Category: Network Theory Objective Questions

Network Theory Questions & Answers for Exams on “Norton’s Theorem Involving Dependent and Independent Sources”.

1. The circuit shown in figure has a load equivalent to _________

A. (frac{4}{3}) Ω
B. (frac{8}{3}) Ω
C. 4 Ω
D. 2 Ω

Answer: B
Clarification: Applying KCL in the given circuit, we get, (frac{V}{4} + frac{V-2I}{2}) = I
Or, (frac{3V-4I}{4}) = I
Or, 3V = 8I
∴ (frac{V}{I} = frac{8}{3}) Ω.

2. In the following circuit, the value of Norton’s resistance between terminals a and b are ___________

A. R_N = 1800 Ω
B. R_N = 270 Ω
C. R_N = 90 Ω
D. R_N = 90 Ω

Answer: D
Clarification: By writing loop equations for the circuit, we get,
V_S = V_X, I_S = I_X
V_S = 600(I₁ – I₂) + 300(I₁ – I₂) + 900 I₁
= (600+300+900) I₁ – 600I₂ – 300I₃
= 1800I₁ – 600I₂ – 300I₃
I₁ = I_S, I₂ = 0.3 V_S
I₃ = 3I_S + 0.2V_S
V_S = 1800IS – 600(0.01V_S) – 300(3I_S + 0.01V_S)
= 1800I_S – 6V_S – 900I_S – 3V_S
10V_S = 900I_S
For Voltage, V_S = R_N I_S + V_OC
Here V_OC = 0
So, Resistance R_N = 90Ω.

3. For the circuit shown in figure below, the value of Norton’s resistance is _________

A. 100 Ω
B. 136.4 Ω
C. 200 Ω
D. 272.8 Ω

Answer: A
Clarification: I_X = 1 A, V_X = V_test
V_test = 100(1-2I_X) + 300(1-2I_X – 0.01V_S) + 800
Or, V_test = 1200 – 800I_X – 3V_test
Or, 4V_test = 1200 – 800 = 400
Or, V_test = 100V
∴ R_N = (frac{V_{test}}{1}) = 100 Ω.

4. For the circuit shown in the figure below, the Norton Resistance looking into X-Y is __________

A. 2 Ω
B. (frac{2}{3})
C. (frac{5}{3})
D. 2 Ω

Answer: D
Clarification: (R_N = frac{V_{OC}}{I_{SC}})
V_N = V_OC
Applying KCL at node A, (frac{2I-V_N}{1} + 2 = I + frac{V_N}{2})
Or, I = (frac{V_N}{1})
Putting, 2V_N – V_N + 2 = V_N + (frac{V_N}{2})
Or, V_N = 4 V.
∴ R_N = 4/2 = 2Ω.

5. In the figure given below, the value of Resistance R by Norton’s Theorem is ___________

A. 40
B. 20
C. 50
D. 80

Answer: B
Clarification: (frac{V_P-100}{10} + frac{V_P}{10}) + 2 = 0
Or, 2V_P – 100 + 20 = 0
∴ V_P = 80/2 = 40V
∴ R = 20Ω (By Norton’s Theorem).

6. In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________

A. 5 Ω
B. 7.5 Ω
C. 5 Ω
D. 7.5 Ω

7. The Norton equivalent impedance Z between the nodes P and Q in the following circuit is __________

A. 1
B. 1 + s + (frac{1}{s})
C. 2 + s + (frac{1}{s})
D. 3 + s + (frac{1}{s})

Answer: A
Clarification: To calculate the Norton resistance, all the current sources get open-circuited and voltage sources get short-circuited.
∴ R_N = ((frac{1}{s}) + 1) || (1+s)
= (frac{left(frac{1}{s} + 1right)×(1+s)}{left(frac{1}{s} + 1right)+(1+s)})
= (frac{frac{1}{s}+1+1+s}{frac{1}{s}+1+1+s}) = 1
So, R_N = 1.

8. In the circuit given below, it is given that V_AB = 4 V for R_L = 10 kΩ and V_AB = 1 V for R_L = 2kΩ. The value of Norton resistance for the network N is ____________

A. 16 kΩ
B. 30 kΩ
C. 3 kΩ
D. 50 kΩ

Answer: B
Clarification: When R_L = 10 kΩ and V_AB = 4 V
Current in the circuit (frac{V_{AB}}{R_L} = frac{4}{10}) = 0.4 mA
Norton voltage is given by V_N = I (R_N + R_L)
= 0.4(R_N + 10)
= 0.4R_N + 4
Similarly, for R_L = 2 kΩ and V_AB = 1 V
So, I = (frac{1}{2}) = 0.5 mA
V_N = 0.5(R_N + 2)
= 0.5 R_N + 1
∴ 0.1R_N = 3
Or, R_N = 30 kΩ.

9. For the circuit given below, the Norton’s resistance across the terminals A and B is _____________

A. 5 Ω
B. 7 kΩ
C. 1.5 kΩ
D. 1.1 kΩ

Answer: B
Clarification: Let V_AB = 1 V
5 V_AB = 5
Or, 1 = 1 × I₁ or, I₁ = 1
Also, 1 = -5 + 1(I – I₁)
∴ I = 7
Hence, R = 7 kΩ.

10. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________

A. 6 Ω and 1.333 A
B. 6 Ω and 0.833 A
C. 32 Ω and 0.156 A
D. 32 Ω and 0.25 A

11. For the circuit given in figure below, the Norton equivalent as viewed from terminals y and y’ is _________

A. 32 Ω and 0.25 A
B. 32 Ω and 0.125 A
C. 6 Ω and 0.833 A
D. 6 Ω and 1.167 A

12. In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________

A. 9.76 W
B. 9.26 W
C. 10.76 W
D. 11.70 W

Answer: B
Clarification: Let us remove the 1 Ω resistor and short x-y.
At Node 1, assuming node potential to be V, (frac{V-10}{5}) + I_SC = 5
But I_SC = (frac{V}{2})
∴ (frac{V-10}{5} + frac{V}{2}) = 5
Or, 0.7 V = 7
That is, V= 10 V
∴ I_SC = (frac{V}{2}) = 5 A
To find R_int, all constant sources are deactivated. R_int = (frac{(5+2)×2}{5+2+2} = frac{14}{9}) = 1.56 Ω
R_int = 1.56 Ω; I_SC = I_N = 5A
Here, I = I_N (frac{R_{int}}{R_{int}+1} = 5 × frac{1.56}{1.56+1}) = 3.04 A
∴ Power loss = (3.04)² × 1 = 9.26 W.

13. The value of R_N from the circuit given below is ________

A. 3 Ω
B. 1.2 Ω
C. 5 Ω
D. 12.12 Ω

Answer: D
Clarification: V_X = 3(frac{V_X}{6}) + 4
Or, V_X = 8 V = V_OC
If terminal is short-circuited, V_X = 0.
I_SC = (frac{4}{3+3}) = 0.66 A
∴ R_N = (frac{V_{OC}}{I_{SC}}) = (frac{8}{0.66}) = 12.12 Ω.

14. The current I, as shown in the figure below, is ________

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

Answer: C
Clarification: The 3 Ω resistance is an extra element because voltage at node B is independent of the 3 Ω resistance.
I₁ = (frac{3}{2+1}) = 1 A (B -> A.
The net current in 2 Ω resistance is I = 1 – I₁
= 2 – 1 = 1 A (A -> B..

15. While computing the Norton equivalent resistance and the Norton equivalent current, which of the following steps are undertaken?
A. Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
B. Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
C. The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched
D. The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched

Answer: D
Clarification: While computing the Norton equivalent voltage consisting of both dependent and independent sources, we first find the equivalent resistance called the Norton resistance by opening the two terminals. Then while computing the Norton current, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched and open-circuiting the dependent current sources keeping the independent current sources untouched.

Network Theory Multiple Choice Questions on “Bandwidth of an RLC Circuit”.

1. The expression of power (P₁) at lower half power frequency is?
A. (I²_maxR)/8
B. (I²_maxR)/4
C. (I²_maxR)/2
D. I²_maxR
Answer: C
Clarification: The upper and lower cut-off frequencies are sometimes called the half-power frequencies,. At these frequencies the power from the source is half of the power delivered at the resonant frequency. The expression of power (P₁) at lower half power frequency is P₁ = (I²_maxR)/2.

2. At upper half power frequency, the expression for power (P₂) is?
A. I²_maxR
B. (I²_maxR)/2
C. (I²_maxR)/4
D. (I²_maxR)/8
Answer: B
Clarification: At upper half power frequency, the expression for power (P₂) is P₂ = (I²_maxR)/2. The response curve is also called the selectivity curve of the circuit.

3. Determine the resonant frequency for the specifications: R = 10Ω, L = 0.1H, C = 10µF.
A. 157
B. 158
C. 159
D. 160
Answer: C
Clarification: The frequency at which the resonance occurs is called resonant frequency. The expression of the resonant frequency is given by f_r = 1/(2π√LC.. On substituting the given values we get resonant frequency = 1/(2π√(0.1×10×10^-6))=159.2 Hz.

4. The expression for lower half power frequency is?
A. (-R+√(x²+4LC.)/4πL
B. (–R-√(x²+4LC.)/4πL
C. (R-√(x²+4LC.)/4πL
D. (R+√(x²+4LC.)/4πL
Answer: A
Clarification: Selectivity indicates how well a resonant circuit responds to a certain frequency and eliminates all other frequencies. At lower power frequency, X_C > X_L(1/2πf₁C.-2πf₁L=R. f₁ = (-R+√(x²+4LC.)/4πL.

5. The expression for upper half power frequency is?
A. (R+√(x²+4LC.)/4πL
B. (R-√(x²+4LC.)/4πL
C. (–R-√(x²+4LC.)/4πL
D. (-R+√(x²+4LC.)/4πL
Answer: A
Clarification: The narrower the bandwidth the greater the selectivity. At upper half power frequency, X_C < X_L => -(1/2πf₂C.+2πf₂L=R. f₂ = (R+√(x²+4LC.)/4πL.

6. The expression for bandwidth is?
A. R/πL
B. R/2πL
C. R/4πL
D. R/8πL
Answer: B
Clarification: The bandwidth of any system is the range of frequencies for which the current or output voltage is equal to 70.7% of its value at the resonant frequency. The expression of bandwidth is BW = f₂ – f₁ = R/2πL.

7. In series circuits, the expression for quality factor is?
A. f_r
B. BW
C. f_r/BW
D. BW/f_r
Answer: C
Clarification: The quality factor is the ratio of the reactive power in the inductor or capacitor to the true power in the resistance in series with the coil or capacitor. The expression of quality factor is Q = f_r/BW.

8. In a series circuit having resistance and inductance, the quality factor is?
A. ωL/R
B. R/ωL
C. ωL
D. R
Answer: A
Clarification: Quality factor Q = ωL/R. A higher value of circuit Q results in smaller bandwidth and a lower value of Q causes a larger bandwidth.

9. If a series circuit contains resistor and capacitor, the expression for quality factor is?
A. C
B. ωRC
C. ωC
D. 1/ωRC
Answer: D
Clarification: The expression of quality factor is Q = 1/ωRC. The ratio of voltage across either L or C to the voltage applied the resonance can be defined as magnification.

10. The quality factor of the coil for a series circuit having R = 10Ω, L = 0.1H, C = 10µF.
A. 1
B. 5
C. 10
D. 15
Answer: C
Clarification: The resonant frequency is given by f_r = 1/(2π√LC.=1/(2π√(0.1×10×10^-6))=159.2 Hz. The relation between quality factor, resonant frequency and bandwidth is Q = f_r/BW = 2πf_rL/R = (6.28×159.2×0.1)/10=10.

Network Theory Multiple Choice Questions on “Problems Involving Standard Time Functions”.

1. A voltage waveform V (t) = 12t² is applied across a 1 H inductor for t≥0, with initial current through it being 0. The current through the inductor for t greater than 0 is ___________
A. 12t
B. 24t
C. 12t³
D. 4t³

Answer: D
Clarification: We know that, I = (frac{1}{L} int_0^t V ,dt)
Here, V (t) = 12t²
∴ I = (int_0^t 12t^2 ,dt)
= (big[frac{12t^3}{3}big]_0^t) = 4t³ A.

2. In the circuit given below, Z₁ = 10∠-60°, Z₂ = 10∠60°, Z₃ = 50∠53.13°. The Thevenin impedance as seen from A-B is _____________

A. 56.5∠45°
B. 60∠30°
C. 70∠30°
D. 34.4∠65°

Answer: A
Clarification: Z_TH = Z_A-B = Z₁ || Z₂ + Z₃
= (frac{Z_1 × Z_2}{Z_1 + Z_2} + Z_3)
= (frac{10∠-60° × 10∠60°}{10∠-60° + 10∠60°}) + 50∠53.13°
= 56.66∠45°.

3. In the circuit given below, the value of Z which would cause parallel resonance at 500 Hz is ___________

A. 125 mH
B. 34.20 μF
C. 2 μF
D. 0.05 μF

Answer: D
Clarification: At resonance the circuit should have unity power factor
Hence, Z should be capacitive.
Now, (frac{1}{jLω} + frac{1}{1/jCω}) = 0
Or, (frac{-1}{jLω}) + jCω = 0
∴ C = (frac{1}{L × ω^2} = frac{1}{2 × (2π × 500)^2})
= 0.05 μF.

4. In the circuit given below, the current source is 1∠0° A, R = 1 Ω, the impedances are Z_C = -j Ω and Z_L = 2j Ω. The Thevenin equivalent looking across A-B is __________

A. (sqrt{2})∠0 V, (1+2j) Ω
B. 2∠45° V, (1-2j) Ω
C. 2∠45° V, (1+j) Ω
D. (sqrt{2})∠45° V, (1+j) Ω

Answer: D
Clarification: Z_TH = 1 + (2j-j) = (1 + j) Ω
∴ V_TH = 1(1+j) = (sqrt{2})∠45° V.

5. A periodic rectangular signal X (t) has the waveform as shown below. The frequency of the fifth harmonic of its spectrum is ______________

A. 40 Hz
B. 200 Hz
C. 250 Hz
D. 1250 Hz

Answer: D
Clarification: Periodic time = 4 ms = 4 × 10^-3
Fundamental frequency = (frac{10^3}{4}) = 250 Hz
∴ Frequency of the fifth harmonic = 250 × 5 = 1250 Hz.

6. A voltage having the waveform of a sine curve is applied across a capacitor. When the frequency of the voltage is increased, what happens to the current through the capacitor?
A. Increases
B. Decreases
C. Remains same
D. Is zero

Answer: A
Clarification: The current through the capacitor is given by,
I_C = ωCV cos (ωt + 90°).
As the frequency is increased, I_C also increases.

7. An inductive coil connected to a 200 V, 50 Hz AC supply with 10 A current flowing through it dissipates 1000 W. Which of the following will have least value?
A. Resistance R
B. Reactance X_L
C. Impedance Z
D. Reactance X_C

Answer: A
Clarification: Power dissipated is given by,
I²R = 1000 W
Hence, R = (frac{1000}{100})
Or, R = 10 Ω
Impedance Z = (frac{V}{I} = frac{200}{10})
Or, Z = 20 Ω
Reactance X_L = (sqrt{Z^2 – R^2})
= (sqrt{20^2 – 10^2})
∴ X_L = 17.3 Ω.

8. An inductive coil connected to a 200 V, 50 Hz AC supply with 10 A current flowing through it dissipates 1000 W. The value of the impedance Z and Reactance X_L are respectively ____________
A. 10 Ω and 15 Ω
B. 20 Ω and 17.3 Ω
C. 17.3 Ω and 20 Ω
D. 15 Ω and 10 Ω

Answer: B
Clarification: Power dissipated is given by,
I²R = 1000 W
Hence, R = (frac{1000}{100})
Or, R = 10 Ω
Impedance Z = (frac{V}{I} = frac{200}{10})
Or, Z = 20 Ω
Reactance X_L = (sqrt{Z^2 – R^2})
= (sqrt{20^2 – 10^2})
∴ X_L = 17.3 Ω.

9. A relay coil having voltage of magnitude 210 V and frequency 50 Hz is connected to a 210 V, 50 Hz supply. If it has resistance of 50 Ω and an inductance of 0.2 H, the apparent power is _____________
A. 549.39 VA
B. 275.6 VA
C. 157 VA
D. 187 VA

Answer: A
Clarification: Z= 50 + j (0.2) (2π) (50) = 50 + 62.8
S = (frac{|V|^2}{Z} = frac{210^2}{50 – 62.8})
Apparent Power |S| = (frac{210^2}{sqrt{50^2 + 62.8^2}})
= 549.39 VA.

10. In the circuit given below, the value of Z_L for maximum power to be transferred is _____________

A. R
B. R + jωL
C. R – jωL
D. jωL

Answer: C
Clarification: The value of load for maximum power transfer is given by the complex conjugate of Z_AB
Z_AB = R + jX_L
= R + jωL
∴ Z_L for maximum power transfer is given by Z_L = R – jωL.

11. A 10 V is connected across a load whose V-I characteristics is given by 7I = V² + 2V. The internal resistance of the battery is of magnitude 1Ω. The current delivered by the battery is ____________
A. 6 A
B. 5 A
C. 7 A
D. 8 A

Answer: B
Clarification: 7I = V² + 2V …………………. (1)
Now, V = 10 – 1 ×I
Putting the value of V in eqt (1), we get,
&I = (10 – I) 2 + 2(10 – I)
Or, I = 100 + I2 – 20I + 20 – 2I
Or, I2 – 29I + 120 = 0…………………. (2)
∴ I = (frac{+29 ± sqrt{29^2 – 4(120)}}{2} = frac{29 ± 19}{2})
I = 5 A, 24 A
Now, I = 24 A is not possible because V will be negative from eqt (2)
∴ I = 5 A.

12. A current I given by I = – 8 + 6(sqrt{2}) (sin (ωt + 30°)) A is passed through three meters. The respective readings (in ampere) will be?
A. 8, 6 and 10
B. 8, 6 and 8
C. – 8, 10 and 10
D. -8, 2 and 2

Answer: C
Clarification: PMMC instrument reads only DC value and since it is a centre zero type, so it will give – 8 values.
So, rms = (sqrt{8^2 + left(frac{6sqrt{2}}{sqrt{2}})^2right)}) = 10 A
Moving iron also reads rms value, so its reading will also be 10 A.

13. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. A DC current of 1 mA produces a full-scale deflection of the instrument. The resistance of the instrument is 100 Ω. A voltage of 100 V (rms) is applied to the input terminals. The value of R required is?
A. 63.56 Ω
B. 89.83 Ω
C. 89.83 kΩ
D. 141.3 kΩ

Answer: C
Clarification: V_OAverage = 0.636 × (sqrt{2})V_rms = 0.8993 V_rms
The deflection with AC is 0.8993 times that with DC for the same value of voltage V
S_AC = 0.8993 S_DC
S_DC of a rectifier type instrument is (frac{1}{I_{fs}}) where I_fs is the current required to produce full scale deflection, I_fs = 1 mA; R_m = 100 Ω; S_DC = 10³ Ω/V
S_AC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R_S = S_AC V – R_m – 2R_d, where R_d is the resistance of diode, for ideal diode R_d = 0
∴ R_S = 899.3 × 100 – 100 = 89.83 kΩ.

14. The discharge of a capacitor through a ballistic galvanometer produces a damped frequency of 0.125 Hz and successive swings of 120, 96 and 76.8 mm. The logarithmic decrement is?
A. 0.225
B. 0.565
C. 0.484
D. 0.7887

Answer: A
Clarification: Logarithmic decrement, δ = ln (frac{x_1}{x_2} = ln frac{120}{96}) = 0.225
Now, δ is related to damping ratio K as, K = (frac{1}{left(1 + (frac{2π}{δ})^2)^{0.5}right)})
∴ K = 0.0357.

15. 12, 1 H inductors are used as edge to form a cube. The inductance between two diagonally opposite corners of the cube is ____________
A. (frac{5}{6}) H
B. (frac{10}{6}) H
C. 2 H
D. (frac{3}{2}) H

Answer: A
Clarification: V_AB = (frac{1}{3}L + frac{1}{6} L + frac{1}{3}L)
V_AB = IL ([frac{5}{6}] )
Now, (frac{V_{AB}}{I} = frac{5L}{6})
Or, L_AB = (frac{5L}{6})
Here, L = 1Ω
∴ L_AB = (frac{5}{6}) H.

Network Theory Problems on “Inner Relationships of Different Parameters”.

1. For the given information Z₁₁ = 3, Z₁₂ = 1, Z₂₁ = 2, Z₂₂ = 1. Find the value of Y₁₁.
A. 1
B. -1
C. 2
D. -2
Answer: A
Clarification: Y₁₁ = Z₂₂/∆z and ∆z=3-2=1 and Z₂₂ = 1. So on substituting we get Y₁₁ = 1/1 = 1.

2. For the given information Z₁₁ = 3, Z₁₂ = 1, Z₂₁ = 2, Z₂₂ = 1. Find the value of Y₁₂.
A. -2
B. 2
C. -1
D. 1
Answer: C
Clarification: Y₁₂ = -Z₁₂/∆z and ∆z=3-2=1 and Z₁₂ = 1. So on substituting we get Y₁₂=-1/1=-1.

3. For the given information Z₁₁ = 3, Z₁₂ = 1, Z₂₁ = 2, Z₂₂ = 1. Find the value of Y₂₁.
A. 2
B. -2
C. 1
D. -1
Answer: B
Clarification: We have the relation Y₂₁=-Z₂₁/∆z. ∆z=3-2=1 and given Z₂₁ = 2. On substituting we get Y₂₁=-2/1=-2.

4. For the given information Z₁₁ = 3, Z₁₂ = 1, Z₂₁ = 2, Z₂₂ = 1. Find the value of Y₂₂.
A. 1
B. 2
C. 3
D. 4
Answer: C
Clarification: The relation between Y₂₂ and Z₁₁ is Y₂₂= Z₁₁/∆z and ∆z=3-2=1 and given Z₁₁ = 3. On substituting we get Y₂₂ = 3/1 = 3.

5. For the given information Z₁₁ = 3, Z₁₂ = 1, Z₂₁ = 2, Z₂₂ = 1. What is the value ∆y?
A. 0
B. 1
C. 2
D. 3
Answer: B
Clarification: ∆y is the determinant of y parameters. The value ∆y is (Y₁₁)(Y₂₂)-(Y₁₂)(Y₂₁). On substituting the values we get ∆y = (1)(3)-(-2)(-1)=1.

6. For the given information Z₁₁ = 3, Z₁₂ = 1, Z₂₁ = 2, Z₂₂ = 1. What is the product of ∆y and ∆z is?
A. 3
B. 2
C. 1
D. 0
Answer: C
Clarification: ∆y is the determinant of y parameters and ∆z is the determinant of z parameters. And we obtained ∆y = 1 and ∆z = 1. So their product = (1) (1) = 1.

7. The relation between Z₁₁ and Y parameters is?
A. Z₁₁ = Y₂₂/∆y
B. Z₁₁ = -Y₂₂/∆y
C. Z₁₁ = Y₁₂/∆y
D. Z₁₁ = (-Y₁₂)/∆y
Answer: A
Clarification: V₁=(Y₂₂/∆y)I₁-(Y₁₂/∆y)I₂. The relation between Z₁₁ and Y parameters is Z₁₁ = Y₂₂/∆y.

8. The relation between Z₁₂ and Y parameters is?
A. Z₁₂ = Y₁₂/∆y
B. Z₁₂ = (-Y₁₂)/∆y
C. Z₁₂ = (-Y₂₂)/∆y
D. Z₁₂ = Y₂₂/∆y
Answer: B
Clarification: V₁=(Y₂₂/∆y)I₁-(Y₁₂/∆y)I₂. The relation between Z₁₂ and Y parameters is Z₁₂ = (-Y₁₂)/∆y.

9. The relation between Z₂₁ and Y parameters is?
A. Z₂₁ = Y₂₁/∆y
B. Z₂₁ = Y₁₂/∆y
C. Z₂₁ = (-Y₂₁)/∆y
D. Z₂₁ = (-Y₁₂)/∆y
Answer: C
Clarification: V₂=(-Y₂₁/∆y)I₁+(Y₁₁/∆y)I₂. The relation between Z₂₁ and Y parameters is Z₂₁ = (-Y₂₁)/∆y.

10. The relation between Z₂₂ and Y parameters is?
A. Z₂₂ = (-Y₁₁)/∆y
B. Z₂₂ = Y₂₁/∆y
C. Z₂₂ = (-Y₂₁)/∆y
D. Z₂₂ = Y₁₁/∆y
Answer: D
Clarification: V₂=(-Y₂₁/∆y)I₁+(Y₁₁/∆y)I₂. The relation between Z₂₂ and Y parameters is Z₂₂ = Y₁₁/∆y.

To practice all areas of Network Theory Problems,

Network Theory Interview Questions and Answers on “Cut-Set and Tree Branch Voltages”.

1. What is the direction of the cut-set?
A. same as the direction of the branch current
B. opposite to the direction of the link current
C. same as the direction of the link current
D. opposite to the direction of the branch current

Answer: A
Clarification: A cut-set is a minimal set of branches of a connected graph such that the removal of these branches causes the graph to be cut into exactly two parts. The direction of the cut-set is same as the direction of the branch current.

2. Consider the graph shown below. The direction of the cut-set of node ‘a’ is?

A. right
B. left
C. upwards
D. downwards

Answer: C
Clarification: The direction of the cut set at node ‘a’ will be the direction of the branch current at node ‘a’. So the direction of the current will be upwards.

3. Consider the graph shown below. The direction of the cut-set at node ‘b’ will be?

A. upwards
B. right
C. downwards
D. left

Answer: B
Clarification: The direction of the current will be towards right. The direction of the cut set at node ‘b’ will be the direction of the branch current at node ‘b’. So the direction of the current will be towards right.

4. In the graph shown below, the direction of the cut-set at node ‘c’ is?

A. downwards
B. upwards
C. left
D. right

Answer: B
Clarification: The direction of the cut set at node ‘c’ will be the direction of the branch current at node ‘c’. So the direction of the current will be upwards.

5. In the graph shown below, the direction of the cut-set at node ‘d’ will be?

A. left
B. downwards
C. right
D. upwards

Answer: C
Clarification: The direction of the cut set at node ‘d’ will be the direction of the branch current at node ‘d’. So the direction of the current will be upwards.

6. The row formed at node ‘a’ in the cut set matrix in the figure shown below is?

A. +1 +1 +1 +1 0 0 0 0
B. +1 0 0 0 +1 0 0 +1
C. -1 0 0 0 -1 0 0 -1
D. -1 -1 0 0 -1 -1 0 0

Answer: B
Clarification: The direction of the cut set at node ‘a’ is towards node ‘a’. So the current direction of I₁ is same as cut set direction. So it is +1. Similarly for all other currents.

7. The row formed at node ‘c’ in the cut set matrix in the following figure?

A. -1 -1 0 0 +1 -1 0 0
B. 0 0 +1 0 0 -1 -1 0
C. +1 0 0 0 +1 0 0 +1
D. -1 0 0 0 -1 0 0 -1

Answer: B
Clarification: The direction of the cut set at node ‘c’ is away from node ‘c’. So the current direction of I₃ is same as cut set direction. So it is +1. Similarly for all other currents.

8. The number of cut set matrices formed from a graph is?
A. N^N-1
B. N^N
C. N^N-2
D. N^N+1

Answer: C
Clarification: For every tree, there will be a unique cut set matrix. So there will be N^N-2 cut set matrices.

9. For every tree there will be _____ number of cut set matrices.
A. 1
B. 2
C. 3
D. 4

Answer: A
Clarification: For every tree, there will a unique cut set matrix. So, number of cut-set matrices for every tree = 1.

10. If a row of the cut set matrix formed by the branch currents of the graph is shown below. Then which of the following is true?

 I1   I2   I3   I4    I5   I6   I7  I8
-1   -1   0   0   +1  -1   0   0

A. -V₁-V₂+V₅-V₆=0
B. -I₁-I₂+I₅-I₆=0
C. -V₁+V₂+V₅-V₆=0
D. -I₁+I₂+I₅-I₆=0

Answer: B
Clarification: KCL equations are derived from cut set matrix and these include currents not voltages. So, -I₁-I₂+I₅-I₆=0.