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Network Theory MCQs – Two-Port Networks MCQs ( Network Theory ) MCQs

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Series-Series Connection of Two Port Network

1. In the circuit given below, the value of R is ___________

a) 2.5 Ω
b) 5.0 Ω
c) 7.5 Ω
d) 10.0 Ω
Answer: c
Explanation: The resultant R when viewed from voltage source = 1008 = 12.5
∴ R = 12.5 – 10 || 10 = 12.5 – 5 = 7.55 Ω.


2. In the circuit given below, the number of chords in the graph is ________________

a) 3
b) 4
c) 5
d) 6
Answer: b
Explanation: Given that, b = 6, n = 3
Number of Links is given by, b – n + 1
= 6 – 3 + 1 = 4.


3. In the circuit given below, the current through the 2 kΩ resistance is _____________

a) Zero
b) 1 mA
c) 2 mA
d) 6 mA
Answer: a
Explanation: We know that when a Wheatstone bridge is balanced, no current will flow through the middle resistance.
Here, R1R2=R3R4
Since, R₁ = R₂ = R₃ = R₄ = 1 kΩ.


4. How many incandescent lamps connected in series would consume the same total power as a single 100 W/220 V incandescent lamp. The rating of each lamp is 200 W/220 V?
a) Not possible
b) 4
c) 3
d) 2
Answer: d
Explanation: In series power = 1P
Now, 1P=1P1+1P2
= 1200+1200
Or, P = 2002 = 100 W.


5. Two networks are connected in series parallel connection. Then, the forward short-circuit current gain of the network is ____________
a) Product of Z-parameter matrices
b) Sum of h-parameter matrices
c) Sum of Z-parameter matrices
d) Product of h-parameter matrices
Answer: b
Explanation: The forward short circuit current gain is given by,
h₂₁ = I2(s)I1(s), when V₂ = 0
So, when the two networks are connected in series parallel combination,
[h₁₁, h₁₂; h₂₁, h₂₂] = [h’₁₁ + h’₁₁, h’₁₂ + h’₁₂; h’₂₁ + h’₂₁, h’₂₂ + h’₂₂]
So, h₂₁ of total network will be sum of h parameter matrices.


6. The condition for a 2port network to be reciprocal is ______________
a) Z₁₁ = Z₂₂
b) BC – AD = -1
c) Y₁₂ = -Y₂₁
d) h₁₂ = h₂₁
Answer: b
Explanation: If the network is reciprocal, then the ratio of the response transform to the excitation transform would not vary after interchanging the position of the excitation.


7. The relation AD – BC = 1, (where A, B, C and D are the elements of a transmission matrix of a network) is valid for ___________
a) Both active and passive networks
b) Passive but not reciprocal networks
c) Active and reciprocal networks
d) Passive and reciprocal networks
Answer: d
Explanation: AD – BC = 1, is the condition for reciprocity for ABCD parameters, which shows that the relation is valid for reciprocal network. The ABCD parameters are obtained for the network which consists of resistance, capacitance and inductance, which indicates that it is a passive network.


8. For a 2 port network, the transmission parameters are given as 10, 9, 11 and 10 corresponds to A, B, C and D. The correct statement among the following is?
a) Network satisfies both reciprocity and symmetry
b) Network satisfies only reciprocity
c) Network satisfies only symmetry
d) Network satisfies neither reciprocity nor symmetry
Answer: a
Explanation: Here, A = 10, B = 9, C = 11, D = 10
∴ A = D
∴ Condition for symmetry is satisfied.
Also, AD – BC = (10) (10) – (9) (11)
= 100 – 99 = 1
Therefore the condition of reciprocity is satisfied.


9. In the circuit given below, the equivalent capacitance is ______________

a) C4
b) 5C13
c) 5C2
d) 3C
Answer: b
Explanation: The equivalent capacitance by applying the concept of series-parallel combination of the capacitance is,
1CEQ=1C+1C+15C/3
= 1C+1C+35C=1C(5+5+35)
Or, C_EQ = 5C13
123 c Energy delivered during talk time
E = ∫ V(t)I(t) dt
Given, I (t) = 2 A = constant = 2 ∫ V(t)dt
= 2 × Shaded area
= 2 × 12 × (10 + 12) × 60 × 10
= 13.2 kJ.


10. In the circuit given below, the 60 V source absorbs power. Then the value of the current source is ____________

a) 10 A
b) 13 A
c) 15 A
d) 18 A
Answer: a
Explanation: Given that, 60 V source is absorbing power, it means that current flow from positive to negative terminal in 60 V source.
Applying KVL, we get, I + I₁ = 12 A ……………… (1)
Current source must have the value of less than 12 A to satisfy equation (1).


11. In the circuit given below, the number of node and branches are ______________

a) 4 and 5
b) 4 and 6
c) 5 and 6
d) 6 and 4
Answer: b
Explanation: In the given graph, there are 4 nodes and 6 branches.
Twig = n – 1 = 4 – 1 = 3
Link = b – n + 1 = 6 – 4 + 1 = 3.


12. A moving coil of a meter has 250 turns and a length and depth of 40 mm and 30 mm respectively. It is positioned in a uniform radial flux density of 450 mT. The coil carries a current of 160 mA. The torque on the coil is?
a) 0.0216 N-m
b) 0.0456 N-m
c) 0.1448 N-m
d) 1 N-m
Answer: a
Explanation: Given, N = 250, L = 40 × 10^-3, d = 30 × 10^-3m, I = 160 × 10^-3A, B = 450 × 10^-3 T
Torque = 250 × 450 × 10^-3 × 40 × 10^-3 × 30 × 10^-3 × 160 × 10^-3
= 200 × 10^-6 N-m = 0.0216 N-m.


13. In the circuit given below, the equivalent inductance is ____________

a) L₁ + L₂ – 2M
b) L₁ + L₂ + 2M
c) L₁ + L₂ – M
d) L₁ + L₂
Answer: a
Explanation: Since, in one inductor current is leaving to dot and in other inductor current is entering to dot.
So, L_EQ = L₁ + L₂ – 2M.


14. In the figure given below, the pole-zero plot corresponds to _____________

a) Low-pass filter
b) High-pass filter
c) Band-pass filter
d) Notch filter
Answer: d
Explanation: In pole zero plot the two transmission zeroes are located on the jω-axis, at the complex conjugate location, and then the magnitude response exhibits a zero transmission at ω – ω_C.


15. In the circuit given below, the maximum power that can be transferred to the resistor R_L is _____________

a) 1 W
b) 10 W
c) 0.25 W
d) 0.5 W
Answer: c
Explanation: For maximum power transfer to the load resistor R_L, R_L must be equal to 100 Ω.
∴ Maximum power = V24RL
= 1024×100 = 0.25 W.

Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters

1. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Transmission parameters can be expressed as ________
a) y₁₁ = DB
b) y₁₁ = C−AB
c) y₁₁ = – 1B
d) y₁₁ = AB
Answer: a
Explanation: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = ABV2–1BV1 …………. (5)
And I₁ = CV₂ – D (ABV2–1BV1)=DBV1+(C−AB)V2 …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = DB
y₁₂ = C−AB
y₂₁ = – 1B
y₂₂ = AB.


2. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Transmission parameters can be expressed as ________
a) y₁₂ = DB
b) y₁₂ = C−AB
c) y₁₂ = – 1B
d) y₁₂ = AB
Answer: b
Explanation: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = ABV2–1BV1 …………. (5)
And I₁ = CV₂ – D (ABV2–1BV1)=DBV1+(C−AB)V2 …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = DB
y₁₂ = C−AB
y₂₁ = – 1B
y₂₂ = AB.


3. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₁ in terms of Transmission parameters can be expressed as ________
a) Y₂₁ = DB
b) Y₂₁ = C−AB
c) Y₂₁ = – 1B
d) Y₂₁ = AB
Answer: c
Explanation: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = ABV2–1BV1 …………. (5)
And I₁ = CV₂ – D (ABV2–1BV1)=DBV1+(C−AB)V2 …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = DB
y₁₂ = C−AB
y₂₁ = – 1B
y₂₂ = AB.


4. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₂ in terms of Transmission parameters can be expressed as ________
a) y₂₂ = DB
b) y₂₂ = C−AB
c) y₂₂ = – 1B
d) y₂₂ = AB
Answer: d
Explanation: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = ABV2–1BV1 …………. (5)
And I₁ = CV₂ – D (ABV2–1BV1)=DBV1+(C−AB)V2 …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = DB
y₁₂ = C−AB
y₂₁ = – 1B
y₂₂ = AB.


5. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₁ in terms of Transmission parameters can be expressed as ____________
a) z₁₁ = AC
b) z₁₁ = ADC–B
c) z₁₁ = 1C
d) z₁₁ = DC
Answer: a
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = 1CI1+DCI2 …………… (5)
And V₁ = A(1CI1+DCI2)–BI2=ACI1+(ADC–B)I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = AC
z₁₂ = ADC–B
z₂₁ = 1C
z₂₂ = DC.


6. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₂ in terms of Transmission parameters can be expressed as ____________
a) z₁₂ = AC
b) z₁₂ = ADC–B
c) z₁₂ = 1C
d) z₁₂ = DC
Answer: b
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = 1CI1+DCI2 …………… (5)
And V₁ = A(1CI1+DCI2)–BI2=ACI1+(ADC–B)I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = AC
z₁₂ = ADC–B
z₂₁ = 1C
z₂₂ = DC.


7. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₁ in terms of Transmission parameters can be expressed as ____________
a) z₂₁ = AC
b) z₂₁ = ADC–B
c) z₂₁ = 1C
d) z₂₁ = DC
Answer: c
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = 1CI1+DCI2 …………… (5)
And V₁ = A(1CI1+DCI2)–BI2=ACI1+(ADC–B)I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = AC
z₁₂ = ADC–B
z₂₁ = 1C
z₂₂ = DC.


8. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₂ in terms of Transmission parameters can be expressed as ____________
a) z₂₂ = AC
b) z₂₂ = ADC–B
c) z₂₂ = 1C
d) z₂₂ = DC
Answer: d
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = 1CI1+DCI2 …………… (5)
And V₁ = A(1CI1+DCI2)–BI2=ACI1+(ADC–B)I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = AC
z₁₂ = ADC–B
z₂₁ = 1C
z₂₂ = DC.


9. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Inverse Transmission parameters can be expressed as ________
a) y₁₁ = A′B′
b) y₁₁ = – 1B′
c) y₁₁ = (C′–D′A′B′)
d) y₁₁ = D′B′
Answer: a
Explanation: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – 1B′V2+A′B′V1 …………. (5)
And I₂ = C’V₁ – D’ (−1B′V2+A′B′V1)=(C′–D′A′B′)V1+D′B′V2 ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = A′B′
y₁₂ = – 1B′
y₂₁ = (C′–D′A′B′)
y₂₂ = D′B′.


10. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Inverse Transmission parameters can be expressed as ________
a) y₁₂ = A′B′
b) y₁₂ = – 1B′
c) y₁₂ = (C′–D′A′B′)
d) y₁₂ = D′B′
Answer: b
Explanation: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – 1B′V2+A′B′V1 …………. (5)
And I₂ = C’V₁ – D’ (−1B′V2+A′B′V1)=(C′–D′A′B′)V1+D′B′V2 ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = A′B′
y₁₂ = – 1B′
y₂₁ = (C′–D′A′B′)
y₂₂ = D′B′.

Two-Port Network MCQs

11. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₁ in terms of Inverse Transmission parameters can be expressed as ________
a) y₂₁ = A′B′
b) y₂₁ = – 1B′
c) y₂₁ = (C′–D′A′B′)
d) y₂₁ = D′B′
Answer: c
Explanation: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – 1B′V2+A′B′V1 …………. (5)
And I₂ = C’V₁ – D’ (−1B′V2+A′B′V1)=(C′–D′A′B′)V1+D′B′V2 ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = A′B′
y₁₂ = – 1B′
y₂₁ = (C′–D′A′B′)
y₂₂ = D′B′.


12. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₂ in terms of Inverse Transmission parameters can be expressed as ________
a) y₂₂ = A′B′
b) y₂₂ = – 1B′
c) y₂₂ = (C′–D′A′B′)
d) y₂₂ = D′B′
Answer: d
Explanation: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – 1B′V2+A′B′V1 …………. (5)
And I₂ = C’V₁ – D’ (−1B′V2+A′B′V1)=(C′–D′A′B′)V1+D′B′V2 ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = A′B′
y₁₂ = – 1B′
y₂₁ = (C′–D′A′B′)
y₂₂ = D′B′.


13. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₁ in terms of Transmission parameters can be expressed as ____________
a) z₁₁ = D′C′
b) z₁₁ = 1C′
c) z₁₁ = (A′D′C′–B′)
d) z₁₁ = A′C′
Answer: a
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (D′C′I1+1C′I2)–B′I1=(A′D′C′–B′)I1+A′C′I2 ………… (5)
And V₁ = D′C′I1+1C′I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = D′C′
z₁₂ = 1C′
z₂₁ = (A′D′C′–B′)
z₂₂ = A′C′.


14. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₂ in terms of Transmission parameters can be expressed as ____________
a) z₁₂ = D′C′
b) z₁₂ = 1C′
c) z₁₂ = (A′D′C′–B′)
d) z₁₂ = A′C′
Answer: b
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (D′C′I1+1C′I2)–B′I1=(A′D′C′–B′)I1+A′C′I2 ………… (5)
And V₁ = D′C′I1+1C′I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = D′C′
z₁₂ = 1C′
z₂₁ = (A′D′C′–B′)
z₂₂ = A′C′.


15. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₂ in terms of Transmission parameters can be expressed as ____________
a) z₂₂ = D′C′
b) z₂₂ = 1C′
c) z₂₂ = (A′D′C′–B′)
d) z₂₂ = A′C′
Answer: d
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (D′C′I1+1C′I2)–B′I1=(A′D′C′–B′)I1+A′C′I2 ………… (5)
And V₁ = D′C′I1+1C′I2 ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = D′C′
z₁₂ = 1C′
z₂₁ = (A′D′C′–B′)
z₂₂ = A′C′.

Relation between Hybrid Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters

1. A periodic voltage v (t) = 1 + 4 sin ωt + 2 cos ωt is applied across a 1Ω resistance. The power dissipated is ____________
a) 1 W
b) 11 W
c) 21 W
d) 24.5 W
Answer: b
Explanation: Given that, v (t) = 1 + 4 sin ωt + 2 cos ωt
So, Power is given by,
Power, P = 121+422√1+222√1
= 11 W.


2. A constant k high pass p section has a characteristic impedance of 300 Ω at f = ∞. At f = f_c, the characteristic impedance will be?
a) 0
b) ∞
c) 300 Ω
d) More than 300 Ω
Answer: b
Explanation: For constant k high pass p section is given by,
Z = R1–(fdf)2√
At f = f_d, denominator term is 0.
So, Z = infinite.


3. In the circuit given below, the current through R is 2 sin 8t. The value of R is ___________

a) (0.18 + j0.72)
b) (0.46 + j1.90)
c) – (0.18 + j1.90)
d) (0.23 – 0.35 j)
Answer: d
Explanation: Here, Inductor is not given, hence ignoring the inductance. Let I₁ and I₂ are currents in the loop then,
I₁ = 2sin8t3
= 0.66 sin 8t
Again, I₂ = −jX4X0.75I13.92−2.56j
= (0.23 – 0.35j) sin 8t
So, R = (0.23 – 0.35 j).


4. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₂₁ in terms of Hybrid parameters can be expressed as ________
a) y₂₁ = (−h21h12h11+h22)
b) y₂₁ = h21h11
c) y₂₁ = –h12h11
d) y₂₁ = 1h11
Answer: b
Explanation: 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 = V1h11–h12V2h11 ………. (5)
And I₂ = h21V1h11+(−h21h12h11+h22)V2 ………. (6)
∴ Comparing (1), (2) and (5), (6), we get,
y₁₁ = 1h11
y₁₂ = –h12h11
y₂₁ = h21h11
y₂₂ = (−h21h12h11+h22).


5. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₂₁ in terms of Hybrid parameters can be expressed as ________
a) z₂₁ = (h11–h21h12h22)
b) z₂₁ = – h21h22
c) z₂₁ = h12h22
d) z₂₁ = 1h22
Answer: b
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.


6. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₁ in terms of Hybrid parameters can be expressed as ________
a) z₁₁ = (h11–h21h12h22)
b) z₁₁ = – h21h22
c) z₁₁ = h12h22
d) z₁₁ = 1h22
Answer: a
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.


7. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₂ in terms of Hybrid parameters can be expressed as ________
a) z₁₂ = (h11–h21h12h22)
b) z₁₂ = – h21h22
c) z₁₂ = h12h22
d) z₁₂ = 1h22
Answer: c
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.


8. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₂₂ in terms of Hybrid parameters can be expressed as ________
a) z₂₂ = (h11–h21h12h22)
b) z₂₂ = – h21h22
c) z₂₂ = h12h22
d) z₂₂ = 1h22
Answer: d
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.


9. Permeability is analogous to _____________
a) Conductivity
b) Resistivity
c) Retentivity
d) Coercivity
Answer: a
Explanation: We know that resistance and reluctance are given by,
R = ρLA
And Reluctance = LμA
So, Permeability is analogous to conductivity.


10. A resistance and an inductance are connected in parallel and fed from 50 Hz ac mains. Each branch takes a current of 5 A. The current supplied by source is ____________
a) 10 A
b) 7.07 A
c) 5 A
d) 0 A
Answer: b
Explanation: The current is given by,
|5 – j5| = 52+52−−−−−−√
= 50−−√=52–√ = 7.07 A.


11. A triangular Pulse of 50 V peak is applied to a capacitor of 0.1 F. The change of the capacitor and its waveform shape is ___________
a) 10 rectangular
b) 5 rectangular
c) 5 triangular
d) 10 triangular
Answer: c
Explanation: We know that,
Q = CV
Or, 0.1 X 50 = 5
And it is a triangular pulse.


12. A 10 μF capacitor is charged from a 5 volt source through a resistance of 10 kΩ. The charging current offer 35 m sec. If the initial voltage on C is – 3 V is ___________
a) 0.56 mA
b) 5.6 mA
c) 6 mA
d) 5 μA
Answer: a
Explanation: Initial current immediately after charging is given by,
VR=5+310000
= 0.8 mA
Now, i = i₀e^-t/RC
= 0.8 mA x et10kX10X10−6
= 0.8 X 10^-3 X e35X10−310−1
= 0.56 mA.


13. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₁ in terms of Inverse Hybrid parameters can be expressed as ________
a) z₁₂ = 1g11
b) z₁₂ = – g12g11
c) z₁₂ = – g21g11
d) z₁₂ = (g22–g21g12g11)
Answer: a
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = I1g11–g12g11I2 ………… (5)
And V₂ = (g22–g21g12g11)I2–g21I1g11 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = 1g11
z₁₂ = – g12g11
z₂₁ = – g21g11
z₂₂ = (g22–g21g12g11).


14. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₂ in terms of Inverse Hybrid parameters can be expressed as ________
a) z₁₂ = 1g11
b) z₁₂ = – g12g11
c) z₁₂ = – g21g11
d) z₁₂ = (g22–g21g12g11)
Answer: b
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = I1g11–g12g11I2 ………… (5)
And V₂ = (g22–g21g12g11)I2–g21I1g11 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = 1g11
z₁₂ = – g12g11
z₂₁ = – g21g11
z₂₂ = (g22–g21g12g11).


15. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₂₂ in terms of Inverse Hybrid parameters can be expressed as ________
a) z₂₂ = 1g11
b) z₂₂ = – g12g11
c) z₂₂ = – g21g11
d) z₂₂ = (g22–g21g12g11)
Answer: d
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = I1g11–g12g11I2 ………… (5)
And V₂ = (g22–g21g12g11)I2–g21I1g11 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = 1g11
z₁₂ = – g12g11
z₂₁ = – g21g11
z₂₂ = (g22–g21g12g11).

Series-Parallel Interconnection of Two Port Network

1. If the diameter of a wire is doubled, the current carrying capacity of the wire is ___________
a) Half
b) Twice
c) Four times
d) One-fourth
Answer: c
Explanation: Since diameter is doubled, area of cross-section becomes four times. Current carrying capacity is proportional to area of cross-section.


2. Consider an RL series circuit having resistance R = 3 Ω, inductance L = 3 H and is excited by 6V. The current after a long time after closing of switch is ____________
a) 1 A
b) 2 A
c) 0 A
d) Infinity
Answer: b
Explanation: At t = ∞ the circuit has effectively two 6Ω resistances in parallel.
So, R_EQ = 6X66+6
= 3612 = 3 Ω
Given voltage = 6 V
So, current = 2 A.


3. The energy stored in a coil is 108 J. The power dissipated instantaneously across the blades of switch after it is opened in 10 ms is ____________
a) 108 W
b) 1080 W
c) 10800 W
d) 108000 W
Answer: c
Explanation: Power dissipated instantaneously across the blades of the switch is given by,
Power, P = EnergyTime
Given that, Energy = 108 J and time = 10 X 10^-3
So, P = 10810X10−3 = 10800 W.


4. A parallel RLC circuit with R₁ = 20, L₁ = 1100 and C₁ = 1200 is scaled giving R₂ = 10⁴, L₂ = 10^-4 and C₂, the value of C₂ is ___________
a) 0.10 nF
b) 0.3 nF
c) 0.2 nF
d) 0.4 nF
Answer: c
Explanation: K₁ = R2R1
= 10420 = 5 X 10² = 500 Ω
And L1L2=kωk1
Or, kωk1=L1L2Xk1
Or, 10−210−4X5X102 = 5 X 10⁴
Or, C₂ = C1kω.k1=0.5X10−25X104X500
= 0.02 X 10^-8 = 0.2 nF.


5. Barletts Bisection Theorem is applicable to ___________
a) Unsymmetrical networks
b) Symmetrical networks
c) Both unsymmetrical and symmetrical networks
d) Neither to unsymmetrical nor to symmetrical networks
Answer: b
Explanation: A symmetrical network can be split into two halves. So the z parameters of the network are symmetrical as well as reciprocal of each other. Hence Barletts Bisection Theorem is applicable to Symmetrical networks.


6. The Thevenin’s equivalent of a network is a 10 V source in series with 2 Ω resistances. If a 3 Ω resistance is connected across the Thevenin’s equivalent is _____________
a) 10 V in series with 1.2 Ω resistance
b) 6 V in series with 1.2 Ω resistance
c) 10 V in series with 5 Ω resistance
d) 6 V in series with 5 Ω resistance
Answer: b
Explanation: The Thevenin equivalent voltage is given by,
V_TH = 10X35
= 6 V
And the Thevenin equivalent Resistance is given by,
R_TH = 3X25 = 1.2 Ω.


7. A magnetic circuit has an iron length of 100 cm and air gap length 10 cm. If μ_r = 200 then which of the following is true?
a) Mmf for iron and air gap are equal
b) Mmf for iron is much less than that for air gap
c) Mmf for iron is much more than that for air gap
d) Mmf for iron and air gap are not equal
Answer: a
Explanation: We know that, MMF for air = B4πX10−7 X 10
Where B is the magnetic field intensity.
Also, MMF for iron = BX100200(4πX10−7)
= BX0.54πX10−7.


8. Two coils X and Y have self-inductances of 5 mH and 10 mH and mutual inductance of 3 mH. If the current in coils X change at a steady rate of 100 A/s, the emf induced in coil Y is ____________
a) 0.3 V
b) 0.5 V
c) 1 V
d) 1.5 V
Answer: a
Explanation: The emf is given by,
V = Mdidt
= 31000 X 100 = 0.3 V
Hence, the emf induced in coil Y is given by 0.3 V.

Two-Port Network MCQs

9. A 50 Hz current has an amplitude of 25 A. The rate of change of current at t = 0.005 after i = 0 and is increasing is ____________
a) 2221.44 A/s
b) 0
c) -2221.44 A/s
d) -3141.6 A/s
Answer: b
Explanation: The current i (t) is given by,
i = 25 sin 314.16 t and didt = 250 X 314.16 cos⁡ωt
Now, at t = 0.005, I = 25 X 314.16 cos (314.16 X 0.005)
= 0.


10. Consider a series RL circuit in which current 12 A is flowing through R and current 16 A is flowing through L. The current supplied by the sinusoidal current source I is ____________
a) 28 A
b) 4 A
c) 20 A
d) Cannot be determined
Answer: c
Explanation: Current I (t) is given by,
I (t) = 162+122−−−−−−−−√
= 256+144−−−−−−−−√
= 400−−−√
= 20 A.


11. Consider a circuit having resistances 16 Ω and 30 Ωis excited by a voltage V. A variable resistance R is connected across the 16 Ω resistance. The power dissipated in 30 Ω resistance will be maximum when value of R is __________
a) 30 Ω
b) 16 Ω
c) 9 Ω
d) 0
Answer: c
Explanation: We know that,
When R = 0, circuit current = V30 A
And Power dissipated = V230 Watts.
This is the maximum possible value which occurs for R = 0 Ω.


12. Consider a cube having resistance R on each of its sides. For this non-planar graph, the number of independent loop equations are _______________
a) 8
b) 12
c) 7
d) 5
Answer: d
Explanation: We know that the number of equations is given by,
L = B – N + 1
Where, B = Number of Branches, N = Number of Nodes
Here, B = 12 and N = 8.
So, L = 12 – 8 + 1 = 5.


13. Given two voltages, 50 ∠0 V and 75 ∠- 60° V. The sum of these voltages is ___________
a) 109 ∠- 60° V
b) 109 ∠- 25° V
c) 109 ∠- 36.6° V
d) 100 ∠- 50.1° V
Answer: c
Explanation: The voltages can be written in the form,
50 + j 0.75∠-60°
= 37.5 – j 64.95
So, sum = (50 + 37.5) – j 64.95
= 87.5 – j 64.95 = 109∠-36.6°.


14. For the circuit given below, the value of z₂₁ parameter is ____________

a) z₂₁ = 0.0667 Ω
b) z₂₁ = 2.773 Ω
c) z₂₁ = 1.667 Ω
d) z₂₁ = 0.999 Ω
Answer: a
Explanation: z₁₁ = V1I1 = 2 + 1 || [2+1 || (2+1)]
z₁₁ = 2 + 1 || (2 + 34) = 2 + 1×1141+114=2+1115 = 2.733
I₀ = 11+3 I’₀ = 14 I’₀
And I’₀ = 1 + 114 I₁ = 415 I₁
Or, I₀ = 14×45I1=115I1
Or, V₂ = I₀ = 115I1
z₂₁ = V2I1=115 = z₁₂ = 0.0667
z₂₂ = V2I2 = 2+1 || (2+1||3) = z₁₁ = 2.733
∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.


15. For the circuit given below, the value of z₁₁ parameter is ____________

a) z₁₁ = 1.775 + j5.739 Ω
b) z₁₁ = 1.775 – j4.26 Ω
c) z₁₁ = -1.775 – j4.26 Ω
d) z₁₁ = 1.775 + j4.26 Ω
Answer: d
Explanation: z₁ = 12(j10)12+j10−j5=j12012+j5
z₂ = j6012+j5
z₃ = 5012+j5
z₁₂ = z₂₁ = z₂ = (−j60)(12−j5)144+25 = -1.775 – j4.26
z₁₁ = z₁ + z₁₂ = (j120)(12−j5)144+25 + z₁₂ = 1.775 + j4.26
z₂₂ = z₃ + z₂₁ = (50)(12−j5)144+25 + z₂₁ = 1.7758 – j5.739
∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.

Advanced Problems on Two Port Network – 1

1. For the circuit given below, the value of the z₁₂ parameter is ___________

a) z₁₂ = 1 Ω
b) z₁₂ = 4 Ω
c) z₁₂ = 1.667 Ω
d) z₁₂ = 2.33 Ω
Answer: a
Explanation: z₁₁ = V1I1 = 1 + 6 || (4+2) = 4Ω
I₀ = 12I1
V₂ = 2I₀ = I₁
z₂₁ = V2I1 = 1Ω
z₂₂ = V2I2 = 2 || (4+6) = 1.667Ω
So, I’₀ = 22+10I2=16I2
V₁ = 6I’₀ = I₂
z₁₂ = V1I2 = 1Ω
Hence, [z] = [4:1; 1:1.667] Ω.


2. For the network of figure, z₁₁ is equal to ___________

a) 53 Ω
b) 32 Ω
c) 2 Ω
d) 23 Ω
Answer: a
Explanation: From the figure, we can infer that,
Z₁₁ = 1 + 1X23
= 1 + 23
= 53 Ω.


3. For the circuit given below, the value of z₁₁ parameter is ____________

a) z₁₁ = 4 + j6 Ω
b) z₁₁ = j6 Ω
c) z₁₁ = -j6 Ω
d) z₁₁ = -j6 + 4 Ω
Answer: a
Explanation: z₁₂ = j6 = z₂₁
z₁₁ – z₁₂ = 4
Or, z₁₁ = z₁₂ + 4 = 4 + j6 Ω
And z₂₂ – z₁₂ = -j10
Or, z₂₂ = z₁₂ + -j10 = -j4 Ω
∴ [z] = [4+j6:j6; j6:-j4] Ω.


4. In a series RLC circuit excited by a voltage 3e^-t u (t), the resistance is equal to 1 Ω and capacitance = 2 F. For the circuit, the values of I (0⁺) and I (∞), are ____________
a) 0 and 1.5 A
b) 1.5 A and 3 A
c) 3 A and 0
d) 3 A and 1.5 A
Answer: c
Explanation: I(s) = 6s+1–3s+0.5
Or, I(t) = 6 e^-t – 3 e^-0.5t
Putting, t = 0, we get, I(0) = 3A
Putting t = ∞, we get, I (∞) = 0.


5. For the circuit given below, the value of z₁₂ parameter is ____________

a) Z₁₂ = 20 Ω
b) Z₁₂ = 25 Ω
c) Z₁₂ = 30 Ω
d) z₁₂ = 24 Ω
Answer: a
Explanation: z₁₁ = V1I1=(20+5)I1I1 = 25Ω
V₀ = 2025V₁ = 20 I₁
-V₀ – 4I₂ + V₂ = 0
Or, V₂ = V₀ + 4I₁ = 20I₁ + 4I₁ = 24 I₁
Or, z₂₁ = V2I1 = 24 Ω
V₂ = (10+20) I₂ = 30 I₂
Or, z₂₂ = V2I1 = 30 Ω
V₁ = 20I₂
Or, z₁₂ = V1I2 = 20 Ω
∴ [z] = [25:20; 24:30] Ω.


6. For the circuit given below, the value of the z₂₂ parameter is ___________

a) z₂₂ = 1 Ω
b) z₂₂ = 4 Ω
c) z₂₂ = 1.667 Ω
d) z₂₂ = 2.33 Ω
Answer: c
Explanation: z₁₁ = V1I1 = 1 + 6 || (4+2) = 4Ω
I₀ = 12I1
V₂ = 2I₀ = I₁
z₂₁ = V2I1 = 1Ω
z₂₂ = V2I2 = 2 || (4+6) = 1.667Ω
So, I’₀ = 22+10I2=16I2
V₁ = 6I’₀ = I₂
z₁₂ = V1I2 = 1Ω
Hence, [z] = [4:1; 1:1.667] Ω.


7. For the circuit given below, the value of z₂₂ parameter is ____________

a) z₂₂ = 0.0667 Ω
b) z₂₂ = 2.773 Ω
c) z₂₂ = 1.667 Ω
d) z₂₂ = 0.999 Ω
Answer: b
Explanation: z₁₁ = V1I1 = 2 + 1 || [2+1 || (2+1)]
z₁₁ = 2 + 1 || (2 + 34) = 2 + 1×1141+114=2+1115 = 2.733
I₀ = 11+3 I’₀ = 14 I’₀
And I’₀ = 1 + 114 I₁ = 415 I₁
Or, I₀ = 14×45I1=115I1
Or, V₂ = I₀ = 115I1
z₂₁ = V2I1=115 = z₁₂ = 0.0667
z₂₂ = V2I2 = 2+1 || (2+1||3) = z₁₁ = 2.733
∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.


8. For the circuit given below, the value of z₂₂ parameter is ____________

a) z₂₂ = 4 + j6 Ω
b) z₂₂ = j6 Ω
c) z₂₂ = -j4 Ω
d) z₂₂ = -j6 + 4 Ω
Answer: c
Explanation: z₁₂ = j6 = z₂₁
z₁₁ – z₁₂ = 4
Or, z₁₁ = z₁₂ + 4 = 4 + j6 Ω
And z₂₂ – z₁₂ = -j10
Or, z₂₂ = z₁₂ + -j10 = -j4 Ω
∴ [z] = [4+j6:j6; j6:-j4] Ω.


9. For the circuit given below, the value of z₂₂ parameter is ____________

a) z₂₂ = 1.775 + j5.739 Ω
b) z₂₂ = 1.775 – j4.26 Ω
c) z₂₂ = -1.775 – j5.739 Ω
d) z₂₂ = 1.775 + j4.26 Ω
Answer: c
Explanation: z₁ = 12(j10)12+j10−j5=j12012+j5
z₂ = j6012+j5
z₃ = 5012+j5
z₁₂ = z₂₁ = z₂ = (−j60)(12−j5)144+25 = -1.775 – j4.26
z₁₁ = z₁ + z₁₂ = (j120)(12−j5)144+25 + z₁₂ = 1.775 + j4.26
z₂₂ = z₃ + z₂₁ = (50)(12−j5)144+25 + z₂₁ = 1.7758 – j5.739
∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.


10. For the circuit given below, the value of z₂₂ parameter is ____________

a) z₂₂ = 20 Ω
b) z₂₂ = 25 Ω
c) z₂₂ = 30 Ω
d) z₂₂ = 24 Ω
Answer: c
Explanation: z₁₁ = V1I1=(20+5)I1I1 = 25Ω
V₀ = 2025V₁ = 20 I₁
-V₀ – 4I₂ + V₂ = 0
Or, V₂ = V₀ + 4I₁ = 20I₁ + 4I₁ = 24 I₁
Or, z₂₁ = V2I1 = 24 Ω
V₂ = (10+20) I₂ = 30 I₂
Or, z₂₂ = V2I1 = 30 Ω
V₁ = 20I₂
Or, z₁₂ = V1I2 = 20 Ω
∴ [z] = [25:20; 24:30] Ω.


11. A capacitor of 220 V, 50 Hz is needed for AC supply. The peak voltage rating of the capacitor is ____________
a) 220 V
b) 460 V
c) 440 V
d) 230 V
Answer: c
Explanation: We know that,
Peak voltage rating = 2 (rms voltage rating)
Given that the RMS voltage rating = 220 V
So, the Peak Voltage Rating = 2 X 220 V
= 440 V.


12. In the circuit given below, the value of the hybrid parameter h₂₁ is _________

a) 10 Ω
b) 0.5 Ω
c) 5 Ω
d) 2.5 Ω
Answer: b
Explanation: Hybrid parameter h₂₁ is given by, h₂₁ = I2I1, when V₂ = 0.
Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,
-5 I₂ – (I₂ – I₁)5 = 0
Or, -I₂ = I₂ – I₁
Or, -2I₂ = -I₁
∴ I2I1=12
Hence h₂₁ = 0.5 Ω.


13. 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
Explanation: 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 = V1h11–h12V2h11 ………. (5)
And I₂ = h21V1h11+(−h21h12h11+h22)V2 ………. (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].


14. In two-port networks the parameter h₂₂ is called _________
a) Short circuit input impedance
b) Short circuit current gain
c) Open circuit reverse voltage gain
d) Open circuit output admittance
Answer: d
Explanation: We know that, h₂₂ = I2V2, when I₁ = 0.
Since the current in the first loop is 0 when the ratio of the current and voltage in second loop is measured, therefore the parameter h₁₂ is called as Open circuit output admittance.


15. 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
Explanation: So, network is non reciprocal because Y₁₂ ≠ Y₂₁ and Y₁₂ are also negative which means either energy storing or providing device is available. So the network is active.
Therefore the network is Non- reciprocal and active.