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

Concept:

For the given T network Iterative impedance can be calculated by terminating the port with the input impedance of the other respective port.

Calculations:

Given that

1) Iterative impedance of port 1 is calculated by

Z_1(in) = R₁ + (R₂ || Z1(in))

Where,

R₁ = 200 Ω , R₂ = 400 Ω

Z1(in) = 200 + (400 || Z1(in))

\(Z_{1(in)} = 200 + \frac{400\times Z_{1(in)}}{400+Z_{1(in)}}\)

on rearranging the above equation we get

Z²1(in) - 200 Z1(in) - 80000 = 0

by solving from the above equation we get

Z_{1 (in)}  = 400 Ω,  - 200 Ω

resistance must be positive so 400 Ω will be the answer.

1) Iterative impedance of port 2 is calculated by

Z₂(in) = (R1 + Z2(in)) || R₂

Where,

R1 = 200 Ω , R2 = 400 Ω

Z2(in) = (200 + Z2(in)) || 400

\(Z_{2(in)} = \frac{(200+Z_{2(in)}) \times 400}{(200+Z_{2(in)})+400}\)

on rearranging the above equation we get

Z21(in) + 200 Z1(in) - 80000 = 0

by solving from the above equation we get

Z1 (in)  = - 400 Ω,  200 Ω

resistance must be positive so 200 Ω will be the answer.

Hence from the above answer option (4) is correct

Explanation:

• The electrical impedance is a measure of the resistance of the circuit to the current when the voltage is applied.
• Impedance applies the principle of resistance to alternating current (AC) circuits, which has both magnitude and phase, unlike resistance, which has only magnitude.
• Z = R +jX
• Impedance is a complex quantity, with the same units as the resistance, the SI unit is the ohm (Ω).
• Its symbol is typically Z, which can be expressed by writing its magnitude which steps in the polar form.

Hence the correct option is Complex quantity.

Reactance is defined as the imaginary part of the electrical impedance and is comparable, but not necessarily equal to, the reciprocal part of the susceptance.

Concept:

The inverse (reciprocal) of the impedance is called the admittance (Y). Therefore, it has the opposite function of impedance. That is, we can say it is the measure of the flow of current which is allowed by a device or a circuit.

Admittance is also a complex number as impedance which is having a real part, Conductance(a), and imaginary part, Susceptance (b).

If Impedance(Z) is given by Z = R + jX

Where,

Z = Impedance in Ohm

R = Resistance in Ohm

X = Reactance in Ohm

Then Admittance(Y) can be expressed as

Y= G + jB

Where,

Y = Admittance in siemens

G = Conductance in siemens \(= \frac{R}{{{R^2} + {X^2}}}\) 

B = Susceptance in siemens \(= \frac{{ - X}}{{{R^2} + {X^2}}}\)

Concept:

Properties of RC network:

• Poles and zeroes are simple, lies only on the -ve real axis of the s-plane.
• Poles and zeroes should be alternate.
• The first critical frequency should be a pole and the pole can be at origin (s = 0)
• The last critical frequency should be a zero and zero can be at s = ∞ 

Properties of RL Network:

• Poles and zeroes are simple, lies only on the -ve real axis of the s-plane.
• Poles and zeroes should be alternate
• The first critical frequency should be zero and this can be at the origin (5 = 0)
• The lost critical frequency should be a pole and this can be at s = ∞ 

Properties of the LC network:

• Poles and zeroes will be complex and lie on the imaginary axis of the s-plane.
• Poles and zeroes will be alternate.

Analysis:

Given function is:

\(Z\left( s \right) = \frac{{\left( {s + 4} \right)\left( {s + 9} \right)}}{{\left( {s + 1} \right)\left( {s + 16} \right)}}\)

The pole-zero plot is given below:

If we compare with the above 3 pole-zero plots, none of them are matching. So, the given driving point impedance belongs to the R-L-C network.

Concept:

Driving point impedance or the input impedance is the impedance seen from the input side.

Example:

A simple RC driving point impedance circuit is as shown below.

Now driving point impedance is calculated as

\(z\left( s \right) = R + \frac{1}{{sC}}\)

After writing in pole-zero form, we get

\(z(s)= \frac{{R\left( {s + \frac{1}{{RC}}} \right)}}{s}\)

Pole at s = 0, zero at \(s = - \frac{1}{{RC}}\)

Analysis:

Given:

R, L in series and C in parallel

\(Z(s) = \frac{(R + sL)\frac{1}{sC}}{sL + R + \frac{1}{sC}}\)

\(Z(s) = \frac{\frac{1}{C}(s+\frac{R}{L})}{s^2+ \frac{R}{L}s+\frac{1}{LC}}\)

Zero = - R/L

R/L = 1 

R = L

Now, 

Poles:

\((s-(\frac{-1}{2} \pm j\frac{\sqrt{3}}{2})) = (s + 1/2 - j \sqrt{3}/{2})(s + 1/2 + j \sqrt{3}/{2})\)

= s² + 0.25 + s + 0.75 

= s² + s + 1

1/LC = 1

Z(0) = 1

1/LC = 1/C

C = 1 F

L = 1 H and R = 1 Ω 

\(\begin{array}{l} {Z_2}\left( s \right) = {Z_1}\left( s \right) + {R_{neg}}\\ = {R_{neg}} + {R_e}\left( {{Z_1}\left( s \right)} \right) + Im\left( {{Z_1}\left( s \right)} \right) \end{array}\)

For \({Z_2}\left( s \right)\) to be positive real

\(\begin{array}{l} {R_e}\left( {{Z_2}\left( s \right)} \right) \ge 0\\ \therefore {R_{neg}} + {R_e}{Z_1}\left( s \right) \ge 0 \end{array}\)

⇒ \({R_e}\left( {{z_1}\left( s \right)} \right) \ge - {R_{neg}}\)

Taking modulus on both sides, we get

\(({R_e}\left( {{Z_1}\left( s \right)} \right) \ge \left| {{R_{neg.}}} \right|\)

So, (1) is correct

\( {\left. {{Z_{11}} = \frac{{{V_1}}}{{{I_1}}}} \right|_{{I_2} = 0}} = \frac{{{R_1}\left( {LS + {R_2}} \right)}}{{{R_1} + {R_2} + LS}}\)

\(\\ = \frac{{{R_1}L\left( {S + \frac{{{R_2}}}{L}} \right)}}{{L\left( {S + \frac{{{R_1} + {R_2}}}{L}} \right)}} = {K_1}\left( {\frac{{S + 1}}{{S + 2}}} \right)\)

\( \frac{{{R_2}}}{L} = 1,\frac{{{R_1} + {R_2}}}{L} = 2,\frac{{{R_1}}}{L} = 1\)

\({\left. {{Z_{22}} = \frac{{{V_2}}}{{{I_2}}}} \right|_{{I_1} = 0}} = \frac{{{R_2}\left( {LS + {R_1}} \right)}}{{{R_1} + {R_2} + LS}}\)

\(\\ = \frac{{{R_2}L\left( {S + \frac{{{R_1}}}{L}} \right)}}{{L\left( {S + \frac{{{R_1} + {R_2}}}{L}} \right)}} = {K_2}\left( {\frac{{S + 1}}{{S + 2}}} \right)\)