Category: Network Theory Objective Questions

Network Theory Questions & Answers for Exams on “Advanced Problems on Reciprocity Theorem”.

1. In Reciprocity Theorem, which of the following ratios is considered?
A. Voltage to current
B. Current to current
C. Voltage to voltage
D. No ratio is considered

Answer: A
Clarification: The Reciprocity Theorem states that if an Emf E in one branch produces a current I in a second branch, then if the same emf E is moved from the first to the second branch, it will produce the same current in the first branch, when the Emf E in the first branch is replaced with a short circuit. Therefore the ratio of Voltage to Current is considered in case of Reciprocity Theorem.

2. The Reciprocity Theorem is valid for ___________
A. Non-Linear Time Invariant circuits
B. Linear Time Invariant circuits
C. Non-Linear Time Variant circuits
D. Linear Time Variant circuits

Answer: B
Clarification: A reciprocal network comprises of linear time-invariant bilateral elements. It is applicable to resistors, capacitors, inductors (including coupled inductors) and transformers. However, both dependent and independent sources ate not permissible.

3. In the circuit given below, the current in the 4-ohm resistor is __________

A. 3.5 A
B. 2.5 A
C. 1.5 A
D. 0.5 A

Answer: C
Clarification: R_th = [(2+4) || 6] + 12 = 15 Ω
I_S = (frac{45}{15}) = 3 A
Now, by current division rule, we get, I = (frac{3 × 6}{12} = frac{18}{12}) = 1.5 A.

4. The Reciprocity Theorem is applicable for __________
A. Single-source networks
B. Multi-source networks
C. Both Single and Multi-source networks
D. Neither Single nor Multi-source networks

Answer: A
Clarification: According to Reciprocity Theorem, the voltage source and the resulting current source may be interchanged without a change in current. Therefore the theorem is applicable only to single-source networks. It therefore cannot be employed in multi-source networks.

5. In the circuit given below, the current through the 12 Ω resistance is _________

A. 1.5 A
B. 2.5 A
C. 3.5 A
D. 4.5 A

Answer: A
Clarification: Equivalent resistance, R_EQ = [(12 || 6) + 2 + 4] = 10 Ω
I_S = (frac{45}{10}) = 4.5 A
Now, by using Current division rule, we get, I = (frac{4.5 × 6}{12+6} = frac{27}{18}) = 1.5 A.

6. A circuit is given in the figure below. We can infer that ________

A. The circuit follows Reciprocity Theorem
B. The circuit follows Millman’s Theorem
C. The circuit follows Superposition Theorem
D. The circuit follows Tellegen Theorem

7. In the circuit given below, the current in the resistance 20 Ω(far enD. is _________

A. 8.43 A
B. 5.67 A
C. 1.43 A
D. 2.47 A

Answer: C
Clarification: Equivalent Resistance R_EQ = 20 + [30 || (20 + (20||20))]
= 20 + [30 || (20 + (frac{20×20}{20+20}))]
= 20 + [30 || (20+10)]
= 20 + [30 || 30]
= 20 + (frac{30 × 30}{30+30})
= 20 + 15 = 35 Ω
The current drawn by the circuit = (frac{200}{35}) = 5.71 A
Now, by using current division rule, we get, I_2Ω = 1.43 A.

8. In the circuit given below, the value of I is __________

A. 2.47 A
B. 5.67 A
C. 8.43 A
D. 1.43 A

Answer: D
Clarification: Equivalent Resistance, R_EQ = [[((30 || 20) + 20) || 20] + 20]
= (Big[Big[left(left(frac{30 × 20}{30+20}right) + 20right) || 20Big] + 20Big])
= [[(12 + 20) || 20] + 20]
= [[32 || 20] + 20]
= (Big[left(frac{32 × 20}{32+20}right) + 20Big])
= [12.31 + 20] = 32.31 Ω
The current drawn by the circuit = (frac{200}{32.31}) = 6.19 A
Now, by using current division rule, we get, I_2Ω = 1.43 A.

9. A circuit is given in the figure below. We can infer that ________

A. The circuit follows Reciprocity Theorem
B. The circuit follows Millman’s Theorem
C. The circuit follows Superposition Theorem
D. The circuit follows Tellegen Theorem

10. In the circuit given below, the current in the 30 Ω resistor is _________

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Answer: B
Clarification: Equivalent Resistance, R_EQ = 20 + [60 || 30]
= 20 + (frac{60 × 30}{60+30})
= 20 + 20 = 40 Ω
Total current from the source, I = (frac{120}{40}) = 3A
Now, by using current division rule, I_3Ω = (frac{3 × 60}{30+60}) = 2 A.

11. In the circuit given below, the current in the 20 Ω resistor is _________

A. 5 A
B. 1 A
C. 1.5 A
D. 2 A

Answer: D
Clarification: Equivalent resistance, R_EQ = [[20 || 60] + 30]
= (Big[frac{20 × 60}{20+60} + 30Big])
= [15 + 30] = 45 Ω
Total current = (frac{120}{45}) = 2.67 A
Current through the 20Ω resistor is, I_20Ω = (frac{2.67 × 60}{60+20}) = 2 A.

12. A circuit is given in the figure below. We can infer that ________

A. The circuit follows Reciprocity Theorem
B. The circuit follows Millman’s Theorem
C. The circuit follows Superposition Theorem
D. The circuit follows Tellegen Theorem

13. In the circuit given below, the value of load R_L, for which maximum power is transferred through it is ___________

A. 2 Ω
B. 3 Ω
C. 1 Ω
D. 6 Ω

Answer: B
Clarification: I + 0.9 = 10 I
Or, I = 0.1 A
V_OC = 3 × 10 I = 30 I
Or, V_OC = 3 V
Now, I_SC = 10 I = 1 A
R_th = 3/1 = 3 Ω.

14. In the circuit given below, the maximum power absorbed by the load resistance R_L is ___________

A. 2200 W
B. 1250 W
C. 1000 W
D. 621 W

Answer: D
Clarification: R_L = (sqrt{R_{TH}^2+X_{TH}^2})
= (sqrt{3^2+4^2}) = 5
Now, 110 = (6 + j8 + 5) I₁ + 5I₂
And 90 = (6 + j8 + 5) I₂ = 5I₁
∴ I₁ = 5.5 – 2.75j and I₂ = 4.5 – 2.2j
Total current in R_L = I₁ + I₂ = (10 – 4.95j) A = 11.15 A
∴ Power absorbed by R_L = I²R
= 11.15² × 5 = 621 W.

15. In the circuit given below, the maximum power delivered to the load is ___________

A. 3 W
B. 5.2 W
C. 3.2 W
D. 4.2 W

Answer: D
Clarification: Equivalent resistance of the circuit is = [{(3 + 2) || 5} + 10]
= (2.5 + 10) = 12.5 Ω
Total current drawn by the circuit is I_T = (frac{50}{12.5}) = 4 A
Current in 3 Ω resistor is I₃ = I_T × (frac{5}{5+5} = frac{4 × 5}{10}) = 2 A
V_TH = V₃ = 3 × 2 = 6V
R_TH = R_AB = [(2 + 5) || 3] = 2.1 Ω
For maximum power transfer R_L = R_TH = 2.1 Ω
∴ Current drawn by R_L is I_L = (frac{6}{2.1+2.1} = frac{6}{4.2}) = 1.42 A
∴ Power delivered to the load = (I_L^2 R_L)
= (1.42)²(2.1) = 4.2 W.

Network Theory Multiple Choice Questions on “Resonant Frequency for a Tank Circuit”.

1. For the tank circuit shown below, find the resonant frequency.

A. 157.35
B. 158.35
C. 159.35
D. 160.35

Answer: B
Clarification: The parallel resonant circuit is generally called a tank circuit because of the fact that the circuit stores energy in the magnetic field of the coil and in the electric field of the capacitor. The resonant frequency f_r = (1/2π) √((1/LC.-(R²/L²)).

2. The expression of ωr in a parallel resonant circuit is?
A. 1/(2√LC.
B. 1/√LC
C. 1/(π√LC.
D. 1/(2π√LC.

Answer: B
Clarification: The stored energy is transferred back and forth between the capacitor and coil and vice-versa. The expression of ω_r in parallel resonant circuit is ω_r = 1/√LC.

3. The expression of bandwidth for the parallel resonant circuit is?
A. 1/RC
B. RC
C. 1/R
D. 1/C

Answer: A
Clarification: The expression of bandwidth for parallel resonant circuit is BW = 1/RC. the circuit is said to be in a resonant condition when the susceptance part of admittance is zero.

4. The quality factor in case of parallel resonant circuit is?
A. C
B. ω_rRC
C. ω_rC
D. 1/ω_rRC

Answer: D
Clarification: The quality factor in case of parallel resonant circuit is Q = 1/ω_rRC. The impedance of parallel resonant circuit is maximum at the resonant frequency and decreases at lower and higher frequencies.

5. The quality factor is the product of 2π and the ratio of ______ to _________
A. maximum energy stored, energy dissipated per cycle
B. energy dissipated per cycle, maximum energy stored
C. maximum energy stored per cycle, energy dissipated
D. energy dissipated, maximum energy stored per cycle

Answer: A
Clarification: At low frequencies X_L is very small and X_C is very large so the total impedance is essentially inductive. The quality factor is the product of 2π and the ratio of maximum energy stored to energy dissipated per cycle.

6. The maximum energy stored in a capacitor is?
A. CV²
B. CV²/2
C. CV²/4
D. CV²/8

Answer: B
Clarification: The maximum energy stored in a capacitor is CV²/2. Maximum energy = CV²/2. As frequency increases the impedance also increases and the inductive reactance dominates until the resonant frequency is reached.

7. The expression of quality factor is?
A. I_L/I
B. I/I_L
C. IL
D. I

Answer: A
Clarification: The expression of quality factor is I_L/I. Quality factor = I_L/I. At the the point X_L = X_C, the impedance is at its maximum.

8. The quality factor is defined as?
A. I
B. I_C
C. I/I_C
D. I_C/I

Answer: D
Clarification: The quality factor is defined as I_C/I. Quality factor = I_C/I. As the frequency goes above resonance capacitive reactance dominates and impedance decreases.

9. In the circuit shown in the figure, an inductance of 0.1H having a Q of 5 is in parallel with a capacitor. Determine the value coil resistance (Ω) of at a resonant frequency of 500 rad/sec.

A. 10
B. 20
C. 30
D. 40

Answer: A
Clarification: Quality factor Q = ω_rL/R. L = 0.1H, Q = 5, ω_r = 500 rad/sec. On solving, R = 10Ω. While plotting the voltage and current variation with frequency, at resonant frequency, the current is maximum.

10. Find the value of capacitance (µF) in the circuit shown below.

A. 10
B. 20
C. 30
D. 40

Answer: D
Clarification: ω²_r = 1/LC. L = 0.1H, ωr = 500 rad/sec. On solving, C = 40 µF. In order to tune a parallel circuit to a lower frequency the capacitance must be increased.

Network Theory Multiple Choice Questions on “Three-Phase Unbalanced Circuits”.

1. If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.
A. 1
B. 2
C. 3
D. zero
Answer: D
Clarification: If the system is a three-wire system, the currents flowing towards the load in the three lines must add to zero at any given instant.

2. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current I_R.
A. (17.32-j10) A
B. (-17.32-j10) A
C. (17.32+j10) A
D. (-17.32+j10) A
Answer: A
Clarification: Taking V_RY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have V_RY = 400∠0⁰V Z₁ = 20∠30⁰Ω = (17.32+j10)Ω I_R = (400∠0^o)/(20∠30^o) = (17.32-j10) A.

3. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current I_Y.
A. (10-j0) A
B. (10+j0) A
C. (-10+j0) A
D. (-10-j0) A
Answer: C
Clarification: The voltage V_YB is V_YB = 400∠-120⁰V. The impedance Z₂ is Z₂ = 40∠60⁰Ω => I_Y = (400∠-120^o)/(40∠60^o)=(-10+j0)A.

4. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current I_B.
A. (34.64-j20) A
B. (34.64+j20) A
C. (-34.64+j20) A
D. (-34.64-j20) A
Answer: D
Clarification: The voltage V_BR is V_BR = 400∠-240⁰V. The impedance Z₃ is Z₃ = 10∠-90⁰Ω => I_B = (400∠240^o)/(10∠-90^o)=(-34.64-j20)A.

5. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I₁.
A. (-51.96-j10) A
B. (-51.96+j10) A
C. (51.96+j10) A
D. (51.96+j10) A
Answer: C
Clarification: The line current I₁ is the difference of I_R and I_B. So the line current I₁ is I₁ = I_R – I_B = (51.96+j10) A.

6. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I₂.
A. (-27.32+j10) A
B. (27.32+j10) A
C. (-27.32-j10) A
D. (27.32-j10) A
Answer: A
Clarification: The line current I₂ is the difference of I_Y and I_R. So the line current I₂ is I₂ = I_Y – I_R = (-27.32+j10) A.

7. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I₃.
A. (24.646+j20) A
B. (-24.646+j20) A
C. (-24.646-j20) A
D. (24.646-j20) A
Answer: C
Clarification: The line current I₃ is the difference of I_B and I_Y. So the line current I₃ is I₃ = I_B – I_Y = (-24.646-j20) A.

8. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the R phase.
A. 6628
B. 6728
C. 6828
D. 6928
Answer: D
Clarification: The term power is defined as the product of square of current and the impedance. So the power in the R phase = 20² x 17.32 = 6928W.

9. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the Y phase.
A. 1000
B. 2000
C. 3000
D. 4000
Answer: B
Clarification: The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 10² x 20 = 2000W.

10. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the B phase.
A. 0
B. 1
C. 3
D. 2
Answer: A
Clarification: The term power is defined as the product of square of current and the impedance. So the power in the B phase = 40² x 0 = 0W.

Network Theory Multiple Choice Questions on “Problems on Initial and Final Value Theorem”.

1. What is the steady state value of F (t), if it is known that F(s) = (frac{2}{s(S+1)(s+2)(s+3)})?
A. (frac{1}{2})
B. (frac{1}{3})
C. (frac{1}{4})
D. Cannot be determined

Answer: B
Clarification: From the equation of F(s), we can infer that, a simple pole is at origin and all other poles are having negative real part.
∴ F(∞) = lim_s→0 s F(s)
= lim_s→0 (frac{2s}{s(S+1)(s+2)(s+3)})
= (frac{2}{(s+1)(s+2)(s+3)})
= (frac{2}{6} = frac{1}{3}).

2. What is the steady state value of F (t), if it is known that F(s) = (frac{1}{(s-1)(s+2)})?
A. 1
B. –(frac{1}{2})
C. (frac{1}{2})
D. Cannot be determined

Answer: D
Clarification: The steady state value of this Laplace transform is cannot be determined since; F(s) has a pole s = 1. Hence the answer is that it cannot be determined.

3. What is the steady state value of F (t), if it is known that F(s) = (frac{1}{(s+2)^2 (s+4)})?
A. (frac{1}{16})
B. Cannot be determined
C. 0
D. (frac{1}{8})

Answer: C
Clarification: The steady state value of F(s) exists since all poles of the given Laplace transform have negative real part.
∴F(∞) = lim_s→0 s F(s)
= lim_s→0 (frac{s}{(s+2)^2 (s+4)})
= 0.

4. What is the steady state value of F (t), if it is known that F(s) = (frac{10}{(s+1)(s^2+1)})?
A. -5
B. 5
C. 10
D. Cannot be determined

Answer: D
Clarification: The steady state value of this Laplace transform is cannot be determined since; F(s) is having two poles on the imaginary axis (j and –j). Hence the answer is that it cannot be determined.

5. What is the steady state value of F (t), if it is known that F(s) = b/(s(s+1)(s+A.), where a>0?
A. (frac{b}{a})
B. (frac{a}{b})
C. 1
D. Cannot be determined

Answer: A
Clarification: F (∞) = lim_s→0 s F(s)
= lim_s→0 (frac{sb}{s(s+1)(s+A.})
= lim_s→0 (frac{b}{(s+1)(s+A.})
= (frac{b}{a}).

6. The inverse Laplace transform of F(s) = (frac{2}{s^2+3s+2}) is ______________
A. -2e^-2t + 2e^-t
B. 2e^-2t + 2e^-t
C. -2e^-2t – 2e^-t
D. 2e^-t + e^-2t

Answer: A
Clarification: s² + 3s + 2 = (s+2) (s+1)
Now, F(s) = (frac{A}{(s+2)} + frac{B}{(s+1)})
Hence, A = (s+2) F(s) |_s=-2
= (frac{2}{s+1})|_s=-2 = -2
And, B = (s+1) F(s) |_s=-1
= (frac{2}{s+2})|_s=-1 = 2
∴ F(s) = (frac{-2}{(s+2)} + frac{2}{(s+1)})
∴ F (t) = L^-1{F(s)}
= -2e^-2t + 2e^-t for t≥0.

7. The inverse Laplace transform of F(s) = (frac{2}{s+c}e^{-bs}) is _____________
A. 2e^{-k (t+B.} u (t+B.
B. 2e^{-k (t-B.} u (t-B.
C. 2e^{k (t-B.} u (t-B.
D. 2e^{k (t-B.} u (t+B.

Answer: B
Clarification: Let G(s) = (frac{2}{s+c})
Or, G (t) = L^-1{G(s)} = 2e^-ct
∴ F (t) = L^-1{G(s) e^-bs}
= 2e^{-k (t-B.} u (t-B..

8. The inverse Laplace transform of F(s) = (frac{3s+5}{s^2+7}) is _____________
A. 3 sin ((sqrt{7})t) – (frac{5}{sqrt{7}}) cos ((sqrt{7}) t)
B. 3 sin ((sqrt{7})t) + (frac{5}{sqrt{7}}) cos ((sqrt{7}) t)
C. 3 cos ((sqrt{7})t) + (frac{5}{sqrt{7}}) sin ((sqrt{7}) t)
D. 3 cos ((sqrt{7})t) – (frac{5}{sqrt{7}}) sin ((sqrt{7}) t)

Answer: C
Clarification: F (t) = L^-1{F(s)}
= L^-1(Big{frac{3s}{s^2+7} + frac{5}{s^2+7}Big})
= L^-1(Big{frac{3s}{s^2 + sqrt{7}^2} + frac{5}{sqrt{7}} frac{sqrt{7}}{s^2 +sqrt{7}^2}Big})
The inverse Laplace transform is = 3 cos ((sqrt{7})t) + (frac{5}{sqrt{7}}) sin ((sqrt{7}) t).

9. The inverse Laplace transform of F(s) = (frac{5}{s^2-9}) is ______________
A. (frac{1}{6}e^{3t} + frac{5}{6e^{-3t}})
B. (frac{1}{6}e^{3t} – frac{5}{6e^{-3t}})
C. (frac{5}{6}e^{3t} + frac{5}{6e^{-3t}})
D. (frac{5}{6}e^{3t} – frac{5}{6e^{-3t}})

Answer: D
Clarification: F (t) = L^-1{F(s)}
= L^-1{(frac{5}{s^2-9})}
= L^-1({frac{A}{s-3} + frac{B}{s+3}})
Hence, A = (s-3) (frac{5}{(s-3)(s+3)}|_{s=3} = frac{5}{6})
And, B = (s+3) (frac{5}{(s-3)(s+3)}|_{s=-3} = -frac{5}{6})
The inverse Laplace transform is (frac{5}{6}e^{3t} – frac{5}{6e^{-3t}}).

10. The inverse Laplace transform of F(s) = (frac{e^{-3s}}{s(s^2+3s+2)}) is ______________
A. {0.5 + 0.5e^-2(t+3) – e^-(t+3)} u (t+3)
B. {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3)
C. {0.5 – 0.5e^-2(t-3) – e^-(t-3)} u (t-3)
D. 0.5 + 0.5e^-2t – e^-t)

Answer: B
Clarification: Let G(s) = (frac{1}{s(s^2+3s+2)})
Or, F(s) = G(s) e^-3s
G (t) = L^-1{G(s)}
= L^-1({frac{A}{s} + frac{B}{s+2} + frac{C}{s+1}})
Solving we get, A = 0.5, B = 0.5, C = -1
So, G (t) = 0.5 + 0.5e^-2t-e^-t
The inverse Laplace transform is F (t) = {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3).

11. Given a sinusoidal voltage that has a peak to peak value of 100 V. The RMS value of the sinusoidal voltage is ___________
A. 50 V
B. 70.7 V
C. 35.35 V
D. 141.41 V

Answer: C
Clarification: Given that,
Peak value = 50 V
We know that, RMS value = (frac{Peak ,Value}{sqrt{2}})
= (frac{50}{sqrt{2}}) = 35.35 V.

12. The Laplace transform of F(t) = sin(2t)cos(2t) is ______________
A. (frac{4}{2(s^2+16)})
B. (frac{1}{s+4} – frac{2}{s+2} + frac{1}{s})
C. (frac{2}{s^3})
D. (frac{1}{2s} + frac{s}{2(s^2+36)})

Answer: A
Clarification: Using the Trigonometric Identity,
We get, sin (2t) cos (2t) = (frac{1}{2}) (sin (4t))
∴L {sin (2t) cos (2t)} = (frac{4}{2(s^2+16)}).

13. Convolution of step signal 100 times that is 100 convolution operations. The Laplace transform is ______________
A. (frac{1}{s^{100}})
B. (frac{1}{s^{50}})
C. 1
D. s¹⁰⁰

Answer: A
Clarification: n times = u (t) * u (t) * …… * u (t)
Laplace transform of the above function = (frac{1}{s^n}), where n is number of convolutions.
∴ Laplace transform for 100 convolutions = (frac{1}{s^{100}}).

14. For the circuit given below, the Time-constant is __________

A. 1.5
B. 1.25
C. 2.5
D. 2.25

Answer: B
Clarification: We know that,
Time constant is given by R_eq.C
The equivalent resistance is given by,
R_eq = R || R
= (frac{R*R}{R+R})
= (frac{R}{2})
= 5 Ω
So, Time-constant = (frac{5X0.5}{2})
= 1.25.

15. In a dual slop integrating type digital voltmeter, the first integrating is carried out for 50 periods of the supply frequency of 50 Hz. If the reference voltage used is 10 V, the total conversion time for an input of 40 V is?
A. 3 s
B. 2 s
C. 4 s
D. 1 s

Answer: C
Clarification: In a dual slope integrating digital voltmeter,
((frac{t_1}{t_2})) V_in = V_ref
Where, t₁ = first integration time = 50 × (frac{1}{50}) = 1
But V_in = 40 V and V_ref = 10 V
∴ t₂ = (frac{V_{in} t_1}{V_{ref}}) = 4 s.

Network Theory Multiple Choice Questions on “Terminated Two-Port Network”.

1. Calculate the Z –parameter Z₁₁ in the circuit shown below.
terminated-two-port-q1″>terminated-two-port-q1″ alt=”network-theory-questions-answers-terminated-two-port-q1″ width=”300″ height=”102″ class=”alignnone size-full wp-image-170509″>
A. 1.5
B. 2.5
C. 3.5
D. 4.5

Answer: B
Clarification: The Z–parameter Z₁₁ is V₁/I₁, port 2 is open circuited. V₁ = (1+1.5)I₁ => V₁/I₁ = 2.5 and on substituting, we get Z₁₁ = 2.5Ω.

2. Determine the Z-parameter Z₁₂ in the circuit shown below.
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A. 1
B. 2
C. 3
D. 4

Answer: A
Clarification: The Z-parameter Z₁₂ is V₂/I₁ |I₂=0. On open circuiting port 2 we obtain the equation, V₁ = (1.5) I₂ => V₁/I₁ = 1.5. On substituting we get Z₁₂ = 1.5Ω.

3. Determine the Z-parameter Z₂₁ in the circuit shown below.

A. 4
B. 3
C. 2
D. 1

Answer: D
Clarification: The Z-parameter Z₂₁ is V₂/I₁ |I₂=0. On open circuiting port 2, we get V₂ = (1.5)I₁ => V₂/I₁ = 1.5. On substituting we get Z₂₁ = 1.5Ω.

4. Determine the Z-parameter Z₂₂ in the circuit shown below.
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A. 1
B. 3
C. 2
D. 4

Answer: C
Clarification: The Z-parameter Z₂₁ is V₂/I₂ |I₁=0. This parameter is obtained by open circuiting port 1. So we get V₂ = ((2+2)||4)I₂ => V₂ = 2(I₂) => V₂/I₂ = 2. On substituting Z₂₁ = 2Ω.

5. Find the value of V₁/I₁ in the circuit shown below.
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A. 1.25
B. 2.25
C. 3.25
D. 4.25

Answer: B
Clarification: We have the relation V₁/I₁=Z₁₁– Z₁₂Z₂₁/(Z_L+Z₂₁) and Z_L is the load impedance and is equal to 2Ω. On solving V₁/I₁=2.5-1/(2+2)=2.25Ω.

6. Determine the input impedance of the network shown below.

A. 4.25
B. 3.25
C. 2.25
D. 1.25

Answer: B
Clarification: From the figure by inspection we can say that the source resistance is 1Ω. So Z_in = (V₁/I₁) + Source resistance. We had V₁/I₁ = 2.25. On substituting Z_in=1+2.25=3.25Ω.

7. Determine the value of source admittance in the circuit shown below.

A. 1
B. 2
C. 3
D. 4

Answer: A
Clarification: From the figure, the value of the admittance parallel to the current source is 1 mho and this is the value of source admittance. So Y_s = 1 mho.

8. Find the value of I₂/V₂ in the circuit shown below.

A. 7/6
B. 6/7
C. 7/12
D. 12/7

Answer: C
Clarification: The relation between I₂/V₂ and Y-parameters is
I₂/V₂=(5/8×1+5/8×1/2-1/16)/(1+1/2)=7/12 mho.

9. The value of the Y-parameter Y₂₂ in the circuit shown below.
A. 12/7
B. 6/7
C. 7/6
D. 7/12

Answer: D
Clarification: The relation between Y₂₂ and I₂/V₂ is Y₂₂ = I₂/V₂. We have the relation I₂/V₂ = (Y₂₂Y_s+Y₂₂Y₁₁-Y₂₁Y₁₂)/(Y_s+Y₁₁). On substituting their values in the equation we get Y₂₂ = 7/12 mho.

10. The value of the Z-parameter Z₂₂ in the circuit shown below.
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A. 6/7
B. 7/12
C. 12/7
D. 7/6

Answer: C
Clarification: The Z-parameter Z₂₂ is inverse of the Y-parameter Y₂₂ i.e., Z₂₂ = 1/Y₂₂. We got Y₂₂ = 7/12. So on substituting we get Z₂₂ = 12/7 mho.

Network Theory Multiple Choice Questions on “Supermesh Analysis”.

1. Consider the circuit shown below. Find the current I₁ (A..

A. 1
B. 1.33
C. 1.66
D. 2

Answer: B
Clarification: Applying Super mesh analysis, the equations will be I₂-I₁=2 -10+2I₁+I₂+4=0. On solving the above equations, I₁=1.33A.

2. Consider the circuit shown below. Find the current I₂ (A..

A. 1.33
B. 2.33
C. 3.33
D. 4.33

Answer: C
Clarification: Applying Super mesh analysis, the equations will be I₂-I₁=2
-10+2I₁+I₂+4=0. On solving the above equations, I₂=3.33A.

3.Consider the circuit shown below. Find the current I₁ (A..

A. -1
B. -2
C. -3
D. -4

Answer: C
Clarification: Applying Super mesh analysis, the equations will be I₁+I₁+10+I₂+I₂=0. I₁+I₂=-5. I₂-I₁=1. On solving, I₁=-3A.

4. Consider the circuit shown below. Find the current I₂ (A..

A. -2
B. -1
C. 2
D. 1

Answer: A
Clarification: Applying Super mesh analysis, the equations will be I₁+I₁+10+I₂+I₂=0. I₁+I₂=-5. I₂-I₁=1. On solving, I₂=-2A.

5. Find the power (W) supplied by the voltage source in the following figure.

A. 0
B. 1
C. 2
D. 3

Answer: A
Clarification: I₃-I₂=2. As I₂=-2A, I₃=0A. Th term power is the product of voltage and current. So, power supplied by source = 10×0=0W.

6. Find the current i₁ in the circuit shown below.

A. 8
B. 9
C. 10
D. 11

Answer: C
Clarification: The current in the first loop is equal to 10A. So the current i₁ in the circuit is i₁ = 10A.

7. Find the current i₂ in the circuit shown below.

A. 6.27
B. 7.27
C. 8.27
D. 9.27

Answer: B
Clarification: For 2nd loop, 10 + 2(i₂-i₃) + 3(i₂-i₁) = 0. For 3rd loop, i₃ + 2(i₃-i₂)=10. As i₁=10A, On solving above equations, we get i₂=7.27A.

8. Find the current i₃ in the circuit shown below.

A. 8.18
B. 9.18
C. 10.18
D. 8.8

Answer: A
Clarification: For 2nd loop, 10 + 2(i₂-i₃) + 3(i₂-i₁) = 0. For 3rd loop, i₃ + 2(i₃-i₂)=10. As i₁=10A, On solving above equations, we get i₃=8.18A.

9. Find the current I₁ in the circuit shown below.

A. 8
B. -8
C. 9
D. -9

Answer: B
Clarification: Applying Super Mesh analysis, (10+5)I₁ – 10(I₂) – 5(I₃) = 50. 2(I₂) + I₃ + 5(I₃-I₁) + 10(I₂-I₁) = 0. I₂ – I₃ = 2. On solving above equations, we get I₁=-8A.

10. Find the current I₂ in the circuit shown below.

A. 5.3
B. -5.3
C. 7.3
D. -7.3

Answer: D
Clarification: Applying Super Mesh analysis, (10+5)I₁-10(I₂)-5(I₃) = 50. 2(I₂) + I₃ + 5(I₃-I₁) + 10(I₂-I₁) = 0. I₂ – I₃ = 2. On solving above equations, we get I₂=-7.3A.