Category: Network Theory Objective Questions

Network Theory Multiple Choice Questions on “Millman’s Theorem”.

1. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then these sources are replaced by?
A. single current source I’ in series with R’
B. single voltage source V’ in series with R’
C. single current source I’ in parallel to R’
D. single voltage source V’ in parallel to R’
Answer: B
Clarification: Millman’s Theorem states that if there are voltage sources V1, V₂,…… Vn with internal resistances R₁, R₂,…..R_n, respectively, are in parallel, then these sources are replaced by single voltage source V’ in series with R’.

2. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent voltage source is?
A. V^‘=(V₁G₁+V₂G₂+⋯.+V_nG_n)
B. V^‘=((V₁G₁+V₂G₂+⋯.+V_nG_n))/((1/G₁+1/G₂+⋯1/G_n))
C. V^‘=((V₁G₁+V₂G₂+⋯.+V_nG_n))/(G₁+G₂+⋯G_n)
D. V^‘=((V₁/G₁+V₂/G₂+⋯.+V_n/G_n))/( G₁+G₂+⋯G_n)
Answer: C
Clarification: The value of equivalent voltage source is V^‘= ((V₁G₁+V₂G₂+⋯.+V_nG_n))/(G₁+G₂+⋯G_n).

3. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent resistance is?
A. R’=G₁+G₂+⋯G_n
B. R’=1/G₁+1/G₂+⋯1/G_n
C. R’=1/((G₁+G₂+⋯G_n))
D. R’=1/(1/G₁+1/G₂+⋯1/G_n)
Answer: C
Clarification: Let the equivalent resistance is R’. The value of equivalent resistance is R’=1/((G₁+G₂+⋯G_n)).

4. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then these sources are replaced by?
A. single voltage source V’ in parallel with G’
B. single current source I’ in series with G’
C. single current source I’ in parallel with G’
D. single voltage source V’ in series with G’
Answer: C
Clarification: Millman’s Theorem states that if there are current sources I₁,I₂,…… In with internal conductances G₁,G₂,…..G_n, respectively, are in series, then these sources are replaced by single current source I’ in parallel with G’.

5. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent current source is?
A. I^‘=((I₁R₁+I₂R₂+⋯.+I_nR_n))/(R₁+R₂+⋯R_n)
B. I’=I₁R₁+I₂R₂+⋯.+I_nR_n
C. I’=((I₁/R₁+I₂/R₂+⋯.+I_n/R_n))/(R₁+R₂+⋯R_n)
D. I’=I₁/R₁+I₂/R₂+⋯.+I_n/R_n
Answer: A
Clarification: The value of equivalent current source is I^‘=((I₁R₁+I₂R₂+⋯.+I_nR_n))/(R₁+R₂+⋯R_n).

6. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent conductance is?
A. G’=R₁+R₂+⋯R_n
B. G’=1/(1/R₁+1/R₂+⋯1/R_n)
C. G’=1/((R₁+R₂+⋯R_n))
D. G’=1/R₁+1/R₂+⋯1/R_n
Answer: C
Clarification: Let the equivalent conductance is G’. The value of equivalent conductance is G’=1/((R₁+R₂+⋯R_n)).

7. Calculate the current through 3Ω resistor in the circuit shown below.
millmans-theorem-q7″>
A. 1
B. 2
C. 3
D. 4
Answer: C
Clarification: Applying Nodal analysis the voltage V is given by (10-V)/2+(20-V)/5=V/3. V=8.7V. Now the current through 3Ω resistor in the circuit is I = V/3 = 8.7/3 = 2.9A ≅ 3A.

8. Find the current through 3Ω resistor in the circuit shown below using Millman’s Theorem.
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A. 4
B. 3
C. 2
D. 1
Answer: B
Clarification: V^‘=((V₁G₁+V₂G₂))/(G₁+G₂)=(10(1/2)+20(1/5))/(1/2+1/5)=12.86V. R’=1/((G₁+G₂))=1/(1/2+1/5)=1.43Ω. Current through 3Ω resistor=I=12.86/(3+1.43)=2.9A≅3A.

9. Consider the circuit shown below. Find the current through 4Ω resistor.
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A. 2
B. 1.5
C. 1
D. 0.5
Answer: B
Clarification: Applying Nodal analysis the voltage V is given by (5-V)/1+(10-V)/3=V/4. V=6V. The current through 4Ω resistor I = V/4 = 6/4 = 1.5A.

10. In the following circuit. Find the current through 4Ω resistor using Millman’s Theorem.
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A. 0.5
B. 1
C. 1.5
D. 2
Answer: C
Clarification: V^‘=((V₁G₁+V₂G₂))/(G₁+G₂)=(5(1/1)+10(1/3))/(1/1+1/3)=6.25V. R’=1/((G₁+G₂))=1/(1/1+1/3)=0.75Ω. I=6.25/(4+0.75)=1.5A.

Network Theory Multiple Choice Questions on “Reactive Power”.

1. The reactive power equation (P_r) is?
A. I_eff² (ωL)sin2(ωt+θ)
B. I_eff² (ωL)cos2(ωt+θ)
C. I_eff² (ωL)sin(ωt+θ)
D. I_eff² (ωL)cos(ωt+θ)
Answer: A
Clarification: If we consider a circuit consisting of a pure inductor, the power in the inductor is reactive power and the reactive power equation (P_r) is P_r =I_eff² (ωL)sin2(ωt+θ).

2. Reactive power is expressed in?
A. Watts (W)
B. Volt Amperes Reactive (VAR)
C. Volt Ampere (VA.
D. No units
Answer: B
Clarification: Reactive power is expressed in Volt Amperes Reactive (VAR) and power is expressed in watts and apparent power is expressed in Volt Ampere (VA..

3. The expression of reactive power (Pr) is?
A. V_effI_msinθ
B. V_mI_msinθ
C. V_effI_effsinθ
D. V_mI_effsinθ
Answer: C
Clarification: The expression of reactive power (P_r) is V_effI_effsinθ Volt Amperes Reactive (VAR). Reactive power = V_effI_effsinθ Volt Amperes Reactive (VAR).

4. The power factor is the ratio of ________ power to the ______ power.
A. average, apparent
B. apparent, reactive
C. reactive, average
D. apparent, average
Answer: A
Clarification: The power factor is the ratio of average power to the apparent power. Power factor = (average power)/(apparent power).

5. The expression of true power (P_true) is?
A. P_asinθ
B. P_acosθ
C. P_atanθ
D. P_asecθ
Answer: B
Clarification: True power is the product of the apparent power and cosθ. The expression of true power (P_true) is P_acosθ. True power = P_acosθ.

6. The average power (P_avg) is expressed as?
A. P_asecθ
B. P_atanθ
C. P_acosθ
D. P_asinθ
Answer: C
Clarification: The average power is the product of the apparent power and cosθ. The average power (P_avg) is expressed as P_acosθ. Average power = P_acosθ.

7. The equation of reactive power is?
A. P_acosθ
B. P_asecθ
C. P_asinθ
D. P_atanθ
Answer: C
Clarification: The reactive power is the product of the apparent power and sinθ. The equation of reactive power is P_asinθ. Reactive power = P_asinθ.

8. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin (ωt-53⁰). Determine the apparent power (VA..
A. 620
B. 625
C. 630
D. 635
Answer: C
Clarification: The expression of apparent power (VA. is P_app= V_effI_eff = (V_m/√2)×(I_m/√2). On substituting the values V_m = 50, I_m = 25, we get apparent power = (50×25)/2 = 625VA.

9. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin (ωt-53⁰). Find the power factor.
A. 0.4
B. 0.5
C. 0.6
D. 0.7
Answer: C
Clarification: In sinusoidal sources the power factor is the cosine of the phase angle between the voltage and the current. The expression of power factor = cosθ = cos53⁰ = 0.6.

10. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin (ωt-53⁰). Determine the average power.
A. 365
B. 370
C. 375
D. 380
Answer: C
Clarification: The average power, P_avg = V_effI_effcosθ. We know the values of V_eff, I_eff are V_eff = 625 and I_eff – 0.6. So the average power = 625 x 0.6 = 375W.

Network Theory Multiple Choice Questions on “Star to Delta and Delta to Star Transformation”.

1. If a resistor Z_R is connected between R and N, Z_BR between R and B, Z_RY between R and Y and Z_BY between B and Y form a delta connection, then after transformation to star, the impedance at R is?
A. (Z_BRZ_BY)/(Z_RY+Z_BY+Z_BR)
B. (Z_RYZ_BR)/(Z_RY+Z_BY+Z_BR)
C. (Z_RYZ_BY)/(Z_RY+Z_BY+Z_BR)
D. (Z_RY)/(Z_RY+Z_BY+Z_BR)

Answer: B
Clarification: After transformation to star, the impedance at R is
(Z_RYZ_BR)/(Z_RY+Z_BY+Z_BR).

2. If a resistor Z_R is connected between R and N, Z_BR between R and B, Z_RY between R and Y and Z_BY between B and Y form a delta connection, then after transformation to star, the impedance at Y is?
A. (Z_RY)/(Z_RY+Z_BY+Z_BR)
B. (Z_BY)/(Z_RY+Z_BY+Z_BR)
C. (Z_RYZ_BY)/(Z_RY+Z_BY+Z_BR)
D. (Z_RYZ_BR)/(Z_RY+Z_BY+Z_BR)

Answer: C
Clarification: After transformation to star, the impedance at Y is
(Z_RYZ_BY)/(Z_RY+Z_BY+Z_BR).

3. If a resistor Z_R is connected between R and N, Z_BR between R and B, Z_RY between R and Y and Z_BY between B and Y form a delta connection, then after transformation to star, the impedance at B is?
A. (Z_BRZ_BY)/(Z_RY+Z_BY+Z_BR)
B. (Z_RYZ_BY)/(Z_RY+Z_BY+Z_BR)
C. (Z_BY)/(Z_RY+Z_BY+Z_BR)
D. (Z_BR)/(Z_RY+Z_BY+Z_BR)

Answer: A
Clarification: After transformation to star, the impedance at Y is
(Z_BRZ_BY)/(Z_RY+Z_BY+Z_BR).

4. If the resistors of star connected system are Z_R, Z_Y, Z_B then the impedance Z_RY in delta connected system will be?
A. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_B
B. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_Y
C. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_R
D. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/(Z_R+Z_Y)

Answer: A
Clarification: After transformation to delta, the impedance Z_RY in delta connected system will be (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_B.

5. If the resistors of star connected system are Z_R, Z_Y, Z_B then the impedance Z_BY in delta connected system will be?
A. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/(Z_B+Z_Y)
B. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_B
C. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_Y
D. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_R

Answer: D
Clarification: After transformation to delta, the impedance Z_BY in delta connected system will be (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_R.

6. If the resistors of star connected system are Z_R, Z_Y, Z_B then the impedance Z_BR in delta connected system will be?
A. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_Y
B. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/_R
C. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_B
D. (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/(Z_B+Z_R)

Answer: A
Clarification: After transformation to delta, the impedance Z_BR in delta connected system will be (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_Y.

7. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z_R = (2+j3) Ω, Z_Y = (1-j2) Ω, Z_B = (3+j4) Ω. Find Z_RY.
A. (5.22-j0.82) Ω
B. (-3.02+j8) Ω
C. (3.8-j0.38) Ω
D. (-5.22+j0.82) Ω

Answer: C
Clarification: Z_RY = (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_B
= (3.8-j0.38) Ω.

8. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z_R = (2+j3) Ω, Z_Y = (1-j2) Ω, Z_B = (3+j4) Ω. Find Z_BY.
A. (-5.22+j0.82) Ω
B. (5.22-j0.82) Ω
C. (3.8-j0.38) Ω
D. (-3.02+j8) Ω

Answer: B
Clarification: Z_BY = (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_R
= (5.22-j0.82)Ω.

9. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z_R = (2+j3) Ω, Z_Y = (1-j2) Ω, Z_B = (3+j4) Ω. Find Z_BR.
A. (5.22-j0.82) Ω
B. (-3.02+j8) Ω
C. (-5.22+j0.82) Ω
D. (3.8-j0.38) Ω

Answer: B
Clarification: Z_BR = (Z_RZ_Y + Z_YZ_B + Z_BZ_R)/Z_Y
= (-3.02+j8) Ω.

10. If a star connected system has equal impedances Z₁, then after converting into delta connected system having equal impedances Z₂, then?
A. Z₂ = Z₁
B. Z₂ = 2Z₁
C. Z₂ = 3Z₁
D. Z₂ = 4Z₁

Answer: C
Clarification: If a star connected system has equal impedances Z₁, then after converting into delta connected system having equal impedances Z₂, then Z₂ = 3Z₁.

Network Theory Multiple Choice Questions on “Transfer Function”.

1. The transfer function of a system having the input as X(s) and output as Y(s) is?
A. Y(s)/X(s)
B. Y(s) * X(s)
C. Y(s) + X(s)
D. Y(s) – X(s)

Answer: A
Clarification: The transfer function is defined as the s-domain ratio of the laplace transfrom of the output to the laplace transfrom of the input. The transfer function of a system having the input as X(s) and output as Y(s) is H(s) = Y(s)/X(s).

2. In the circuit shown below, if current is defined as the response signal of the circuit, then determine the transfer function.
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A. H(s)=C/(S² LC+RCS+1)
B. H(s)=SC/(S² LC-RCS+1)
C. H(s)=SC/(S² LC+RCS+1)
D. H(s)=SC/(S² LC+RCS-1)

Answer: C
Clarification: If the current is defined as the response signal of the circuit, then the transfer function H (s) = I/V = 1/(R+sL+1/sC. = sC/(s²Lc+RCs+1) where I corresponds to the output Y(s) and V corresponds to the input X(s).

3. In the circuit shown below, if voltage across the capacitor is defined as the output signal of the circuit, then the transfer function is?

A. H(s)=1/(S² LC-RCS+1)
B. H(s)=1/(S² LC+RCS+1)
C. H(s)=1/(S² LC+RCS-1)
D. H(s)=1/(S² LC-RCS-1)

Answer: B
Clarification: If the voltage across the capacitor is defined as the output signal of the circuit, the transfer function is H(s) = V_o/V = (1/sC./(R+sL+1/sC.=1/(S²LC+RCS+1).

4. Let us assume x (t) = A cos(ωt + φ), then the Laplace transform of x (t) is?
A. X(S)=A(Scos Ø-ω sinØ)/(S²-ω²)
B. X(S)=A(Scos Ø+ω sinØ)/(S²+ω²)
C. X(S)=A(Scos Ø+ω sinØ)/(S²-ω²)
D. X(S)=A(Scos Ø-ω sinØ)/(S²+ω²)

Answer: D
Clarification: We use the transfer function to relate the study state response to the excitation source. And we had assumed that x (t) = A cos(ωt + φ). On expanding and taking the laplace transform we get X(s) = AcosØs/(s²+ω²)-AsinØs/(s²+ω²) = A(Scos Ø-ω sinØ)/(S²+ω²).

5. Let us assume x (t) = A cos(ωt + φ), what is the s-domain expression?
A. Y(s)=H(s) A(Scos Ø-ω sinØ)/(S²-ω²)
B. Y(s)=H(s) A(Scos Ø+ω sinØ)/(S²+ω²)
C. Y(s)=H(s) A(Scos Ø-ω sinØ)/(S²+ω²)
D. Y(s)=H(s) A(Scos Ø+ω sinØ)/(S²-ω²)

Answer: C
Clarification: We had the equation Y(s) = H(s)X(s) to find the steady state solution of y(t). The s-domain expression for the response for the input is Y(s)=H(s) A(Scos Ø-ω sinØ)/(S²+ω²).

6. Let us assume x (t) = A cos(ωt + φ), on taking the partial fractions for the response we get?
A. Y(s)=k₁/(s-jω)+(k₁^‘)/(s+jω)+Σterms generated by the poles of H(s)
B. Y(s)=k₁/(s+jω)+(k₁^‘)/(s+jω)+Σterms generated by the poles of H(s)
C. Y(s)=k₁/(s+jω)+(k₁^‘)/(s-jω)+Σterms generated by the poles of H(s)
D. Y(s)=k₁/(s-jω)+(k₁^‘)/(s-jω)+Σterms generated by the poles of H(s)

Answer: A
Clarification: By taking partial fractions, Y(s)=k₁/(s-jω)+(k₁^‘)/(s+jω)+Σterms generated by the poles of H(s). The first two terms result from the complex conjugate poles of the deriving source.

7. Let us assume x (t) = A cos(ωt + φ), what is the value of k₁?
A. 1/2 H(jω)Ae^jØ
B. H(jω)Ae^-jØ
C. H(jω)Ae^jØ
D. 1/2 H(jω)Ae^-jØ

Answer: D
Clarification: The first two terms on the right hand side of Y(s) determine the steady state response. k₁=H(s) (A(scosØ-ωsin⁡Ø))/(s+jω)|s=jω = 1/2 H(jω)Ae^-jØ.

8. The relation between H (jω) and θ (ω) is?
A. H(jω)=e^{-jθ (ω)}
B. H(jω)=|H(jω)|e^{-jθ (ω)}
C. H(jω)=|H(jω)|e^{jθ (ω)}
D. H(jω)=e^{jθ (ω)}

Answer: C
Clarification: In general, H(jω) is a complex quantity, thus H(jω) = |H(jω)|e^jθ(ω) where |H(jω)| is the magnitude and the phase angle is θ(ω) of the transfer function vary with frequency ω.

9. Let us assume x (t) = A cos(ωt + φ), what is the value of k₁ by considering θ (ω) is?
A. |H(jω)|e^{j[θ (ω)+Ø]}
B. A/2|H(jω)|e^{j[θ (ω)+Ø]}
C. |H(jω)|e^{-j[θ (ω)+Ø]}
D. A/2 |H(jω)|e^{-j[θ (ω)+Ø]}

Answer: B
Clarification: The expression of k₁ becomes K₁ = A/2|H(jω)|e^{j[θ (ω)+Ø]}. We obtain the steady state solution for y(t) by taking the inverse transform ignoring the terms generated by the poles of H(s).

10. Let us assume x (t) = A cos(ωt + φ), What is the final steady state solution for y (t)?
A. A|H(jω)|cos⁡[ωt+Ø+ θ (ω)]
B. A|H(jω)|cos⁡[ωt-Ø+ θ (ω)]
C. A|H(jω)|cos⁡[ωt-Ø- θ (ω)]
D. A|H(jω)|cos⁡[ωt+Ø- θ (ω)]

Answer: A
Clarification: We obtain the steady state solution for y (t) by taking the inverse transform of Y(s) ignoring the terms generated by the poles of H (s). Thus y_ss(t) = A|H(jω)|cos⁡[ωt+Ø+ θ (ω)] which indicates how to use the transfer function to find the steady state response of a circuit.

Network Theory Multiple Choice Questions on “Inverse Hybrid (g) Parameter”.

1. For the circuit given below, the value of the Inverse hybrid parameter g₁₁ is ___________

A. 0.067 Ω
B. 0.025 Ω
C. 0.3 Ω
D. 0.25 Ω

Answer: A
Clarification: Inverse Hybrid parameter g₁₁ is given by, g₁₁ = (frac{I_1}{V_1}), when I₂ = 0.
Therefore short circuiting the terminal Y-Y’, we get,
V₁ = I₁ ((10||10) + 10)
= I₁ (left(left(frac{10×10}{10+10}right)+10right))
= 15I₁
∴ (frac{I_1}{V_1} = frac{1}{15}) = 0.067 Ω
Hence g₁₁ = 15 Ω.

2. For the circuit given below, the value of the Inverse hybrid parameter g₂₁ is ___________

A. 0.6 Ω
B. 0.5 Ω
C. 0.3 Ω
D. 0.2 Ω

Answer: B
Clarification: Inverse Hybrid parameter g₂₁ is given by, g₂₁ = (frac{V_2}{V_1}), when I₂ = 0.
Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,
V₁ = I₁ (10 + 10)
V₂ = I₁ 10
∴ (frac{V_2}{V_1} = frac{I_1 10}{I_1 20}) = 0.5
Hence g₂₁ = 0.5 Ω.

3. For the 2 port network as shown below, the Z-matrix is ___________

A. [Z₁; Z₁ + Z₂; Z₁ + Z₂; Z₃]
B. [Z₁; Z₁; Z₁ + Z₂; Z₂]
C. [Z₁; Z₂; Z₂; Z₁ + Z₂]
D. [Z₁; Z₁; Z₁; Z₁ + Z₂]

Answer: D
Clarification: z₁₁ = (frac{V_1}{I_1}), when I₂ = 0
z₂₂ = (frac{V_2}{I_2}), when I₁ = 0
z₁₂ = (frac{V_1}{I_2}), when I₁ = 0
z₂₁ = (frac{V_2}{I_1}), when I₂ = 0
Now, in the given circuit putting I₁ = 0, we get,
z₁₂ = Z₁ and z₂₂ = Z₁ + Z₂
And putting I₂ = 0, we get,
z₂₁ = Z₁ and z₁₁ = Z₁.

4. Which one of the following parameters does not exist for the two-port network in the circuit given below?

A. h
B. Y
C. Z
D. g

Answer: C
Clarification: Y-parameter = (frac{1}{Z})[1; -1; -1; 1]
And from the definition of the Y parameters, ∆Y = 0. Therefore the Y-parameter exists.
Since ∆Y = 0, so by property of reciprocity, ∆h = 0 and ∆g = 0.
Hence both hybrid and inverse hybrid parameters exist.
But the Z-parameters cannot exist here because if one terminal is opened the circuit will become invalid.
∴ Z- parameters do not exists.

5. In the circuit given below, the value of the Inverse hybrid parameter g₁₁ is _________

A. 10 Ω
B. 0.133 Ω
C. 5 Ω
D. 2.5 Ω

Answer: B
Clarification: Inverse Hybrid parameter g₁₁ is given by, g₁₁ = (frac{I_1}{V_1}), when I₂ = 0.
Therefore short circuiting the terminal Y-Y’, we get,
V₁ = I₁ ((5 || 5) + 5)
= I₁ (left(left(frac{5×5}{5+5}right)+5right))
= 7.5I₁
∴ (frac{I_1}{V_1} = frac{1}{7.5}) = 0.133
Hence g₁₁ = 7.5 Ω.

6. In the circuit given below, the value of the Inverse hybrid parameter g₂₁ is _________

A. 10 Ω
B. 0.5 Ω
C. 5 Ω
D. 2.5 Ω

Answer: B
Clarification: Inverse Hybrid parameter g₂₁ is given by, g₂₁ = (frac{V_2}{V_1}), when I₂ = 0.
Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,
V₁ = I₁ (5 + 5)
V₂ = I₁ 5
∴ (frac{V_2}{V_1} = frac{I_1 5}{I_1 10}) = 0.5
Hence g₂₁ = 0.5 Ω.

7. The short-circuit admittance matrix of a two port network is as follows.
[0; -0.5; 0.5; 0]
Then the 2 port network is ____________
A. Non-reciprocal and passive
B. Non-reciprocal and active
C. Reciprocal and passive
D. Reciprocal and active

Answer: B
Clarification: The network is non reciprocal because Y₁₂ ≠ Y₂₁ and Y₁₂ is also negative which means either energy storing or providing device is available. So network is active. Therefore the network is Non- reciprocal and active.

8. If a two port network is passive, then we have, with the usual notation, the relationship as _________
A. h₂₁ = h₁₂
B. h₁₂ = -h₂₁
C. h₁₁ = h₂₂
D. h₁₁ h₂₂ – h₁₂ h₂₂ = 1

Answer: D
Clarification: We know that, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (1)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
I₁ = (frac{V_1}{h_{11}} – frac{h_{12} V_2}{h_{11}}) ………. (5)
And I₂ = (frac{h_{21} V_1}{h_{11}} + left(- frac{h_{21} h_{12}}{h_{11}} + h_{22}right) V_2) ………. (6)
Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y=0, we get,
h₁₁ h₂₂ – h₁₂ h₂₂ = 1 [hence proved].

9. For the circuit given below, the value of the Inverse hybrid parameter g₂₂ is ___________

A. 7.5 Ω
B. 5 Ω
C. 6.25 Ω
D. 3 Ω

Answer: A
Clarification: Inverse Hybrid parameter g₂₂ is given by, g₂₂ = (frac{V_2}{I_2}), when V₁ = 0.
Therefore short circuiting the terminal X-X’ we get,
-5 I₂ – 5 I₁ + V₂ = 0
-5 I₁ – 5(I₁ – I₂) = 0
Or, 2I₁ = I₂
Putting the above equation in the first equation, we get,
-7.5 I₂ = -V₂
Or, (frac{V_2}{I_2}) = 7.5
Hence g₂₂ = 7.5 Ω.

10. In two-port networks the parameter g₁₁ is called _________
A. Short circuit input impedance
B. Short circuit current ratio
C. Open circuit voltage ratio
D. Open circuit input admittance

Answer: D
Clarification: We know that, g₁₁ = (frac{I_1}{V_1}), when I₂ = 0.
Since the second voltage terminal is short circuited when the ratio of the current and voltage is measured, therefore the parameter g₁₁ is called as Open circuit input admittance.

11. In two-port networks the parameter g₂₁ is called _________
A. Short circuit input impedance
B. Short circuit current ratio
C. Open circuit voltage ratio
D. Open circuit input admittance

Answer: C
Clarification: We know that, g₂₁ = (frac{V_2}{V_1}), when I₂ = 0.
Since the second output terminal is short circuited when the ratio of the two voltages is measured, therefore the parameter g₂₁ is called as Open circuit voltage ratio.

12. In two-port networks the parameter g₁₂ is called _________
A. Short circuit input impedance
B. Short circuit current gain
C. Open circuit reverse voltage gain
D. Open circuit output admittance

Answer: C
Clarification: We know that, g₁₂ = (frac{I_1}{I_2}), when V₁ = 0.
Since the primary terminal is short circuited and the ratio of the two currents is measured, therefore the parameter g₁₂ is called as Short circuit current ratio.

13. In two-port networks the parameter g₂₂ is called _________
A. Short circuit input impedance
B. Short circuit current ratio
C. Open circuit voltage ratio
D. Open circuit input admittance

Answer: B
Clarification: We know that, g₂₂ = (frac{V_2}{I_2}), when V₁ = 0.
Since the primary voltage terminal is short circuited and the ratio of the voltage and current in second loop is measured, therefore the parameter g₂₂ is called as Short circuit current ratio.

14. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₁₂ in terms of Inverse Hybrid parameters can be expressed as ________
A. y₁₂ = (left(g_{11} – frac{g_{12} g_{21}}{g_{22}}right))
B. y₁₂ = (frac{g_{12}}{g_{22}} )
C. y₁₂ = –(frac{g_{21}}{g_{22}} )
D. y₁₂ = (frac{1}{g_{22}})

Answer: B
Clarification: We know that, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (1)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
I₁ = (left(g_{11} – frac{g_{12} g_{21}}{g_{22}}right)V_1 + frac{g_{12}}{g_{22}} V_2)………. (5)
And I₂ = –(frac{g_{21} V_1}{g_{22}} + frac{V_2}{g_{22}})………. (6)
∴ Comparing (1), (2) and (5), (6), we get,
y₁₁ = (left(g_{11} – frac{g_{12} g_{21}}{g_{22}}right))
y₁₂ = (frac{g_{12}}{g_{22}} )
y₂₁ = –(frac{g_{21}}{g_{22}} )
y₂₂ = (frac{1}{g_{22}}).

15. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₂₁ in terms of Inverse Hybrid parameters can be expressed as ________
A. y₂₁ = (left(g_{11} – frac{g_{12} g_{21}}{g_{22}}right))
B. y₂₁ = (frac{g_{12}}{g_{22}} )
C. y₂₁ = –(frac{g_{21}}{g_{22}} )
D. y₂₁ = (frac{1}{g_{22}})

Answer: C
Clarification: We know that, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (1)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
I₁ = (left(g_{11} – frac{g_{12} g_{21}}{g_{22}}right)V_1 + frac{g_{12}}{g_{22}} V_2)………. (5)
And I₂ = –(frac{g_{21} V_1}{g_{22}} + frac{V_2}{g_{22}})………. (6)
∴ Comparing (1), (2) and (5), (6), we get,
y₁₁ = (left(g_{11} – frac{g_{12} g_{21}}{g_{22}}right))
y₁₂ = (frac{g_{12}}{g_{22}} )
y₂₁ = –(frac{g_{21}}{g_{22}} )
y₂₂ = (frac{1}{g_{22}}).

Network Theory Multiple Choice Questions on “Tree and Co-Tree”.

1. A graph is said to be a directed graph if ________ of the graph has direction.
A. 1 branch
B. 2 branches
C. 3 branches
D. every branch

Answer: D
Clarification: If every branch of the graph has direction, then the graph is said to be a directed graph. If the graph does not have any direction then that graph is called undirected graph.

2. The number of branches incident at the node of a graph is called?
A. degree of the node
B. order of the node
C. status of the node
D. number of the node

Answer: A
Clarification: Nodes can be incident to one or more elements. The number of branches incident at the node of a graph is called degree of the node.

3. If no two branches of the graph cross each other, then the graph is called?
A. directed graph
B. undirected graph
C. planar graph
D. non-planar graph

Answer: C
Clarification: If a graph can be drawn on a plane surface such that no two branches of the graph cross each other, then the graph is called planar graph.

4. Consider the graph given below. Which of the following is a not a tree to the graph?

Answer: D
Clarification: Tree is sub graph which consists of all node of original graph but no closed paths. So, ‘d’ is not a tree to the graph.

5. Number of twigs in a tree are? (where, n-number of nodes)
A. n
B. n+1
C. n-1
D. n-2

Answer: C
Clarification: Twig is a branch in a tree. Number of twigs in a tree are n-1. If there are 4 nodes in a tree then number of possible twigs are 3.

6. Loops which contain only one link are independent are called?
A. open loops
B. closed loops
C. basic loops
D. none of the mentioned

Answer: C
Clarification: The addition of subsequent link forms one or more additional loops. Loops that contain only one link are independent are called basic loops.

7. If the incident matrix of a graph is given below. The corresponding graph is?

     a    b   c   d   e   f
     1   +1   0  +1   0   0   +1
     2   -1  -1   0  +1   0    0
     3    0  +1   0   0   +1  -1
     4    0   0  -1  -1   -1   0

A.

Answer: B

     a    b   c  d    e   f
     1   +1   0  +1   0   0   +1
     2   -1  -1   0  +1   0    0
     3    0  +1   0   0   +1  -1
     4    0   0  -1  -1   -1   0

8. If A represents incidence matrix, I represents branch current vectors, then?
A. AI = 1
B. AI = 0
C. AI = 2
D. AI= 3

Answer: B
Clarification: If A represents incidence matrix, I represents branch current vectors, then the relation is AI= 0 that is its characteristic equation must be equated to zero

9. If a graph consists of 5 nodes, then the number of twigs in the tree is?
A. 1
B. 2
C. 3
D. 4

Answer: D
Clarification: Number of twigs = n-1. As given number of nodes are 5 then n = 5. On substituting in the equation, number of twigs = 5 -1 = 4.

10. If there are 4 branches, 3 nodes then number of links in a co-tree are?
A. 2
B. 4
C. 6
D. 8

Answer: A
Clarification: Number of links = b-n+1. Given number of branches = 4 and number of nodes = 3. On substituting in the equation, number of links in a co-tree = 4 – 3 + 1 = 2.