impedance, for maximum current amplitude and average power the impedance of the circuit must be a minimum.
The minimum value of a square term can be zero only and as R is a constant, for minimum impedance we must have

Or
Or

D) Graph P as a function of for V=100V, R=200 ohms, L=2 H, and C= 0.5 micro F. Discuss the behavior of I and P in the limits w=0 w-->infinity
For the given data we get the average power P as a function of (I think) as

Or
Or
The qualitative graph is shown bellow

For  =0 the capacitive reactance becomes infinite while inductive reactance is zero and thus the total impedance of the circuit is infinitely large. Here capacitor behaves as non-conducting and the current in the circuit becomes zero makes power dissipated equal to zero.
As  increases the capacitor starts conducting and the current increases. With increase in  the inductive reactance increases but the total impedance decreases and current increases till = 1000 rad/s.
At the reactance becomes zero the impedance Z is minimum and the current and power dissipated in the circuit is maximum given by
P = V2/(2R) = 25 W
When the inductive reactance is more than the capacitive reactance and the impedance increases results in decrease in the power dissipated.
When w is infinite the capacitive reactance is zero but inductive reactance is infinite and thus the impedance is infinite. Gives the current and the power in the circuit equal to zero.

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