Resonance : what is definition of Resonance meaning types Series resonant circuit , Parallel resonant circuit

define Resonance : what is definition of Resonance meaning types Series resonant circuit , Parallel resonant circuit examples solved question answers formula ?
Resonance
Resonance  A two-terminal network, in general offers a complex impedance consisting of resistive components. if a sinusoidal voltage is applied to such a network, the current is out of phase with the applied voltage. under special circumstances, however the impedance offered by the network is purely resistive. the phenomenon is called resonance and the frequency at which resonance takes place is called the frequency of resonance.
Classification of Resonance
The resonance may be classified into two groups

  1. Series resonant circuit
  2. Parallel resonant circuit

Series Resonance
Figure shows a series RLC circuit. A sinusoidal voltage V sends a current I through the circuit. the circuit is said to be resonant when the resultant reactance of the circuit is zero. the impedance  of the circuit at frequency 0 is
Z = R + J (0L – 1/0C) ……………….(i)
Then the current
I = V/R + J (0L – 1/0C) …………………(ii)
At resonance, the circuit must have unity power factor, i.e., 0L – 1/0C = 0
Hence,                   00L = 1/00C ……………….(3)
00 = 1/LC ………………….(4)
F0 = 1/2 LC ……………(5)
Where f0 is the frequency of resonance in hertz. Therefore, at resonance the current is
I0 = V/R ……………..(6)
The impedance of an inductor is ZL = J0L and is shown as a straight line in figure and that for the capacitor ZC = – J/00C, is a rectangular hyperbola. the impedance of L and C in series is
ZLC = (ZL + ZC)
and is shown in curve (c) of figure
For f < fo, ZLC becomes capacitive and for f > f0, ZLC becomes inductive, where fo is the resonant frequency.
The impedance of the entire circuit
Z = R + J(0L – 1/0C)……………(7)
|Z| = R2 + (0L – 1/0C)2
The curve is shown as (d) in figure.
Variation of current and voltage with frequency
The impedance of figure is
Z = R + J(0L – 1/0C)
The current is
I = V/R + J(0L – 1/0C)
Which at resonance becomes IO = V/R
Hence, the current is maximum at resonance.
The voltage across capacitor C is
VC = 1/JOC = 1/J0C |V/R + J (OL – 1/0C)|……………..(8)
Hence, magnitude
|VC| = V/OC R2 + (OL – 1/OC)2 ……………..(9)
The frequency fc at which VC is maximum may be obtained by equating dVC2 /doo to zero. this results in,
fc = (1/2) 1/LC – R2/2L2 ………….(10)
The voltage across inductor L
VL = V(JOOL)/R + J (OOL – 1/00C) ………….(11)
The magnitude
|VC| = VOL/R2 + (OL – 1/OC)2 …………….(12)
The frequency fl at which VL is maximum may be obtained by equating dVl2/do to zero.
thus,   fl = 1/2 (LC – C2R2/2 ………………(13)
Obviously, fl > fc
The variations VL and VC are of equal magnitude and opposite phase at resonance, as shown in figure. if R is extremely small, fl and fc tend to equal fo.
Features

  1. At resonant frequency the input impedance is purely resistive and has minimum value.
  2. At o0 the input voltage and the current are in phase.
  3. The current drawn by the network from the input voltage source is maximum.
  4. At resonance the voltage across the inductor and the voltage across the capacitor are equal in magnitude and are phase shifted by 1800.
  5. At 00 the network is purely resistive so that the power factor is unity.
  6. Above o0 the network is inductive so that the power factor is lagging.
  7. Below 00 the network is capacitive and therefore the power factor is leading.
  8. The resonant frequency represents the rate at which the electrical energy in the capacitor is transformed to the magnetic energy in the inductor and vice-versa.
  9. The quality factor (Q-factor) reqresents the voltage amplification factor and has a value much greater than unity.

Quality factor
Q0 = |VL|/V = |VC|/V
|VL| = |J0OL. IO| = 00 LI0
|VC|= 1/J00C . I0| = I0/00C
|V| = |I0R| = I0R
Q0 = 00L/R = 1/00RC = 1/R L/C
Frequency response of series resonant circuit
|I| = V/ZIN = V/|R + J(0L – 1/0C)|
|I| = V/R2 + (0L – 1/0C)2
0 = BW = 02 – 01
Q0 = 00/0
00 = 01 02
02 = 0/2 + 00
01 = 00 – 0/2
01 and 02 = cut-off requencies or = half power frequency or = 3 dB frequency
Bandwidth of series resonant circuit
QO = 00/0    0 = 00/Q0
0 = 1/LC   1/R 1/L/C, 0 = R/L

  1. The BW of the circuit represents the range of frequencies over which the current drawn by the circuit decreases to 0.707 of it’s maximum value.
  2. The bandwidth of the network depends upon R and L present in the circuit.

Example 1. A 50 resistor is connected in series with an inductor having internal resistance, a capacitor and 100 v variable frequency supply as shown in figure. At a frequency of 200 Hz, a maximum current of 0.7 Aflows through the circuit and voltage across the capacitor is 200 v. determine the circuit constants.
Sol. At resonance, current in the circuit is maximum
I = 0.7 A
Voltage across capacitor is VC = IXC
VC = 200, I = 0.7
XC = 1/0C
0C = 0.7/200
C = 2.785 uF
At resonance
XL = XC
XC = 1/0C = 285.7
XL = 0L = 285.7
L = 0.23 H
At resonance, the total impedance
Z = R + 50
R + 50 = V/I = 100/0.7
R + 50 = 142.86
R = 92.86
Parallel Resonance
Yin = 1/R + 1/J0L + J0C
YIN = 1/R + J(0C – 1/0L)
At 0 = 00; j-term = 0
00C – 1/00L = 0
00 = 1/LC rad/s
f0 = 1/2 LC  Hz
YIN |0 = 0 = Y0 = 1/R + J(0) (minimum value)
Z0 = 1/Y0 = R (maximum value)
I0 = V Y0 = V/R (minimum value)

  1. YIN |0 = 0 = 1/R (Real)
  2. 0 > 00 (capacitive network)
  3. 0 < 00 (inductiven network)

|IL| = |IC| and are out of phase by 1800 for 0 = 00
Features

  1. The resonant frequency is same as that of a series resonant circuit.
  2. At resonance the network is resistive and has maximum inpur impedance.
  3. The input voltage and current are in phase and therefore the power factor is unity.
  4. At 00 the current drawn by the network has a minimum value.
  5. At 00 the current through the inductor and through the capacitor are equal in magnitude and are phase shifted by 1800.
  6. At 00 the network is resistive and the power factor is unity.
  7. Above oo the network is capacitive and the power factor is leading.
  8. Below 00 the network is inductive and the power factor is lagging.
  9. The qualify factor represents current amplification factor and is always much greater than 1.

Quality factor
Q0 = |IL|/I0 = |IC|/IO
|IL| = V/J00L|= V/J00L
|IC| = |V/1/J00C|= V00C
I0 = V/R
Q0 = R/O0L = O0RC = R C/L
General results
Q0 = O0/O
0 = 02 – 01
00 = 0102
02 = 00 + 1/2
01 = 00 – 1/2
Bandwidth
0 = 00/Q0
0 = 1/RC

  1. The BW of a parallel resonant circuit depends only upon the values of R and C.
  2. The frequency of resonance is the geometric mean of the two half-power frequency.
  3. The resonant curve is a measure of selection of a particular frequency with specified gain.

Higher is a value of Q-factor, better is the selection of a particular frequency with specified gain and lower is the value of BW of network.
Example 2. Find the value of L at which the circuit resonates at a frequency of 1000 rad/s in the circuit shown in figure.
Sol.  Y = 1/10 – J12 + 1/5 + JXL
Y = 10/102 + 122 + 5/25 + X2L + J(12/102 + 122 – XL/ 25 + XL2)
XL/25 + XL2 = 12/102 + 122
X2L – 20.3 XL + 25 = 0
XL = 18.98 or 1.32
XL = 00L    L = 18.98 mH  1.32 mH
Intro Exercise – 6

  1. In a series RLC high Q circuit, the current peaks at a frequency

(a) equal to the resonant frequency
(b) greater than the resonant frequency
(c) less than the resonant frequency
(d) none of the above

  1. A series LCR circuit, consisting of R = 10, |XL| = 20 and |XC| = 20, is connected across an AC supply of 200 V (rms). the rms voltage across the capacitor is

(a) 200 < – 900 v
(b) 200 < + 900 v
(c) 400 < + 900 v
(d) 400 < – 900 v

  1. A series RLC circuit has a Q of 100 and an impedance of (100 + j0) at its resonant angular frequency of 107 rad/s. the values of R and L are

(a) 100, 10-3 h
(b) 10, 102 h
(c) 1000, 10 h
(d) 100, 100 h

  1. The current, I (t) through a 10 resistor in series is equal to [(3 + 4 sin(100 t + 450) + 4sin (300t + 600)] A. the rms value of the current and the power dissipated in the circuit are

(a) 41 A, 410 W, respectively
(b) 35 A, 410 W, respectively
(c) 5 A, 250 W, respectively
(d) 11 A, 250 W, respectively

  1. A DC voltage source is connected across a series RLC circuit. under steady conditions, the applied DC voltage drops entirely across the

(a) R only
(b) L only
(c) C only
(d) R and L combination

  1. The parallel RLC circuit shown in figure is in resonance. in this circuit

(a) |IR| < 1 MA
(b) |IR + IL| > 1 mA
(c) |IR + IC| < 1 mA
(d) |IR + IC| > 5 mA

  1. For an RLC series circuit R = 20, L = 0.6 H, the value of C will be

[CD = Critically damped, OD = Over damped, UD = under damed]
CD               OD               UD
(a) C = 6 mF C > 6 mF C < 6 mF
(b) C = 6 mF C < 6 mF C > 6 mF
(c) C > 6 mF C = 6 mF C < 6 mF
(d) C < 6 mF C = 6 mF C > 6 mF

  1. The circuit shown in figure is critically damped. the value R is

(a) 40
(b) 60
(c) 120
(d) 180

  1. In a series RLC circuit, R = 2 K, L = 1 H and C = 1/400uF. The resonant frequency is

(a) 2 x 104 Hz
(b) 1/x 104 Hz
(c) 104 Hz
(d) 2 x 104 Hz

  1. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. if each of R, L and C is doubled from its original value, the new Q of the circuit is

(a) 25
(b) 50
(c) 100
(d) 200

  1. For a circuit, the 3 dB frequencies are given as 372.1 Hz and 340.3 Hz then at resonance the center frequency will be

(a) 356.19 Hz
(b) 355.89 Hz
(c) 355.84 Hz
(d) 370.68 Hz

  1. Consider the circuit as shown below :

The Q-factor of the inductor is equal to
(a) 4.74
(b) 4.472
(c) 4.358
(d) 4.853

  1. A circuit which resonates at 1 MHz has a Q of 100. bandwidth between half-power point is

(a) 10 kHz
(b) 100 kHz
(c) 10 kHz
(d) 100 kHz

  1. In a series resonant circuit, VC = 150 V, VL = 150 V and VR = 50 V. what is the value of the source voltage?

(a) zero
(b) 50 V
(c) 350 V
(d) 200 V

  1. A lossy capacitor in series circuit model consists of R = 20 and C = 25 pF. the equivalent parallel model at 50 kHz will have values of R and C as

(a) 1 G, 10 pF
(b) 1.01 G, 20 pF
(c) 810.5, 25 pF
(d) 810 G, 25 pF

  1. The value of input frequency which is required to cause a gain equal to 1.5. the value is

(a) 20 rad/s
(b) 20 Hz
(c) 10 rad/s
(d) no such value exists

  1. For the circuit shown below

The frequency at which the voltage function results in a zero current is
(a) 100
(b) 12
(c) 20
(d) none of these

  1. If L = 2.5 mH, QO = 5 and C = 0.01 uF. The value of neper frequency is

(a) 2 x 105 Np/s
(b) 2 x 104 Np/s
(c) 2.5 x 105 Np/s
(d) 2.5 x 104 Np/s

  1. A particular band-pass function has a network function as H(s) = 3s/s2 + 4s + 3′ then its quality factor Q is defined by

(a) 3/4
(b) 2/3
(c) 3/2
(d) none of these

  1. In the given circuit, the voltage across R(VR) and inductor L(VL) are respectively

(a) 3.03 v, 10.5 v
(b) 2.52 v, 9.52 v
(c) 3.03 v, 9.52 v
(d) 2.52 v, 10.5 v
Answers with Solutions

  1. (a)

in a series RLC circuit, high is circuit, the current peaks at a frequency equal to the resonant frequency because at this condition circuit impedance minimum and current maximum.

  1. (d)

A series RLC circuit, R = 10 |XL| = 20, |VC| = 20 Circuit in resonance, so, current flowing in the circuit
I = VRMS/R = I(RMS)
I(RMS) = 20 A
rms voltage across capacitor = 20 x 20 = 400 V but voltage is lagging to current, so voltage aross C = 400 < – 900 V

  1. (a)

A-series RLC circuit has
Q = 100, Z = 100 + J0 = R,
00 = 107 rad/s, R = 100
Q0 = 00L/R   Q0 = 100 = 107 x L/100
L = 10-3 H

  1. (c)

current through the resistor
I (T) = 3 + 4 sin (100t + 450) + 4 sin (300t + 600)
RMS value of current = I21 (rms) + I22(rms) + I23 (ram)
= 32 + (4/2)2 + (4/2)2
I (rms) = 9 + 8 + 8
I (rms) = 5 A
power dissipated in circuit = I2 (rms) x R
= 25 x 10
= 250 W

  1. (c)

impedance offered by C  XC = 1/0C = 1/0.C = 0
Impedance offered by L   XL = 0L = 0.L = 0
So, DC voltage drop entirely across C

  1. (a)

At frequency of resonance the network is 100 k like a resistive network, input voltage and current are in phase; input admitiance has minimum value. so, that input impedance has maximum value, current drawn by network from mains has minimum value.
I = v 1/R2 + (0C – 1/0L)2
but   (0C – 1/0L) = 0 ………at resonance
so,    V = IRR
Current through inductor = – V/J0L, because current in lagging, with voltage.
so, |IR + IL| < 1; similarly |IR + IC| > 1 and |IR|< 1

  1. (a)

For series RLC circuit,  Condition of over damped
when,        C > 4L/R2
C > 4 x 0.6/400 or C > 6 mF
Critically damped when C = 6 mF
under damped when C < 6 mF.

  1. (b)

For series critically damped circuit
REQ = 4L/C = 4 x 4/10 m
R0 = 40
R ||120 = 40
R x 120/R + 120 = 40
3R = R + 120
2R = 120
R = 60

  1. (b)

we know that resonant frequency of a series RLC circuit is
fo = 1/2 lc
fo = 1/2  1 x 1/400 x 10-6 = 103 x 20/2 = 104 Hz
fo = 104 Hz

  1. (b)

Q = fo/BW
fo = 1/2 LC
BW = R/L (characteristic equation = s2 + R/L s + 1/LC)
Q = 1/R L/C
When, R, L, and C are doubled
Q = 1/2 Q
Q = 50

  1. (c)

0H = 2FH = 2337.97 rad/s
0l = 2fl = 2138.17 rad/s
00 = 0h ,0l
00 = 2235.3 rad/s
fo = 355.84 Hz

  1. (c)

we know that
QIN = L/CR2 – 1
QIN = 1/2 x 52 x 10-3 – 1
QIN = 1/50 x 10-3 -1 = 19
QIN = 4.358

  1. (a)

Q = f/f
f = f/q = 106/100
f = 10 kHz

  1. (b)

in a series resonant circuit at resonance
VC = VL
XC = XL
and both the voltage are cancelled out.
VS = 50 V

  1. (c)

we know that
QS = 1/0CSRS
f = 50 x 103
CS = 25 pF, R = 20
QS = 1/2 (50 x 103) (20) (25 x 10-12)
QS = 6366.2
For this large value of QS is
RP = RS Q2S = 810.5
CP = CS = 25 pF

  1. (d)

H (0) = VO/V1 = 1/1 + JoRC
Gain = 1/1 + 02R2C2
For  any value of 0, R, C
Gain < 1

  1. (d)

H(s) = I(s)/V(s) = 1/Z(s)
Z(s) = 2.5 + (5/3) (20/s)/(5/3) + (20/s)
Z(s) = 20s/s2 + 12 + 2.5
H(s) = (0.4) x s2 + 12/(s + 2) (s + 6)
the numerator is zero when s = + j12
hence, voltage function at this frequency result in a current of zero.
i.e.,  0 = 12 = rad/s

  1. (b)

the neper frequency is given by
a = 0o/2QO
00 = 1/LC
0O = 1/2.5 x 10-3 x 0.01 x 10-6
00 = 1/0.525 x 10-9 = 104/0.50 = 200 x 103 rad/s
a = 2 x 105/2 x 5
a = 2 x 104 Np/s

  1. (d)

H(s) = Ks/s2 + as + 3
then quality factor is given by = b/a
given, H (s) = 3s/s2 + 4s + 3
a = 4
b = 3
Q = 3/4

  1. (c)

XL = 2fl = 2 x 10
XL = 3.14 K
I = V/Z = V/R2 + (JXL)2
I = V/R2 + (0L)2
I = 10/ (1)2 + (3.14)2 = 10/3296
I = 3.03 mA
VL = I x XL = 3.03 x 3.14
VL = 9.52 V