R'V' Srinivasa Murthy

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• R.V. Srinivasa Murthy
• Assistant Professor
• Dept. of E C
• A.P.S. College of Engineering
• Bangalore

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Parallel Resonance

• A parallel resonant circuit is one in which a
  coil and a capacitance are connected in parallel
  across a variable frequency A.C. Supply. The
  response of a parallel resonant circuit is
  somewhat different from that of a series resonant
  circuit.

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1) A coil in parallel with a pure capacitor

• Consider the parallel circuit shown. Let ZL be
  the impedance of the coil given by ZL R j ?
  L
• Then YL 1 1 R-j ? L
• ZL R j ? L R2 ?2L2
• Similarly let ZC -j / ?C
  
•   YC 1 j ? C
• ZC
• Total admittance of the circuit YYL YC
• Y R-j ? L j ? C
• R2 ?2L2
  
• R j ? C - ? L
• R2 ?2L2
  R2 ?2L2

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At resonance the impedance ( or admittance) of
the circuit is purely resistive( or conductive)
.For this to be true the j part of equation (1)
should be zero.

•   ?o C - ?o L 0
  R2 ?02 L2
  ?0 C ?0 L
  
  R2 ?02 L2
    R2 ?02 L2 L
  C
  ?02 L/C - R2 1 - R2
  L2 LC
  L2 f0 1/ 2? v (1/LC R2 / L2 )

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Impedance at resonance We know that at
resonance the susceptive part of the admittance
is zero.

• Hence Y0 R
  R2 ?02 L2 But R2
  ?02 L2 L/C
• So Y0 RC/L or Zo L/RC
• where Zo is called the dynamic resistance.
• When coil resistance R is small, dynamic
  resistance of the parallel circuit becomes high.
  Hence the current at resonance is minimum. Hence
  this type of circuit is called rejector circuit.

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Frequency response characterisiticsThe
frequency response curve of a parallel resonant
circuit is as shown in the figure.

• We find that current is minimum at resonance.
• The half-power points are given by the points at
  which the current is v2 Ir.
• From the above characteristic it is clear that
  the characteristic is exactly opposite to that of
  series resonant.

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Quality factor ( Q-factor) The quality factor
of a parallel resonant circuit is defined as the
current magnification

• Q Current through capacitance at
  resonance Total Current at resonance
• IC0 / I0 V / ( 1/ ?0C ) V / Z0
  Z0?0C (L / RC) ?0C ?0L / RHence the
  expression for the Q- factor for both series and
  parallel resonant circuit are the same
• Also Band width f0 / Q

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A coil and a Practical Capacitor in parallel
Consider a parallel resonant circuit in which
the resistance of the capacitance is also
considered

• Impedance of the coil ZL RL j ?L
  YL 1 / ZL 1/ RLj ?L
  
  RL j ?L / RL2 ?2L2

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Impedance of the Capacitor ZC RC j /
?C RC j / ?C Rc2 1/?2C2

• Therefore total admittance Y YL YC
  (RL j ?L / RL2 ?2 L2) RC j /?C
  RC2 1/?2C2
• At resonance the susceptance part of the total
  admittance is zero, which gives
• 1/?oC ?oL
  
  RC2 1/?02C2 RL2 ?02L2
• 1/LC RL2 ?o2L2 ?o2 RC2 1/?02C2
  
• ?o2 ( RC2 L/C ) RL2/ LC 1/ C2

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?o2 1/LC(RL2 L/C ) ( RC2
L/C) ?0 1/v LC RL2 L/C v
( RC2 L/C) fo 1 ( RL2
L/C)   2? v LC v
( RC2 L/C)At Resonance the admittance is
purely conductive given by Y0 RL
RC RL2 ?20L2
RC2 1/?20C2
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Example 1 Determine the value of RC shown in
figure to yield Resonance

• Solution Let ZL be the impedance of the
  inductive branch Then ZL 10 j 10
• YL 1/ (10 j 10) 10 j 10 10 - j 20
• 102 102 200
  Let ZC be the impedance of
  the capacitive branch then ZC RC j 2 YC
  1 RC j 2 RC j 2
  RC2 4 
• Total admittance of the circuit Y YL YC
• For the circuit under Resonance the Susceptance
  part is zero ( 2/ RC2 4) - (10 / 200) 0

RC 6 ohms
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Example 2 An Impedance coil of 25 ohms
Resistance and 25 mH inductance is connected in
parallel with a variable capacitor.

• For what value of Capacitor C will the circuit
  resonate.
• If 90 volts,400 Hz source is used, what will be
  the line Current under these conditions?

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Solution ?0 2 ? f0 2? (400) 
?0 2 1 - R2
LC L 2 6.316 x
106 1 - R2
LC L 2   1 ?0 2
R2 LC L 2

6.316 x 106 252 / (25 x
10-3)2 7.316 x 106 C 5.467
?F Z0 L/RC (25 x 10-3)/ 25 x 5.467 x
10-6 182.89
ohms  I0 V0/ Z0 90/ 182.89 0.492 ampere
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Example 3 In a parallel resonant circuit R, L
and C are all in parallel Half power
frequencies are 103 and 118 rad / sec
respectively.

• The magnitude of impedance at 105 rad/ sec is 10
  ohms.
• Find R,L and C
• Solution The given circuit is an ideal
• parallel resonant circuit.
• Total admittance Y YR YL YC
• YR 1/R
• YL 1/ j XL
• YC 1/ -j XC
• So Y 1/R j (1/ XC - 1/ XL)

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At Resonance the j part of the equation is zero
(1/ XC - 1/ XL) 0 or XC XL
1/ 2 ? f0 C 2 ? f0 L

• Or f0 1/ 2?v LC
• Q IC / I or IL / I is the current
  magnification
• At Resonance Y0 1/ R or Z0 R
• Since (?2 - ?1) ?0/ Q 118 103 105
  / Q Q 7
• But Q ?0L / R 7 105 x L / 10 L 0.67 Henry
• ?0 1/v LC 105 1 / v 0.67 x C C 1.35 x 10-4
  Farad

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Effect of Generator resistance on Bandwidth and
Selectivity

• Consider parallel resonant circuit driven by a
  generator having internal resistance Rg as shown
  in figure.
• At anti- resonance the parallel resonant circuit
  behaves like a resistive network Zr L / CR
• Q
  ?rL / R 1 / ?rRC

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And B.W f2 f1 fr / QBut now as the
parallel circuit is driven by a generator having
internal resistance(Rg)

• Which comes in parallel with Zr at resonance.
  Hence the total impedance becomes Zr Rg
  Zr Rg ( L / CR)
  
  
  
  RgL L / C(
  R L/ CRg) -----------(1)
  RgCRL
• From equation (1) it is clear that driving a
  parallel resonant circuit with a Generator having
  internal Rg is nothing but to increase a
  resistance in series with an inductor by value L
  /RgC
• Hence Quality factor of parallel circuit driven
  by generator with internal Resistance Rg is given
  by Q ?rL (?rL / R)
• 
  R L / RgC ( 1 L/CR)
• Rg
  

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Q Q0 ( 1 Zr /
Rg)Bandwidth B.W fr / Q fr ( 1 Zr / Rg
) QB.W B.W ( 1 Zr / Rg)
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Example 4 A coil of 10 henry and Resistance of
10 ohm is in parallel with 100 picofarad
capacitor. The combination is connected across a
generator of 100 Volts, having internal
resistance 100 k? Determine (i)Voltage across
the circuit (ii) Band-width
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Solution fr 1/ 2? v (1/LC R2 /
L2 )
1/
2?v( 1 / 10 x 100 x 10-12 ) (102 / 102)

• 5.032 kH
• Dynamic resistance Z r L / CR
• 10 / 100 x 10-12 x 10 1010 ohms
• Q ?rL / R
• 2? x 5.032 x 103 x10 31617
• 10

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• Voltage across parallel circuit at Resonance
• 100 x 1010 100 Volts
• 1010 100 x103
• (ii) Band width fr / Q 1 Zr / Rg)
  5.032 x 103 1 1010 / 100 x 103
• 31617
• 15.91 kHz