Category: Network Theory Objective Questions

Network Theory Multiple Choice Questions on “Series Circuits”.

1. If we apply a sinusoidal input to RL circuit, the current in the circuit is __________ and the voltage across the elements is _______________
A. square, square
B. square, sinusoid
C. sinusoid, square
D. sinusoid, sinusoid

Answer: D
Clarification: If we apply a sinusoidal input to RL circuit, the current in the circuit is square and the voltage across the elements is sinusoid. In the analysis of the RL series circuit, we can find the impedance, current, phase angle and voltage drops.

2. The circuit shown below consists of a 1kΩ resistor connected in series with a 50mH coil, a 10V rms, 10 KHz signal is applied. Find impedance Z in rectangular form.

A. (1000+j0.05) Ω
B. (100+j0.5) Ω
C. (1000+j3140) Ω
D. (100+j3140) Ω

Answer: C
Clarification: Inductive Reactance X_L = ωL = 2πfL = (6.28)(10⁴)(50×10^-3) = 3140Ω. In rectangular form, total impedance Z = (1000+j3140) Ω.

3. Find the current I (mA. in the circuit shown below.

A. 3.03
B. 30.3
C. 303
D. 0.303

Answer: A
Clarification: Total impedance Z = (1000+j3140) Ω. Magnitude = 3295.4Ω
Current I=V_s/Z = 10/3295.4=3.03mA.

4. Find the phase angle θ in the circuit shown below.

A. 62.33⁰
B. 72.33⁰
C. 82.33⁰
D. 92.33⁰

Answer: B
Clarification: Phase angle θ = tan^-1⁡(X_L/R). The values of X_L, R are X_L = 3140Ω and R = 1000Ω. On substituting the values in the equation, phase angle θ =tan^-1⁡(3140/1000)=72.33⁰.

5. In the circuit shown below, find the voltage across resistance.

A. 0.303
B. 303
C. 3.03
D. 30.3

Answer: C
Clarification: Voltage across resistance = IR. The values of I = 3.03 mA and R = 10000Ω. On substituting the values in the equation, the voltage across resistance = 3.03 x 10^-3×1000 = 3.03V.

6. In the circuit shown below, find voltage across inductive reactance.

A. 9.5
B. 10
C. 9
D. 10.5

Answer: A
Clarification: Voltage across inductor = IX_L. The values of I = 3.03 mA and X_L = 10000Ω. On substituting the values in the equation, the voltage across inductor = 3.03×10^-3×1000 = 9.51V.

7. Determine the source voltage if voltage across resistance is 70V and the voltage across inductor is 20V as shown in the figure.

A. 71
B. 72
C. 73
D. 74

Answer: C
Clarification: If voltage across resistance is 70V and the voltage across inductor is 20V, source voltage V_s=√(70²+20²) = 72.8≅73V.

8. Find the phase angle in the circuit shown below.

A. 15
B. 16
C. 17
D. 18

Answer: B
Clarification: The phase angle is the angle between the source voltage and the current. Phase angle θ=tan^-1(V_L/V_R). The values of V_L = 20V and V_R = 70V. On substituting the values in the equation, phase angle in the circuit = tan^-1(20/70)=15.94^o≅ 16^o.

9. An AC voltage source supplies a 500Hz, 10V rms signal to a 2kΩ resistor in series with a 0.1µF capacitor as shown in the following figure. Find the total impedance.

A. 3750.6Ω
B. 3760.6Ω
C. 3780.6Ω
D. 3790.6Ω

Answer: B
Clarification: The capacitive reactance X_C = 1/2πfC = 1/(6.28×500×0.1×10^-6))=3184.7Ω. In rectangular form, X = (2000-j3184.7)Ω. Magnitude = 3760.6Ω.

10. Determine the phase angle in the circuit shown below.

A. 58
B. 68
C. -58
D. -68

Answer: C
Clarification: The phase angle in the circuit is phase angle θ = tan^-1⁡(-X_C/R) =tan^-1⁡((-3184.7)/2000)=-57.87^o≅-58^o.

11. Find the current I (mA. in the circuit shown below.

A. 2.66
B. 3.66
C. 4.66
D. 5.66

Answer: A
Clarification: The term current is the ratio of voltage to the impedance. The current I (mA. in the circuit is current I = V_S / Z = 10/3760.6 = 2.66mA

12. Find the voltage across the capacitor in the circuit shown below.

A. 7
B. 7.5
C. 8
D. 8.5

Answer: D
Clarification: The voltage across the capacitor in the circuit is capacitor voltage = 2.66×10^-3×3184.7 =8.47V.

13. Determine the voltage across the resistor in the circuit shown below.

A. 3
B. 4
C. 5
D. 6

Answer: C
Clarification: The voltage across the resistor in the circuit resistive voltage = 2.66×10^-3×3184.7 = 5.32V.

14. In the circuit shown below determine the total impedance.

A. 161
B. 162
C. 163
D. 164

Answer: B
Clarification: Reactance across capacitor = 1/(6.28×50×10×10^-6) = 318.5Ω.
Reactance across inductor = 6.28×0.5×50=157Ω. In rectangular form, Z = (10+j157-j318.5) Ω = (10-j161.5)Ω. Magnitude=161.8Ω.

15. Find the current in the circuit shown below.

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer: C
Clarification: The term current is the ratio of voltage to the impedance. The current in the circuit is current I=V_S/Z = 50/161.8 = 0.3A.

Network Theory Multiple Choice Questions on “Problems Involving Coupling Coefficient”.

1. If the 3-phase balanced source in the given figure delivers 1500 W at a leading power factor of 0.844, then the value of Z_L is approximately __________
A. 90∠32.44°
B. 80∠32.44°
C. 80∠-32.44°
D. 90∠-32.44°

Answer: D
Clarification: 3V_P I_P cosθ = 1500
Or, 3((frac{V_L}{sqrt{3}}) (frac{V_L}{sqrt{3} Z_L})) cos θ = 1500
Or, Z_L = (frac{V_L^2}{1500} = frac{400^2 (0.844)}{1500}) = 90 Ω
And θ = ∠-arc cos⁡(0.844)
= ∠-32.44°.

2. An RLC series circuit has a resistance R of 20Ω and a current which lags behind the applied voltage by 45°. If the voltage across the inductor is twice the voltage across the capacitor, the value of inductive resistance is ____________
A. 10 Ω
B. 20 Ω
C. 40 Ω
D. 60 Ω

Answer: C
Clarification: Z = 20 + j20
V = V_R = j (V_L – V_C)
Given, V_L = 2 V_C
Or, Z_L = 2 Z_C
Or, Z_L – Z_C = 20
Or, 2 Z_C – Z_C = 20
Or, Z_C = 20 Ω
Or, Z_L = 2Z_C = 40 Ω.

3. In the Wheatstone bridge shown below, if the resistance in each arm is increased by 0.05%, then the value of V_out will be?

A. 50 mV
B. Zero
C. 5mV
D. 0.1mV

Answer: B
Clarification: In Wheatstone bridge, balance condition is
R₁R₃ = R₂R₄
Here, R₁ = 5, R₂ = 10, R₃ = 16, R₄ = 8
And when the Wheatstone bridge is balanced then, at V_out voltage will be Zero.

4. The voltage drop across a standard resistor of 0.2 Ω is balanced at 83 cm. The magnitude of the current, if the standard cell emf of 1.53 V is balanced at 42 m is ___________
A. 13.04 A
B. 10 A
C. 14.95 A
D. 12.56 A

Answer: C
Clarification: Voltage drop per unit length = (frac{1.53}{42}) = 0.036 V/cm
Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V
∴ Current through resistor, I = (frac{2.99}{0.2}) = 14.95 A.

5. The readings of polar type potentiometer are
I = 12.4∠27.5°
V = 31.5∠38.4°
Then, Reactance of the coil will be?
A. 2.51 Ω
B. 2.56 Ω
C. 2.54 Ω
D. 2.59 Ω

Answer: C
Clarification: Here, V = 31.5∠38.4°
I = 12.4∠27.5°
Z = (frac{31.5∠38.4°}{12.4∠27.5°}) = 2.54∠10.9°
But Z = R + jX = 2.49 + j0.48
∴ Reactance X = 2.54 Ω.

6. The simultaneous applications of signals x (t) and y (t) to the horizontal and vertical plates respectively, of an oscilloscope, produce a vertical figure of 8 displays. If P and Q are constants and x(t) = P sin (4t + 30), then y(t) is equal to _________
A. Q sin (4t – 30)
B. Q sin (2t + 15)
C. Q sin (8t + 60)
D. Q sin (4t + 30)

Answer: B
Clarification: (frac{f_y}{f_x} = frac{x-peak}{y-peak})
Here, x-peak = 1 and y-peak = 2
∴ y(t) = Q sin (2t + 15).

7. A resistor of 10 kΩ with a tolerance of 5% is connected in parallel with 5 kΩ resistors of 10% tolerance. The tolerance limit is __________
A. 9%
B. 12.4%
C. 8.33%
D. 7.87%

Answer: C
Clarification: Here, R₁ and R₂ are in parallel.
Then, (frac{1}{R} = frac{1}{R_1} + frac{1}{R_2})
Or, R = (frac{50}{15}) kΩ
∴ (frac{△R}{R} = frac{△R_1}{R_1^2} + frac{△R_2}{R_2^2})
And △R₁ = 0.5 × 10³, △R₂ = 0.5 × 10³
∴ (frac{△R}{R} = frac{10 × 10^3}{3 × 10 × 10^3} × frac{0.5 × 10^3}{10 × 10^3} + frac{10}{3} × frac{10^3}{5 × 10^3} × frac{0.5 × 10^3}{5 × 10^3})
= (frac{0.5}{30} + frac{1}{15} = frac{2.5}{30}) = 8.33%.

8. A 200/1 Current Transformer (CT) is wound with 400 turns on the secondary on a toroidal core. When it carries a current of 180 A on the primary, the ratio is found to be -0.5%. If the number of secondary turns is reduced by 1, the new ratio error (in %) will be?
A. 0.0
B. -0.5
C. -1.0
D. -2.0

Answer: C
Clarification: Turn compensation only alters ratio error n=400
Ratio error = -0.5% = – (frac{0.5}{100}) × 400 = -2
So, Actual ratio = R = n+1 = 401
Nominal Ratio K_N = (frac{400}{1}) = 400
Now, if the number of turns are reduced by one, n = 399, R = 400
Ratio error = (frac{K_N-R}{R} = frac{200-200}{200}) = 0.

9. In the Two wattmeter method of measuring power in a balanced three-phase circuit, one wattmeter shows zero and the other positive maximum. The load power factor is ___________
A. 0
B. 0.5
C. 0.866
D. 1.0

Answer: B
Clarification: The load power factor is = 0.5. Since at this power factor one wattmeter shows zero and the other shows a positive maximum value of power.

10. The meter constant of a single-phase, 230 V induction watt-meter is 600 rev/kW-h. The speed of the meter disc for a current of 15 A at 0.8 power factor lagging will be?
A. 30.3 rpm
B. 25.02 rpm
C. 27.6 rpm
D. 33.1 rpm

Answer: C
Clarification: Meter constant = (frac{Number, of, revolution}{Energy} = frac{600 × 230 × 15 × 0.8}{1000}) = 1656
∴ Speed in rpm = (frac{1656}{60}) = 27.6 rpm.

11. In the figure given below, a 220 V 50 Hz supplies a 3-phase balanced source. The pressure Coil (PC. and Current Coil (CC. of a watt-meter are connected to the load as shown. The watt-meter reading is _________

A. Zero
B. 1600 W
C. 242 W
D. 400 W

Answer: C
Clarification: Watt-meter reading = Current through CC × Voltage across PC × cos (phase angle).
I_BR = I_CC = (frac{220∠120°}{100°}) = 2.2∠120°
V_YB = V_PC = 220∠-120°
w = 2.2∠120° × 220∠-120° × cos 240° = – 242 W.

12. In the Owen’s bridge shown in below figure, Z₁ = 200∠60°, Z₂ = 400∠-90°, Z₃ = 300∠0°, Z₄ = 400∠30°. Then,

A. Bridge is balanced with given impedance values
B. Bridge can be balanced, if Z₄ = 600∠60°
C. Bridge can be balanced, if Z₃ = 400∠0°
D. Bridge cannot be balanced with the given configuration

Answer: D
Clarification: For Bridge to be balanced, the product of impedances of the opposite arm should be equal in magnitude as well as phase angle. Here Z₃ Z₂ ≠ Z₁ Z₄ for whatever chosen value. Therefore the Bridge cannot be balanced.

13. In Maxwell’s capacitance bridge for calculating unknown inductance, the various values at balance are, R₁ = 300 Ω, R₂ = 700 Ω, R₃ = 1500 Ω, C₄ = 0.8 μF. The values of R₁, L₁ and Q factor, if the frequency is 1100 Hz are ____________
A. 240 Ω, 0.12 H, 3.14
B. 140 Ω, 0.168 H, 8.29
C. 140 Ω, 0.12 H, 5.92
D. 240 Ω, 0.36 H, 8.29

Answer: B
Clarification: From Maxwell’s capacitance, we have
R₁ = (frac{R_2 R_3}{R_4} = frac{300 ×700}{1500}) = 140 Ω
L₁ = R₂ R₃ C₄
= 300 × 700 × 0.8 × 10^-6 = 0.168 H
∴ Q = (frac{ωL_1}{R_1})
= (frac{2 × π × 1100 × 0.168}{140}) = 8.29.

14. In the figure below, the values of the resistance R₁ and inductance L₁ of a coil are to be calculated after the bridge is balanced. The values are _________________

A. 375 Ω and 75 mH
B. 75 Ω and 150 mH
C. 37.5 Ω and 75 mH
D. 75 Ω and 75 mH

Answer: A
Clarification: Applying the usual balance condition relation,
Z₁ Z₄ = Z₂ Z₃
We have, (R₁ + jL₁ ω) (frac{R_4/jωC_4}{R_4+1/jωC_4}) = R₂ R₃
Or, R₁ R₄ + jL₁ ωR₄ = R₂ R₃ + j R₂ R₃ R₄ C₄ ω
∴ R₁ = 2000 × (frac{750}{4000}) = 375 Ω
∴ L₁ = 2000 × 750 × 0.5 × 10^-6 = 75 mH.

15. The four arms of an AC bridge network are as follows:
Arm AB: unknown impedance
Arm BC: standard capacitor C₂ of 1000pf
Arm CD: a non-inductive resistance of R of 100 Ω in parallel to a capacitor of 0.01 μF
Arm DA: a non-inductive resistance of 1000 Ω
The supply frequency is 50 Hz and connected across terminals B and D. If the bridge is balanced with the above value, determine the value of unknown Impedance.
A. 10 kΩ
B. 100 kΩ
C. 250 kΩ
D. 20 kΩ

Answer: A
Clarification: For the balance conditions,
Z₁ Z₃ = Z₂ Z₄
1000 × (frac{1}{jω × 1000 × 10^{-12}} = (R + jX) frac{100}{1 + j100 × ω × 0.01 × 10^{-6}})
Or, (frac{10^{12}}{jω} = (R + jX) left(frac{100}{1 + jω + 10^{-6}}right))
Or, (frac{- j 10^{10}}{ω}) – 10⁴ = R + jX
Comparing the real part, we get,
R = 10 kΩ.

Network Theory Multiple Choice Questions on “Sinusoidal Response of an R-L-C Circuit”.

1. The particular current obtained from the solution of i in the sinusoidal response of R-L-C circuit is?
A. i_p = V/√(R²+(1/ωC+ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC+ωL)/R))
B. i_p = V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R))
C. i_p = V/√(R²+(1/ωC+ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R))
D. i_p = V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC+ωL)/R))

Answer: B
Clarification: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i_p = V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R)).

2. In the sinusoidal response of R-L-C circuit, the complementary function of the solution of i is?
A. i_c = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t
B. i_c = c₁ e^(K₁-K₂)t + c₁ e^(K₁-K₂)t
C. i_c = c₁ e^(K₁+K₂)t + c₁ e^(K₂-K₁)t
D. i_c = c₁ e^(K₁+K₂)t + c₁ e^(K₁+K₂)t

Answer: A
Clarification: From the R-L circuit, we get the characteristic equation as
(D²+R/L D+1/LC.=0. The complementary function of the solution i is i_c = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t.

3. The complete solution of the current in the sinusoidal response of R-L-C circuit is?
A. i = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t – V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R))
B. i = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t – V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ-tan^-1)⁡((1/ωC-ωL)/R))
C. i = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t + V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R))
D. i = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t + V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ-tan^-1)⁡((1/ωC-ωL)/R))

Answer: C
Clarification: The complete solution for the current becomes i = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t + V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R)).

4. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.
A. -38.5±j1290
B. 38.5±j1290
C. 37.5±j1290
D. -37.5±j1290

5. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complementary current.
A. i_c = e^-37.5t(c₁cos1290t + c₂sin1290t)
B. i_c = e^-37.5t(c₁cos1290t – c₂sin1290t)
C. i_c = e^37.5t(c₁cos1290t – c₂sin1290t)
D. i_c = e^37.5t(c₁cos1290t + c₂sin1290t)

Answer: A
Clarification: The roots of the charactesistic equation are D₁ = -37.5+j1290 and D₂ = -37.5-j1290. The complementary current obtained is i_c = e^-37.5t(c₁cos1290t + c₂sin1290t).

6. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the particular solution.

A. i_p = 0.6cos(500t + π/4 + 88.5⁰)
B. i_p = 0.6cos(500t + π/4 + 89.5⁰)
C. i_p = 0.7cos(500t + π/4 + 89.5⁰)
D. i_p = 0.7cos(500t + π/4 + 88.5⁰)

Answer: D
Clarification: Particular solution is i_p = V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R)). i_p = 0.7cos(500t + π/4 + 88.5⁰).

7. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complete solution of current.

A. i = e^-37.5t(c₁cos1290t + c₂sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)
B. i = e^-37.5t(c₁cos1290t + c₂sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)
C. i = e^-37.5t(c₁cos1290t + c₂sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)
D. i = e^-37.5t(c₁cos1290t + c₂sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)

Answer: A
Clarification: The complete solution is the sum of the complementary function and the particular integral. So i = e^-37.5t(c₁cos1290t + c₂sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

8. The value of the c₁ in the following equation is?
i = e^-37.5t(c₁cos1290t + c₂sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
A. -0.5
B. 0.5
C. 0.6
D. -0.6

Answer: B
Clarification: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, c₁ = -0.71cos (133.5⁰) = 0.49.

9. The value of the c₂ in the following equation is?
i = e^-37.5t(c₁cos1290t + c₂sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
A. 2.3
B. -2.3
C. 1.3
D. -1.3

Answer: C
Clarification: Differentiating the current equation, we have di/dt = e^-37.5t (-1290c₁sin1290t + 1290c₂cos1290t) – 37.5e^-37.5t(c₁cos1290t+c₂sin1290t) – 0.71x500sin(500t+45^o+88.5^o). At t = 0, di/dt = 1414. On solving, we get c₂ = 1.31.

10. The complete solution of current obtained by substituting the values of c₁ and c₂ in the following equation is?
i = e^-37.5t(c₁cos1290t + c₂sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
A. i = e^-37.5t(0.49cos1290t – 1.3sin1290t) + 0.7cos(500t + 133.5⁰)
B. i = e^-37.5t(0.49cos1290t – 1.3sin1290t) – 0.7cos(500t + 133.5⁰)
C. i = e^-37.5t(0.49cos1290t + 1.3sin1290t) – 0.7cos(500t + 133.5⁰)
D. i = e^-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰)

Answer: D
Clarification: The complete solution is the sum of the complementary function and the particular integral. So i = e^-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰).

Network Theory Multiple Choice Questions on “Properties of Transfer Functions”.

1. The coefficients of numerator polynomial and the denominator polynomial in a transfer function must be?
A. real
B. complex
C. at least one real coefficient
D. at least one complex coefficient
Answer: A
Clarification: The coefficients of P(s), the numerator polynomial and of Q(s), the denominator polynomial in a transfer function must be real. Therefore all poles and zeros if complex must occur in conjugate pairs.

2. In a transfer function, the degree of numerator polynomial is ___________ than the degree of the denominator polynomial.
A. greater than
B. less than
C. equal to
D. less than or equal to
Answer: D
Clarification: In a transfer function, the degree of numerator polynomial is less than or equal to than the degree of the denominator polynomial. And the degree of the numerator polynomial of Z₂₁(s) or Y₂₁(s) is less than or equal to the degree of the denominator polynomial plus one.

3. The real parts of all poles and zeros in a driving point function must be?
A. zero
B. negative
C. zero or negative
D. positive
Answer: C
Clarification: The real parts of all poles and zeros in a driving point function must be zero or negative but should not be positive and the complex or imaginary poles and zeros must occur in conjugate pairs.

4. If the real part of driving point function is zero, then the pole and zero must be?
A. complex
B. simple
C. one complex pole
D. one complex zero
Answer: B
Clarification: If the real part of driving point function is zero, then the pole and zero must be simple but should not contain any complex pole or complex zero.

5. The degree of the numerator polynomial and denominator polynomial in a driving point function may differ by?
A. 0
B. 1
C. 0 or 1
D. 2
Answer: C
Clarification: The degree of numerator polynomial and denominator polynomial in a driving point function may differ by zero or one. And the polynomials P(s) and Q(s) may not have any missing terms between the highest and lowest degrees unless all even or odd terms are missing.

6. The lowest degree in numerator polynomial and denominator polynomial in a driving point function may differ by at most __________
A. 0
B. 1
C. 2
D. 3
Answer: B
Clarification: The lowest degree in numerator polynomial and denominator polynomial in a driving point function may differ by at most one and the coefficients in the polynomials P(s) and Q(s) of network function must be real and positive.

7. The coefficients in the denominator polynomial of the transfer function must be?
A. positive
B. negative
C. positive or zero
D. negative or zero
Answer: A
Clarification: The coefficients in the denominator polynomial of the transfer function must be positive but should not be negative and the coefficients in the polynomials P(s) and Q(s) of transfer function must be real.

8. The coefficients in the numerator polynomial of the transfer function may be?
A. must be negative
B. must be positive
C. may be positive
D. may be negative
Answer: D
Clarification: The coefficients in the numerator polynomial of the transfer function may be negative and the complex or imaginary poles and zeros must occur in conjugate pairs.

9. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?
A. all odd terms are missing
B. all even terms are missing
C. all even or odd terms are missing
D. all even and odd terms are missing
Answer: C
Clarification: The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing and the polynomial P(s) may have missing terms between the lowest and the highest degree.

10. The degree of numerator polynomial in a transfer function may be as small as _________ independent of the degree of the denominator polynomial.
A. 1
B. 2
C. 0
D. 3
Answer: C
Clarification: The degree of numerator polynomial in a transfer function may be as small as zero, independent of the degree of the denominator polynomial and for the voltage transfer ratio and the current transfer ratio, the maximum degree of P(s) must be equal to the degree of Q(s).

Network Theory Multiple Choice Questions on “Circuit Elements and Kirchhoff’s Laws”.

1. Potential difference in electrical terminology is known as?
A. Voltage
B. Current
C. Resistance
D. Conductance

Answer: A
Clarification: Potential difference in electrical terminology is known as Voltage and is denoted either by V or v. It is expressed in terms of energy per unit charge.

2. The circuit in which current has a complete path to flow is called ______ circuit.
A. short
B. open
C. closed
D. open loop

Answer: C
Clarification: The circuit in which current has a complete path to flow is called a closed circuit. When the current path is broken so that current cannot flow, the circuit is called an open circuit.

3. If the voltage-current characteristics is a straight line through the origin, then the element is said to be?
A. Linear element
B. Non-linear element
C. Unilateral element
D. Bilateral element

Answer: A
Clarification: If the voltage-current characteristic is a straight line through the origin, then the element is said to be Linear element. The difference in potential energy of charges is called Potential difference.

4. The voltage across R₁ resistor in the circuit shown below is?

A. 10
B. 5
C. 2.5
D. 1.25

Answer: B
Clarification: According to voltage divider rule, 10v is divide equally across resistors R₁ and R₂. So the voltage across R₁ will be 5v.

5. The energy stored in the inductor is?
A. Li²/4
B. Li²/2
C. Li²
D. Li²/8

Answer: B
Clarification: The energy stored in the inductor the area under the power of the inductor and is given by W = ʃpdt = ʃLidi = Li²/2.

6. How many types of dependent or controlled sources are there?
A. 1
B. 2
C. 3
D. 4

Answer: D
Clarification: There are 4 dependent or controlled sources. They are VCVS(Voltage Controlled Voltage Source), VCCS(Voltage Controlled Current Source, CCVS(Current Controlled Voltage Source, CCCS(Current Controlled Current Source).

7. Find the voltage V_x in the given circuit.

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

Answer: A
Clarification: From the circuit applying Kirchhoff’s voltage law, we can write 50 = 15 + 10 + 15 + V_x => V_x = 10V.

8. If the resistances 1Ω, 2Ω, 3Ω, 4Ω are parallel, then the equivalent resistance is?
A. 0.46Ω
B. 0.48Ω
C. 0.5Ω
D. 0.52Ω

Answer: B
Clarification: The equivalent resistance 1/R_t = (1/R₁)+(1/R₂)+(1/R₃)+(1/R₄). And R₁, R₂, R₃, R₄ are 1Ω, 2Ω, 3Ω, 4Ω respectively. => R_t = 0.48Ω.

9. Find total current(mA. in the circuit.

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

Answer: A
Clarification: R₂ is parallel to R₃. So equivalent resistance of R₂ and R₃ is 1K. The total resistance in the circuit is (1+1+1)K = 3K.Current in the circuit is 3V/3KΩ = 1mA.

10. If the resistances 3Ω, 5Ω, 7Ω, 9Ω are in series, then their equivalent resistance(Ω) is?
A. 9
B. 20
C. 24
D. 32

Answer: C
Clarification: If the resistances are in series, then equivalent resistance is the sum of all the resistances that are in series. Equivalent resistance is (3+5+7+9)Ω = 24Ω.

Network Theory Multiple Choice Questions on “Reciprocity Theorem”.

1. To check for the Reciprocity Theorem we consider ______ of response to excitation.
A. ratio
B. addition
C. product
D. subtraction

Answer: A
Clarification: For the Reciprocity Theorem to satisfy the ratio of response to the excitation of the circuit should be equal to the ratio of response to excitation after the source is replaced.

2. For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be?
A. different
B. same
C. before source is replaced is greater than after the source is replaced
D. before source is replaced is less than after the source is replaced

Answer: B
Clarification: For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be same and if that condition satisfies the reciprocity theorem is valid for the given circuit.

3. The circuit which satisfies Reciprocity Theorem is called?
A. Short circuit
B. Open circuit
C. Linear circuit
D. Non-linear circuit

Answer: C
Clarification: The circuit which satisfies Reciprocity Theorem is called linear circuit. A linear circuit is an electronic circuit in which, for a sinusoidal input voltage of frequency f, any steady-state output of the circuit (the current through any component, or the voltage between any two points) is also sinusoidal with frequency f.

4. Find the current through the 2Ω(c-D. resistor in the circuit shown below.

A. 0.143
B. 1.43
C. 14.3
D. 143

Answer: B
Clarification: Total resistance in the circuit = 2+[3||(2+2│├|2)]=3.5Ω. The current drawn by the circuit (I_t)=20/3.5=5.71Ω. The current drawn by 2Ω resistor = 1.43A.

5. In the following circuit, the current drawn by 2Ω resistor (a-B. after the source is replaced is?

A. 143
B. 14.3
C. 1.43
D. 0.143

6. The following circuit satisfies Reciprocity Theorem.

A. True
B. False

Answer: A
Clarification: The ratio of response to excitation before the source is replaced is equal to 0.0715. And the ratio of response to excitation before the source is replaced is equal to 0.0715. So, the circuit satisfies the Reciprocity theorem.

7. Find the current through 3Ω resistor in the circuit shown below.

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

Answer: B
Clarification: The 6Ω resistor is parallel to 3Ω resistor and the resultant is in series with 2Ω resistor. Total current from source = 12/(2+(6│|3))=3A. Current through 3Ω resistor = 3 × 6/(6+3)=2A.

8. Find the current through 2Ω resistor after source is replaced in the below circuit.

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

9. The following circuit satisfies the reciprocity theorem.

A. False
B. True

Answer: B
Clarification: The ratio of response to excitation before the source is replaced is equal to 0.167. And the ratio of response to excitation before the source is replaced is equal to 0.167. So, the circuit satisfies the Reciprocity theorem.

10. While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of?
A. voltage to voltage
B. current to current
C. voltage to current
D. None of the above

Answer: C
Clarification: While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of voltage to current or current to voltage.