Resonant Frequency MCQ Quiz in বাংলা - Objective Question with Answer for Resonant Frequency - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

For the series RLC circuit ω_r is geometrical mean of ωL and ωU

\({{\rm{ω}}_{{\rm{U}}}} = \frac{{{\rm{ω}}_r^2}}{{{{\rm{ω}}_{\rm{L}}}}}\;\)

As 50 = ω_L,100 = ω_r

\({{\rm{ω}}_{{\rm{U}}}} = \frac{{{\rm{100}}^2}}{{{{\rm{50}}_{\rm{}}}}}\;\)= 200 rad/sec

now bandwidth will be B.W = ω_U - ω_L = 150 rad/s

Quality factor = Q = \(\rm \dfrac{Resonant\ frequency}{Bandwidth}\)

Q = \(\dfrac{100}{150}\)= 0.66

Concept:
RLC series circuit:

An RLC circuit is an electrical circuit consisting of Inductor (L), Capacitor (C), Resistor (R) it can be connected either parallel or series.

When the RLC circuit is set to resonate (XL = XC), the resonant frequency is expressed as 

 \(f = \frac{1}{{2π }}\sqrt {\frac{1}{{LC}}}\)

Quality factor:

The quality factor Q is defined as the ratio of the resonant frequency to the bandwidth.

\(Q=\frac{{{f}_{r}}}{BW}\)

Mathematically, for a coil, the quality factor is given by:

 \(Q=\frac{{{\omega }_{0}}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}\)

Where,

XL & XC = Impedance of inductor and capacitor respectively

L, R & C = Inductance, resistance, and capacitance respectively

fr = frequency

ω0 = angular resonance frequency

Calculation:

If all the component values get doubled, then

\(\Rightarrow {f_{new}} = \frac{1}{{2\pi \sqrt {2L \times 2C} }} = 0.5\;{f_0}\)

Concept:

For a series RLC circuit, the resonant frequency is given by:

\(f_r=\frac{1}{2π\sqrt{LC}}\)

\(L=\frac{1}{(2π)^2{f_r^2C}}\)   ---(1)

Also, the bandwidth for a series RLC circuit is given by:

\(BW=\frac{R}{L}\)

Using Equation (1), the above expression becomes:

\(BW=\frac{R}{\frac{1}{(2π)^2{f_r^2C}}}​​\)

\(BW=R× (2π)^2{f_r^2C}\)   ---(2)

Calculation:

Given: f_r = 2.5 MHz, BW = 20 rad/sec, and C = 115 pF

Putting these values in Equation (2), we can write:

20 × 10³ = R × 4π² (2.5 × 10⁶)² (115× 10^-12)

20 × 10³ = R × 4 (9.6)(6.25)(115)

20 × 103 = R × 4 (9.6)(6.25)(115)

R = 0.72 Ω 

Explanation:

At Resonance,

The circuit is said to be in resonance when the source current is in phase with source voltage.

By KVL,

V = V­_R∠0 + V­_L∠90 + V­_C∠-90

As V­_L = V­_C

Net reactive voltage is zero at resonance.

Therefore,

X_L = X_C

As Z = R + j(X_L – X_C)

Z_min = R

Hence,

Current will be maximum at resonance,

\({I_{max}} = \frac{V}{{{Z_{\min }}}} = \frac{V}{R}\)

Hence.

The reason given for the following assertion is correct.

Y =  Total admittance of the circuit

 \(Y = \frac{1}{R} + jω C + \frac{1}{R_L + jω L}\)

\(Y = \frac{1}{R} + jω C + \frac{R_L - jω L}{R^2_L + ω^2 L^2} \)

\(Y = \frac{1}{R} + \frac{R_L}{R^2_L + ω^2 L^2} + j(ω C -\frac{ω L}{R^2_L + ω^2 L^2})\)

At resonance (ωo), the imaginary part of Y must be zero.

\(ω_o C = \frac{ω_o L}{R^2_L + ω^2_o L^2}\)

\(R^2_L + ω^2_o L^2 = \frac{Lω^2_o L^2}{C}\)

\(ω_o = \sqrt{\frac{1}{LC} - \frac{R^2_L}{L^2} }\)            ......1

Here substituting the values of L, R, and C, in equation 1 

So we get,

ω_o  = 1 rad/sec

The admittance at resonance is

\(Y = \frac{1}{R} + \frac{R_L}{R^2_L +\omega^2_o L^2}\)      [since the imaginary part is zero]

Here substituting the values of L, R, and  ωo and get the value of Y

Y = 1.5 mho

• In an electrical circuit, the condition that exists when the inductive reactance and the capacitive reactance are of equal magnitude, causing electrical energy to oscillate between the magnetic field of the inductor and the electric field of the capacitor, is termed as resonance.
• In a series RLC circuit at resonance, inductive reactance (X_L) is equal to capacitive reactance (X_C). The circuit becomes purely resistive, i.e. at resonance: X_L = X_C

          The impedance at resonance becomes:

​          \(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} = R\) 

• The impedance - frequency curve is shown below:

The formula for determining resonant frequency (fr)of series RLC circuit is \(f_r = \dfrac{1}{2\pi \sqrt {LC}}\)

Derivation:

For a series LCR circuit, Impedance (Z) of the circuit is given by:

\(\)\(\Rightarrow Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)

Inductive reactance,

⇒ XL = Lω  

Capacitive reactance

​\(\Rightarrow X_c=\frac{1}{Cω}\)

Resonance will take place when XL = XC.

​⇒ XL = XC

\(\Rightarrow Lω =\frac{1}{Cω}\)

\(\Rightarrow ω =\frac{1}{\sqrt{LC}}\)

As we know, ω = 2πf

Where f = frequency

\(\Rightarrow f =\frac{1}{2\pi\sqrt{LC}}\)

The net impedance of the circuit will be:

Z_net = Z₃ + Z₁||Z₂

Given Z₁ = 26 + 13j and Z₂ = 13 – 39j

\({Z_1}||{Z_2} = \frac{{\left( {26 + 13j} \right)\left( {13 - 39j} \right)}}{{26 + 13j + 13 - 39j}}\)

\( = \frac{{338 + 169j - 1014j + 507}}{{39 - 26j}}\)

\( = \frac{{\left( {845 - 845j} \right)\left( {39 + 26j} \right)}}{{2197}}\)

= 0.384 (1 - j)(39 + 26j)

= 0.384 (39 + 26j – 39j + 26)

= 25 – 5j

∴  Z_net = Z₃ + 25 – 5j

For resonance to occur, Z_net should be purely resistive, i.e. the imaginary part should be zero.

Let Z₃ = R + jX

Z_net = R + jX + 25 – 5j

Z_net = (R + 25) + j(X - 5)

Equating the imaginary part to zero, we get:

X – 5 = 0

X = 5 Ω

∴ The value of Z₃ that will produce resonance at the terminal will be:

Z₃ = R + j5

R could be any value.

Applying 1A, at input port \({V_1} = 10\;V\)

Voltage across 1A source

\({V_{test}} = 10 + j\omega {10^{ - 3}} - \frac{j}{{\omega 50 \times {{10}^{ - 9}}}}\left( {5 + 1} \right)\)

At resonance \({I_m}\left( {{Z_{in}}} \right) = 0\)

\(\Rightarrow {\omega _0}{10^{ - 3}} = \frac{6}{{{\omega _0}50 \times {{10}^{ - 9}}}}\)

\(\Rightarrow \;{\omega _0} = 346\;kHz\)

The inductive reactance (X_L) is directly proportional to frequency and capacitive reactance is inversely proportional to frequency.

\({X_L} \propto f\;and\;{X_c} \propto \frac{1}{f}\)

At below resonance frequency, X_L is low and X_C is high → circuit behave like capacitive circuit

At above resonance frequency, X_L is high and X_C is low → circuit behave like inductive circuit