Y Parameters MCQ Quiz - Objective Question with Answer for Y Parameters - Download Free PDF

Concept

The Y parameters are expressed as:

$I_1=Y_{11}{V}_1+Y_{12}{V}_2$

$I_2=Y_{21}{V}_1+Y_{22}{V}_2$

The Z parameters are expressed as:

$V_1=Z_{11}{I}_1+Z_{12}{I}_2$

$V_2=Z_{21}{I}_1+Z_{22}{I}_2$

The relationship between Y and Z parameters is:

$Y=Z^{-1}$

where, I₁ = Input current

I₂ = Output current

V₁ = Input voltage

V₂ = Output voltage

Note: It is easier to find the Z parameters for the T network and Y parameters for the π network

Calculation

Applying KVL at the input side:

$V_1=24I_1+8(I_1+I_2)$

$V_1=32I_1+8I_2$

Applying KVL at the output side:

$V_2=8I_2+8(I_1+I_2)$

$V_2=8I_1+16I_2$

The Z parameter matrix is as:

$Z=\left[\begin{array}{cc}32& 8\\ 8& 16\end{array}\right]$

$Y={\left[\begin{array}{cc}32& 8\\ 8& 16\end{array}\right]}^{-1}$

$Y=\frac{1}{(32\times 16-8\times 8)}\left[\begin{array}{cc}16& -8\\ -8& 32\end{array}\right]$

$Y=\left[\begin{array}{cc}\frac{1}{28}& -\frac{1}{56}\\ -\frac{1}{56}& \frac{1}{14}\end{array}\right]$

Concept:

The short circuit impedance parameter or the Y parameter is given by

 $\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]$=$\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]$$\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

I₁= Y₁₁V₁ +Y₁₂V₂

I₂=Y₂₁V₁ + Y₂₂V₂

We also have the circuit 

$\left[\begin{array}{c}Y\end{array}\right]$=$\left[\begin{array}{cc}{Y}_A+Y_C& -Y_C\\ -Y_C& Y_B+Y_C\end{array}\right]$ Ʊ

here,

Y₁₁=Y_A+Y_C , Y₁₂ =Y₂₁ =-Y_C , Y₂₂ =Y_B+Y_C

Calculation

Here Y_A =0.6 Ʊ,Y_B=0.4 Ʊ ,Y_C=0.3 Ʊ

Y₁₁ = 0.9 Ʊ

Y₁₂ = -0.3 Ʊ

Y₂₁ = -0.3 Ʊ

Y₂₂ = 0.7 Ʊ

Hence the correct answer is option 1

Concept:

The equations of the Y parameter are:

I1 = Y11 V1 + Y12 V2

I2 = Y21 V1 + Y22 V2 

where, Y₁₁ = Input admittance

Y₂₂ = Output admittance

Y₁₂ = Y₂₁ = Transfer admittance

Calculation:

Applying KCL at input side:

-I₁ + 2V₁ +4(V₁ - V₂) = 0

I₁ = 6V₁ - 4V₂

Applying KCL at output side:

-I₂ + 3V₂ +4(V₂ - V₁) = 0

I₂ = -4V₁ + 7V₂

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=$ $\left[\begin{array}{cc}6& -4\\ -4& 7\end{array}\right]$$\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

Transfer admittance is given by:

Y12 = Y21 = -4

Concept:

Y parameters: The set of two equations which represents Y parameters are given below

I1 = Y11 V1 + Y12 V2

I2 = Y21 V1 + Y22 V2

We can represent the above two equations in matrix form as

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$    -------- (1)

Z parameters: The set of two equations that represents Z parameters are given below

V1 = Z11 I1 + Z12 I2

V2 = Z21 I11 + Z22 I2

We can represent the above two equations in matrix form as

$\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]=\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]$

We can modify the above equation 

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]={\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]}^{-1}\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$     -------- (2)

By equating equations (1) and (2) we will get

$\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]={\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]}^{-1}$

$\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]=\frac{\left[\begin{array}{cc}{Z}_{22}& -Z_{12}\\ -Z_{21}& Z_{11}\end{array}\right]}{\mathrm{\Delta }Z}$       -------- (3)

Where 

$\mathrm{\Delta }Z=Z_{11}{Z}_{22}-Z_{12}{Z}_{21}$

Calculation : 

The given Z matrix is $\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 0.6\end{array}\right]$

So Z11 = 0.9, Z12 = 0.2, Z21 =  0.2, and Z22 = 0.6

From equation (3) we can get

$Y_{22}=\frac{Z_{11}}{\mathrm{\Delta }Z}$

= $0.9/(0.9\times 0.6-0.2\times 0.2)=0.9/0.5=1.8$

Concept:

Y parameters

These are also called the admittance parameters.

The Y parameters for the two-port network are shown as:

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

Calculation:

When V₂ = 0 or short circuit port 2

Then, I₁ = Y₁₁ V₁

I₂ = Y₂₁ V₁

V₁ and I₁ relation can be drawn from current division rule as

$V_1=0.5\times (\frac{0.5}{0.5+0.5})I_1$

V₁ = 0.25I₁

I₁ = 4V₁

Y11 = 4 S

similarly, V1 and I2 relation can be drawn as

V₁ = 0.5(-I₂)

I₂ = -2V₁

Y21 = -2 S

By applying the same procedure for port 1

When V1 = 0 or short circuit port 1

Then, I1 = Y12 V2

I2 = Y22 V2

V2 and I2 relation can be drawn from current division rule as

$V_2=0.5\times (\frac{0.5}{0.5+0.5})I_2$

V2 = 0.25I2

I2 = 4V2

Y22 = 4 S

similarly, V1 and I2 relation can be drawn as

V2 = 0.5(-I1)

I1 = -2V2

Y12 = -2 S

$\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]=\left[\begin{array}{cc}4& -2\\ -2& 4\end{array}\right]S$

Hence option 1 is correct

For the given π network 

Y parameter can be calculated by

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_1+Y_2& -Y_2\\ -Y_2& Y_2+Y_3\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

Substituting the value of Y₁, Y_2, and Y₃

$=\left[\begin{array}{cc}\left(2+2\right)& \left(-2\right)\\ \left(-2\right)& \left(2+2\right)\end{array}\right]$

$=\left[\begin{array}{cc}4& -2\\ -2& 4\end{array}\right]S$

Concept:

Y parameters

These are also called the admittance parameters.

The Y parameters for the two-port network are shown as:

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

Calculation:

When V₂ = 0 or short circuit port 2

Then, I₁ = Y₁₁ V₁

I₂ = Y₂₁ V₁

V₁ and I₁ relation can be drawn from current division rule as

$V_1=0.5\times (\frac{0.5}{0.5+0.5})I_1$

V₁ = 0.25I₁

I₁ = 4V₁

Y11 = 4 S

similarly, V1 and I2 relation can be drawn as

V₁ = 0.5(-I₂)

I₂ = -2V₁

Y21 = -2 S

By applying the same procedure for port 1

When V1 = 0 or short circuit port 1

Then, I1 = Y12 V2

I2 = Y22 V2

V2 and I2 relation can be drawn from current division rule as

$V_2=0.5\times (\frac{0.5}{0.5+0.5})I_2$

V2 = 0.25I2

I2 = 4V2

Y22 = 4 S

similarly, V1 and I2 relation can be drawn as

V2 = 0.5(-I1)

I1 = -2V2

Y12 = -2 S

$\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]=\left[\begin{array}{cc}4& -2\\ -2& 4\end{array}\right]S$

Hence option 1 is correct

For the given π network 

Y parameter can be calculated by

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_1+Y_2& -Y_2\\ -Y_2& Y_2+Y_3\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

Substituting the value of Y₁, Y_2, and Y₃

$=\left[\begin{array}{cc}\left(2+2\right)& \left(-2\right)\\ \left(-2\right)& \left(2+2\right)\end{array}\right]$

$=\left[\begin{array}{cc}4& -2\\ -2& 4\end{array}\right]S$

Concept:

I₁ = Y₁₁ V₁ + Y₁₂ V₂

I₂ = Y₂₁ V₁ + Y₂₂ V₂

When port 1 is short cirucited, i.e. V₁ = 0,

I₁ = Y₁₂ V₂ and I₂ = Y₂₂ V₂

Calculation:

The given equations are:

I₁ = 4I₂ and V₂ = 0.25 I₂

⇒ I₂ = 4V₂ ⇒ Y₂₂ = 4

I₁ = 4 (4V₂) = 16 V₂

Concept:

Y-parameter: It is also known as short circuit parameter i.e.

The Y-parameter for the two-port network is:

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

Or I₁ = Y₁₁V₁ + Y₁₂V₂

I₂ = Y₂₁V₁ + Y₂₂V₂

Analysis:

Given circuit is:

Calculation of Y₁₁ and  Y₂₂

Let V₂ = 0

Then I = 2V₂ = 0 (open)

I₁ = - I₂

V₁ = 4 I₁

i.e,

$I_1=\frac{1}{4}{V}_1$

$I_2=-\frac{1}{4}{V}_1$

So, $Y_{11}=\frac{1}{4}$

$Y_{21}=-\frac{1}{4}$

Calculation of Y₁₂ and Y₂₂

Let V₁ = 0:

Applying KCL at (a), we get:

I₁ + 2V₂ + I₂ = 0        ----(1)

At (a):

$\frac{Va}{2}-2V_2+\frac{V_a-V_2}{2}=0$

$V_a=\frac{5}{2}{V}_2$

$I_1=-\frac{V_a}{2}=-\frac{5}{4}{V}_2$

$\frac{V_a-V_2}{2}=-I_2=\frac{\frac{5}{2}{V}_2-V_2}{2}$

$-I_2=+\frac{3}{4}{V}_2$

$I_2=-\frac{3}{4}{V}_2$

$I_1=-\frac{5}{4}{V}_2$

$I_2=-\frac{3}{4}{V}_2$

$Y_{12}=-\frac{5}{4}V$

$Y_{22}=-\frac{3}{4}V$

Thus

Short trick: Students can tick the answer just by calculating Y₁₁ and Y₂₁ by elimination method.

Concept:

Y parameters

These are also called the admittance parameters.

The Y parameters for the two-port network are shown as:

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

I₁ = Y₁₁V₁ + Y₁₂V₂

I₂ = Y₂₁V₁ + Y₂₂V₂

Calculation:

Consider the circuit with nodes specified as V₁ and V₂

Writing the node equation at the node1

$-I_1+\frac{V_1}{1}-3V_2+\frac{V_1-V_2}{1}=0$

I₁ = 2V₁ – 4V₂  ---- (i)

Writing the node equation at the node2

$-I_2+\frac{V_2-V_1}{1}+\frac{V_2}{1}=0$

I₂ = - V₁ + 2V₂ ---- (ii)

From the equations (i) and (ii) the Y parameters for the given network are:

$Y=\left[\begin{array}{cc}2& -4\\ -1& 2\end{array}\right]$

Concept:

The equations of the Y parameter are:

I1 = Y11 V1 + Y12 V2

I2 = Y21 V1 + Y22 V2 

where, Y₁₁ = Input admittance

Y₂₂ = Output admittance

Y₁₂ = Y₂₁ = Transfer admittance

Calculation:

Applying KCL at input side:

-I₁ + 2V₁ +4(V₁ - V₂) = 0

I₁ = 6V₁ - 4V₂

Applying KCL at output side:

-I₂ + 3V₂ +4(V₂ - V₁) = 0

I₂ = -4V₁ + 7V₂

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=$ $\left[\begin{array}{cc}6& -4\\ -4& 7\end{array}\right]$$\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

Transfer admittance is given by:

Y12 = Y21 = -4

Concept

The Y parameters are expressed as:

$I_1=Y_{11}{V}_1+Y_{12}{V}_2$

$I_2=Y_{21}{V}_1+Y_{22}{V}_2$

The Z parameters are expressed as:

$V_1=Z_{11}{I}_1+Z_{12}{I}_2$

$V_2=Z_{21}{I}_1+Z_{22}{I}_2$

The relationship between Y and Z parameters is:

$Y=Z^{-1}$

where, I₁ = Input current

I₂ = Output current

V₁ = Input voltage

V₂ = Output voltage

Note: It is easier to find the Z parameters for the T network and Y parameters for the π network

Calculation

Applying KVL at the input side:

$V_1=24I_1+8(I_1+I_2)$

$V_1=32I_1+8I_2$

Applying KVL at the output side:

$V_2=8I_2+8(I_1+I_2)$

$V_2=8I_1+16I_2$

The Z parameter matrix is as:

$Z=\left[\begin{array}{cc}32& 8\\ 8& 16\end{array}\right]$

$Y={\left[\begin{array}{cc}32& 8\\ 8& 16\end{array}\right]}^{-1}$

$Y=\frac{1}{(32\times 16-8\times 8)}\left[\begin{array}{cc}16& -8\\ -8& 32\end{array}\right]$

$Y=\left[\begin{array}{cc}\frac{1}{28}& -\frac{1}{56}\\ -\frac{1}{56}& \frac{1}{14}\end{array}\right]$

Concept:

When two or more networks are connected in parallel, the final Y-parameters of the equivalent network will simply be the addition of the individual Y-parameters, i.e.

Y_net = Y₁ + Y₂ + …. Y_n

Calculations:

The given network is a parallel combination of two networks as shown:

We have to find the y-parameter matrix of the resistor circuit.

Since, y parameter shows relation between current with voltages as:

I₁ = y₁₁ V₁ + y₁₂ + V₂

I₂ = y₂₁ V₁ + y₂₂ + V₂

With V₂ = 0 (short-circuited)

$\frac{V_1}{I_1}=1\mathrm{\Omega },$

$whichimpliesthat\frac{I_1}{V_1}=y_{11}=1$

Also, $\frac{V_1}{I_1}=-1,$

$whichimpliesthat\frac{I_2}{V_1}=y_{21}=-1$

With V₁ = 0 now,

$\frac{V_2}{I_1}=-1;\frac{I_1}{V_2}=y_{12}=-1$

And $\frac{V_2}{I_2}=1\mathrm{\Omega };\frac{I_2}{V_2}=y_{22}=1$

So, $y_2=\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$

The net y parameter of the parallel circuit will be;

y = y₁ + y₂

$=\left[\begin{array}{cc}5& 2\\ 2& 3\end{array}\right]+\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$

$y=\left[\begin{array}{cc}6& 1\\ 0& 4\end{array}\right]$

Y (Admittance) Parameters:

They are also called the short circuit parameters, as they are calculated under short circuit conditions, i.e. at V1 = 0 and V2 = 0.

Expressed in Matrix Form as:

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]$

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

With the output short-circuited, i.e. V2 = 0, the two parameters obtained are:

$Y_{11}={\frac{I_1}{V_1}|}_{V_2=0}$

$Y_{21}={\frac{I_2}{V_1}|}_{V_2=0}$

With the input short-circuited, i.e. V1 = 0, the two parameters obtained are:

$Y_{12}={\frac{I_1}{V_2}|}_{V_1=0}$

$Y_{22}={\frac{I_2}{V_2}|}_{V_1=0}$

Analysis:

V1 = 60I1 + 20I2

V2 = 20I1 + 40I2

This is in the form of Z - parameters i.e. 

Z₁₁ = 60, Z₁₂ = 20, Z₂₁ = 20, Z₂₂ = 40

$\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]=\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]$

$Y={\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]}^{-1}$

$Y=\frac{1}{Z_{11}{Z}_{22}-Z_{12}{Z}_{21}}\left[\begin{array}{cc}{Z}_{22}& -Z_{12}\\ -Z_{21}& Z_{11}\end{array}\right]$

$$$Y=\frac{1}{60\times 40-20\times 20}\left[\begin{array}{cc}40& -20\\ -20& 60\end{array}\right]$

Y₁₁ = 20 × 10^-3

Y₁₂ = -10 × 10^-3

Y₂₁ = -10 × 10^-3

Y₂₂ = 30 × 10^-3

For parallel connection, y-parameters can be added. As the given circuit is same from both the sides, the circuit is symmetric and reciprocal, both.

So, y₁₁ = y₂₂

and y₂₁ = y₁₂

We divide the given circuit into two and analyze both of them individually.

Analyzing the 1^st Circuit as shown:

Now,

$y_{11}={\frac{I_1}{V_1}|}_{V_2=0}\text{ and }{y}_{21}={\frac{I_2}{V_1}|}_{V_2=0}$

With V₂ = 0, the circuit is redrawn as:

$So,\frac{V_1}{i_1}=\frac{1}{3}+\frac{1}{3}+\frac{2}{9}=\frac{3+3+2}{9}=\frac{8}{9}$

$So,\frac{i_1}{V_1}={\left(y_{11}\right)}_{1^{st}ckt.}=\frac{9}{8}$     ----(1)

Applying current division,

$\Rightarrow \frac{i_1\times \left(\frac{1}{3}\right)}{\frac{1}{3}+\frac{2}{3}}=-i_2$

$\Rightarrow i_1=-i_2\left(\frac{3}{1}\right)(\frac{1}{3}+\frac{2}{3})$

i₁ = -3i₂     ----(2)

Using Equation - (1),

$\frac{i_1}{V_1}=-\frac{3i_2}{V_1}=\frac{9}{8}$

$So,\frac{i_2}{V_1}={\left(y_{21}\right)}_{2^{nd}ckt.}=\frac{-3}{8}$

Analyzing the 2^nd Circuit now:

The equivalent impedance circuit is as shown:

$Again,y_{11}={\frac{I_1}{V_1}|}_{V_2=0}and{y}_{21}={\frac{I_1}{V_1}|}_{V_2=0}$

With V₂ = 0, the circuit is redrawn as:

The parallel combination of 8/3s and 16/3s will give an equivalent impedance of:

$\frac{\frac{8}{3s}\times \frac{16}{3s}}{\frac{8}{3s}+\frac{16}{3s}}=\frac{8\times 16\left(3s\right)}{9s^2\times (8+16)}=\frac{8\times 16}{3s\left(24\right)}=\frac{16}{9s}$

So, $\frac{V_1}{i_1}=\frac{16}{9s}$

$\Rightarrow {{\left(y_{11}\right)}_{2^{nd}ckt}}_{}=\frac{i_1}{V_1}=\frac{9s}{16}$     ----(3)

Applying Current division,

$\Rightarrow \frac{i_1\times 8/3s}{\frac{8}{3s}+\frac{16}{3s}}=-i_2$

$\Rightarrow \frac{8i_1\left(3s\right)}{3s\left(24\right)}=-i_2$

I₁ = -3i₂

Using Equation (3),

$\Rightarrow -\frac{3i_2}{V_1}=\frac{9s}{16}$

$\frac{i_2}{V_1}={\left(y_{21}\right)}_{2^{nd}ckt.}=-\frac{3s}{16}$