Previous Year Questions- Two Port Network and Network Functions | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q1: Two passive two-port network P and Q are connected as shown in the figure. The impedance matrix of network P is  The admittance matrix of network  Let the ABCD matrix of the two-port network R in the figure be  The value of β in Ω is _______ (rounded off to 2 decimal places)      (2024)
(a) -25.36
(b) -19.8
(c) -22.36
(d) -15.25
Ans: (b)
Sol: From Z_p matrix
Convert Z-parameters in to ABCD parameters
From equation (ii),
Sub equation (iii) in equation (i),
From equation (iii) and (iv),
Convert Y-parameters to ABCD parameters From equation (vi),  
Sub equation (vii) in equation (v),
From equation (vii) and equation (viii),

Q2: The admittance parameters of the passive resistive two-port network shown in the figure are
y₁₁ = 5 S, y₂₂ = 1 S, y₁₂ = y₂₁ = −2.5 S  
The power delivered to the load resistor R_L in Watt is ____ (Round off to 2 decimal places).     (2023)
(a) 238
(b) 452.25
(c) 632.12
(d) 145.25
Ans: (a)
Sol: Equivalent circuit

Q3: In the two-port network shown, the h₁₁ parameter (where, h₁₁=  V₁/I₁, when V₂ = 0) in ohms is _____________ (up to 2 decimal places).      (2018)
(a) 0.25
(b) 0.5
(c) 0.75
(d) 0.85
Ans: (b)
Sol: By KCL,
Substitute equation (ii) in equation (i) [/latex]

Q4: Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1.
A₁, B₁, C₁, D₁, A₂, B₂, C₂ and D₂ are the generalized circuit constants. If the Thevenin equivalent circuit at port 3 consists of a voltage source  V_T and impedance Z_T connected in series, then      (SET-1 (2017))
(a)
(b)
(c)
(d)
Ans: (d)
Sol: For two port network, we can write,

Q5: The z-parameters of the two port network shown in the figure are
 z₁₁ = 40Ω, z₁₂ = 60Ω, z₂₁ = 80Ω and  z₂₂ = 100Ω.
The average power delivered to R_L = 20Ω, in watts, is _______.      (SET-2 (2016))
(a) 18.25
(b) 24.35
(c) 28.45
(d) 35.55
Ans: (d)
Sol: From equation (i) and (iii), we get
Using equation (ii) and (iv), we get
From the figure,
Power dissipated in

Q6: The driving point input impedance seen from the source V_s of the circuit shown below, in Ω, is ______.     (SET-2 (2016))
(a) 10
(b) 20
(c) 30
(d) 40
Ans: (b)
Sol: To find impedance seen by V_s
Applying KCL at node A,

Q7: The driving point impedance Z(s) for the circuit shown below is     (SET-3 (2014))
(a)
(b)
(c)
(d)
Ans: (a)
Sol: Driving point impedance, Z(s) is ,

Q8: With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:
(i) 1 Ω connected at port B draws a current of 3 A
(ii) 2.5 Ω connected at port B draws a current of 2 A
For the same network, with 6 V dc connected at port A, 1 Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is      (2012)
(a) 6 V
(b) 7 V
(c) 8 V
(d) 9 V
Ans: (b)
Sol: (i)

Q9: With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:
(i) 1 Ω connected at port B draws a current of 3 A
(ii) 2.5 Ω connected at port B draws a current of 2 A
With 10 V dc connected at port A, the current drawn by 7 Ω connected at port B is     (2012)
(a) 3/7A
(b) 5/7A
(c) 1A
(d) 9/7A
Ans: (c)
Sol: -ve sign is signifies that current is drawn.

Q10: The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a, b) and (c, d) respectively. It has an impedance matrix Z with parameters denoted by Z_ij. A 1 Ω resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box ) is      (2010)
(a)  
(b)
(c)
(d)
Ans: (c)
Sol: As, 1Ω resistor is connected in series with the network as port-1.
V₂ does not get affected,  
Modified Z-parameter

Q11: The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are     (2006)
(a) z parameters,
(b) h parameters,
(c) g parameters,
(d) z parameters,
Ans: (c)
Sol: Since port-1 is open circuit, I₁ = 0
Port-2 is short-circuit, V₂ = 0
$𝐼1=0$
Q12: The parameters of the circuit shown in the figure are
R_i = 1 MΩ, R_o = 10 Ω, A = 10⁶ V/V.
If v_i = 1μV, then output voltage, input impedance and output impedance respectively are     (2006)
(a) 1V,  ∞, 10Ω
(b) 1 V, 0, 10Ω
(c) 1 V, 0, ∞
(d) 10 V, ∞, 10 Ω  
Ans: (a)
Sol: Output voltage V₀ = AV_i
V₀ = 10⁶ × 1 × 10⁻⁶ = 1V
To calculate input impedance, V_dc source is connected at input port  
$\mathrm{Ans:\; \Omega }$Input impedance
as loop is not closed, I_i = 0
So,
To calculate output impedance, V_dc source is connected at output port.
Output impedance,

Q13: Two networks are connected in cascade as shown in the figure. With usual notations the equivalent A, B, C and D constants are obtained. Given that, C = 0.025∠45°, the value of Z₂ is     (2005)(a) 10∠30°Ω
(b) 40∠−45°Ω
(c) 1 Ω
(d) 0 Ω
Ans: (b)
Sol: Putting I₂ = 0 in equation (i)

Q14: For the two port network shown in the figure, the Z-matrix is given by    (2005)
(a)  
(b)
(c)
(d)
Ans: (d)
Sol: From above, we can write,

Q15: The Z-matrix of a 2-port network as given by
The element Y₂₂ of the corresponding Y-matrix of the same network is given by      (2004)
(a) 1.2
(b) 0.4
(c) -0.4
(d) 1.8
Ans: (d)
Sol: 
Q16: The h-parameters for a two-port network are defined by  For the two-port network shown in figure, the value of h₁₂ is given by        (2003)
(a) 0.125
(b) 0.167
(c) 0.625
(d) 0.25
Ans: (d)
Sol: h₁₂ is ratio of E₁ to E₂ for the input open-circuited condition.
 Assuming I₁ = 0

Q17: A two port network, shown in figure is described by the following equations
The admittance parameters Y₁₁, Y₁₂, Y₂₁, Y₂₂ for the network shown are     (2002)
(a) 0.5 mho, 1 mho, 2 mho and 1 mho respectively
(b) 1/3 mho, -1/6 mho, -1/6 mho and 1/3 mho respectively
(c) 0.5 mho, 0.5 mho, 1.5 mho and 2 mho respectively
(d) -2/5 mho, -3/7 mho, 3/7 mho and 2/5 mho respectively
Ans: (b)
Sol: Using KVL,
E₁ = 2I₁ + 2(I₁+I₂)
Again using KVL,

Q18: A passive 2-port network is in a steady-state. Compared to its input, the steady state output can never offer      (2001)
(a) higher voltage
(b) lower impedance
(c) greater power
(d) better regulation
Ans: (c)
Sol: For a passive two port network, output powe can never be grater than input power.