Download Series RLC at resonance

Series RLC Circuit at Resonance.
A circuit is at RESONANCE when Xc = XL ie 2πfL = 1
2πfC
Also at resonance Vc =VL
Now if Xc =XL then Z = R,
For a supply voltage Vs = 25V the current is V
Z
VL = Vc = 1 x 100 = 100V ( ouch)
ie f2 =
1
4π2 CL
ie 25V = 1A
25Ω
Ex1 For the RLC circuit above plot a graph of Impedance Z against frequency
Frequency
Hz
500Hz
1kHz
1.3kHz
1.59kHz
1.8kHz
2.5kHz
Impedance 288
Ω
Applications of resonance effect
1. Most common application is tuning. For example, when we tune a radio to
a particular station, the LC circuits are set at resonance for that particular
carrier frequency.
2. A series resonant circuit provides voltage magnification
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Q Factor at Resonance
The ratio (capacitor voltage VC, or inductor voltage VL at resonance)/ voltage
source, is a measure of the quality of a resonance circuit.
This is known as Q factor of the circuit or the voltage magnification factor
Q= VL or Vc since VL = Vc ie Q = 100 = 4
VR VR
25
Q= XL = 2 . pi. f . L
R
R
Q=XC =
1
R
2. pi. f. R
Q= 2.pi.fc.L
R
Q= 1√ L
R C
Exercises
1. An AC voltage of 10V is applied across an RLC series circuit, where R=
20Ω, C = 47nF, L = 0.1mH.
Calculate: (1) the resonant frequency for the circuit,
(2) the Q Factor at resonance.
2. What is the resonant frequency and quality factor Q of a series circuit RLC,
R= 56ohms, L= 12 microhenrys and C is 40 picofarads?
3. Determine the resonant frequency and magnification factor Q of a series
RLC circuit in which R= 47Ω, L=12 mH and C= 65pF?
4. Find the value of capacitance C in a series RLC circuit when the circuit
resonant is 16.5 MHz and the inductance L is 4 µH?
5 Determine the value of a capacitor C in a series RLC circuit if the
inductance has 4 µH and the resonant frequency is at 3.56 MHz?
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6 Find the value of inductance L in a series RLC circuit when the circuit
resonant is 10 MHz and the capacitance C is 10 pF?
7 Calculate the value of an inductor L in a series RLC circuit, if the resonant
frequency is 23.7MHz and the capacitor has 7pF?
8 Draw a phasor diagram for the circuit. From the phasor diagram, determine the
(a) Applied Voltage (b) Power Factor
Confirm your answers by calculation
If the supply frequency is 100Hz, determine the values of R, L, C
9 A series RLC circuit which resonates at fr =500KHz has L= 100 µH, R= 25
Ω and C = 1000pF. Find the Q factor of the circuit. Calculate the new
value of C for resonance at 500 kHz when the value of L is doubled and
determine also the new Q factor.
10 If a series RLC circuit has L= 85 µH, C= 290pF, R= 100 Ω and VS =10 V.
Determine:
a) The resonant frequency (fc).
b) Calculate, also the circuit currents at 0.25 fc, 0.5 fc, 0.8 fc, 1.25 fc, 2 fc
and 4 fc.
c) Find for each resonant frequency (fc): XL, XC, XL - XC and Z.
d) Enter all quantities in a table.
e) Plot a graph of currents versus resonant frequencies.
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AC Parallel Circuits
Impedances in parallel
Y=1/Z
Y : admittance
Bc =1 /Xc
Bc : capacitive susceptance
BL=1/XL
BL: inductive susceptance
G= 1/R
G: Conductance
RC Circuits
The parallel RC circuit is generally of less interest than the series circuit. This is
largely because the output voltage Vout is equal to the input voltage Vin.
This circuit does not act as a filter on the input signal unless fed by a current
source.
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For a step input (which is effectively a 0 Hz or DC signal), the derivative of the
input is an impulse at t = 0. Thus, the capacitor reaches full charge very quickly
and becomes an open circuit— the well-known DC behaviour of a capacitor.
Parallel RL Circuits
In ac, a parallel RL circuit offers significant impedance to the flow of current. This
impedance will change with frequency, since that helps determine XL, but for any
given frequency, it will not change over time.
Current (I) is the sum of the currents IR and IL through R and L, keeping in mind
that the coil opposes any change in current through itself, so its current lags
behind its voltage by 90°. Therefore, our basic equation for current must be:
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I = V/Z= V/R –j(V/XL)
jωL=jXL
1/Z = 1/R +1/jXL
Z=(R x jXL) / (R + jXL)
= R . XL2/ (R2 + j XL2) + j[ R2. XL / (R2 + XL2)]
Summary
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Impedances (Z) are managed just like resistances (R) in parallel circuit
analysis: parallel impedances diminish to form the total impedance, using
the reciprocal formula. Just be sure to perform all calculations in complex
(not scalar) form! ZTotal = 1/(1/Z1 + 1/Z2 + . . . 1/Zn)
Ohm's Law for AC circuits: V = IZ ; I = V/Z ; Z = V/I
When resistors and inductors are mixed together in parallel circuits (just
as in series circuits), the total impedance will have a phase angle
somewhere between 0o and +90o. The circuit current will have a phase
angle somewhere between 0o and -90o.
Parallel AC circuits exhibit the same fundamental properties as parallel
DC circuits: voltage is uniform throughout the circuit, branch currents add
to form the total current, and impedances diminish (through the reciprocal
formula) to form the total impedance.
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