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 Z_L = J₀L and is shown as a straight line in figure and that for the capacitor Z_C = – J/₀₀C, is a rectangular hyperbola. the impedance of L and C in series is
Z_LC = (Z_L + Z_C)
and is shown in curve (c) of figure
For f < f_o, Z_LC becomes capacitive and for f > f₀, Z_LC becomes inductive, where f_o 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 I_O = V/R
Hence, the current is maximum at resonance.
The voltage across capacitor C is
V_C = 1/J_OC = 1/J₀C |V/R + J (_OL – 1/₀C)|……………..(8)
Hence, magnitude
|V_C| = V/_OC R² + (_OL – 1/_OC)² ……………..(9)
The frequency f_c at which V_C is maximum may be obtained by equating dV_C² /d_oo to zero. this results in,
f_c = (1/2) 1/LC – R²/2L² ………….(10)
The voltage across inductor L
V_L = V(J_OOL)/R + J (_OOL – 1/₀₀C) ………….(11)
The magnitude
|V_C| = V_OL/R² + (_OL – 1/_OC)² …………….(12)
The frequency f_l at which V_L is maximum may be obtained by equating dV_l²/d_o to zero.
thus,   f_l = 1/2 (LC – C²R²/2 ………………(13)
Obviously, f_l > f_c
The variations V_L and V_C are of equal magnitude and opposite phase at resonance, as shown in figure. if R is extremely small, f_l and f_c tend to equal f_o.
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 = |V_L|/V = |V_C|/V
|V_L| = |J0_OL. I_O| = 0₀ LI₀
|V_C|= 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/Z_IN = 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
Q_O = 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 V_C = IX_C
V_C = 200, I = 0.7
X_C = 1/0C
0C = 0.7/200
C = 2.785 uF
At resonance
X_L = X_C
X_C = 1/0C = 285.7
X_L = 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
Y_in = 1/R + 1/J0L + J0C
Y_IN = 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
Y_IN |_{0 = 0} = Y₀ = 1/R + J(0) (minimum value)
Z₀ = 1/Y₀ = R (maximum value)
I₀ = V Y₀ = V/R (minimum value)

1. Y_IN |_{0 = 0} = 1/R (Real)
2. 0 > 0₀ (capacitive network)
3. 0 < 0₀ (inductiven network)

|I_L| = |I_C| 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 o_o 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 = |I_L|/I₀ = |I_C|/I_O
|I_L| = V/J0₀L|= V/J0₀L
|I_C| = |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 + JX_L
Y = 10/10² + 12² + 5/25 + X²_L + J(12/10² + 12² – X_L/ 25 + X_L²)
X_L/25 + X_L² = 12/10² + 12²
X²_L – 20.3 X_L + 25 = 0
X_L = 18.98 or 1.32
X_L = 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

2. A series LCR circuit, consisting of R = 10, |X_L| = 20 and |X_C| = 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

3. 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

4. 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

5. 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

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

(a) |I_R| < 1 MA
(b) |I_R + I_L| > 1 mA
(c) |I_R + I_C| < 1 mA
(d) |I_R + I_C| > 5 mA

7. 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

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

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

9. 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

10. 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

11. 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

12. 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

13. 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

14. In a series resonant circuit, V_C = 150 V, V_L = 150 V and V_R = 50 V. what is the value of the source voltage?

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

15. 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

16. 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

17. 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

18. If L = 2.5 mH, Q_O = 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

19. 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

20. In the given circuit, the voltage across R(V_R) and inductor L(V_L) 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.

2. (d)

A series RLC circuit, R = 10 |X_L| = 20, |V_C| = 20 Circuit in resonance, so, current flowing in the circuit
I = V_RMS/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

3. (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

4. (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

5. (c)

impedance offered by C  X_C = 1/0_C = 1/0.C = 0
Impedance offered by L   X_L = 0L = 0.L = 0
So, DC voltage drop entirely across C

6. (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 = I_RR
Current through inductor = – V/J0L, because current in lagging, with voltage.
so, |I_R + I_L| < 1; similarly |I_R + I_C| > 1 and |I_R|< 1

7. (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.

8. (b)

For series critically damped circuit
R_EQ = 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

9. (b)

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

10. (b)

Q = f_o/BW
f_o = 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

11. (c)

0_H = 2F_H = 2337.97 rad/s
0_l = 2f_l = 2138.17 rad/s
0₀ = 0_h ,0_l
0₀ = 2235.3 rad/s
f_o = 355.84 Hz

12. (c)

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

13. (a)

Q = f/f
f = f/q = 10⁶/100
f = 10 kHz

14. (b)

in a series resonant circuit at resonance
V_C = V_L
X_C = X_L
and both the voltage are cancelled out.
V_S = 50 V

15. (c)

we know that
Q_S = 1/0C_SR_S
f = 50 x 10³
C_S = 25 pF, R = 20
Q_S = 1/2 (50 x 10³) (20) (25 x 10^-12)
Q_S = 6366.2
For this large value of Q_S is
R_P = R_S Q₂^S = 810.5
C_P = C_S = 25 pF

16. (d)

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

17. (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

18. (b)

the neper frequency is given by
a = 0_o/2Q_O
0₀ = 1/LC
0_O = 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

19. (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

20. (c)

X_L = 2f_l = 2 x 10
X_L = 3.14 K
I = V/Z = V/R² + (JX_L)²
I = V/R² + (0L)²
I = 10/ (1)² + (3.14)² = 10/3296
I = 3.03 mA
V_L = I x X_L = 3.03 x 3.14
V_L = 9.52 V