250+ TOP MCQs on Problems Involving Dot Conventions and Answers

Network Theory Multiple Choice Questions on “Problems Involving Dot Conventions”.

1. The current through an electrical conductor is 1A when the temperature of the conductor 0°C and 0.7 A when the temperature is 100°C. The current when the temperature of the conductor is 1200°C is ___________
A. 0.08 A
B. 0.16 A
C. 0.32 A
D. 0.64 A

Answer: B
Clarification: (frac{1}{0.7} = frac{R_O (1+αt)}{R_O})
= 1 + α100
∴α = 0.0043 per °C
Current at 1200 °C is given by, (frac{1}{I} = frac{R_O (1+α1200)}{R_O})
= 1 + α1200
= 1 + 0.0043 × 1200 = 6.16
∴ I = (frac{1}{6.16}) = 0.16 A.

2. In the circuit given below, the equivalent capacitance is ____________

A. 1.625 F
B. 1.583 F
C. 0.583 F
D. 0.615 F

Answer: D
Clarification: C_CB = (left(frac{C_2 C_3}{C_2+ C_3}right)) + C₅ = 1.5 F
Now, C_AB =(left(frac{C_1 C_{CB}}{C_1+ C_{CB}}right)) + C₆ = 1.6 F
C_XY = (frac{C_{AB} × C_4}{C_{AB} + C_4}) = 0.615 F.

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

A. 3.5 μF
B. 1.2 μF
C. 2.4 μF
D. 4.05 μF

Answer: D
Clarification: The 2.5 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 1.5 μF.
Hence, C₁ = (frac{1.5(2.5+1)}{1.5+2.5+1})
= (frac{5.25}{5}) = 1.05
Now, C₁ is in parallel with the 3 μF capacitor.
∴ C_EQ = 1.05 + 3 = 4.05 μF.

4. In the circuit given below, the voltage across A and B is?

A. 13.04 V
B. 17.84 V
C. 12 V
D. 10.96 V

Answer: B
Clarification: Loop current I₁ = (frac{6}{10}) = 0.6 A
I₂ = (frac{12}{14}) = 0.86 A
V_AB = (0.6) (4) + 12 + (0.86) (4)
= 2.4 + 12 + 3.44
= 17.84 V.

5. In the figure given below, the voltage source provides the circuit with a voltage V. The number of non-planar graph of independent loop equations is ______________

A. 8
B. 12
C. 7
D. 5

Answer: D
Clarification: The total number of independent loop equations are given by L = B – N + 1 where,
L = number of loop equations
B = number of branches = 12
N = number of nodes = 8
∴ L = 12 – 8 + 1 = 5.

6. When a DC voltage is applied to an inductor, the current through it is found to build up in accordance with I = 20(1-e^-50t). After the lapse of 0.02 s, the voltage is equal to 2 V. What is the value of inductance?
A. 2 mH
B. 5.43 mH
C. 1.54 mH
D. 0.74 mH

Answer: B
Clarification: V_L = L(frac{dI}{dt})
Where, I = 20(1-e^-50t)
Therefore, V_L = L(frac{d 20(1-e^{-50t})}{dt})
= L × 20 × 50e^-50t
At t = 0.02 s, V_L = 2 V
∴ L = (frac{2}{20 × 50 × e^{-50×0.02}})
= 5.43 μH.

7. An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V. It is disconnected and then immersed in oil with a relative permittivity of 2.5. The energy after immersion is?
A. 275 μJ
B. 250 μJ
C. 225 μJ
D. 625 μJ

Answer: B
Clarification: E = (frac{1}{2}) CV²
Or, C = (frac{ε_0 ε_r A}{d} = ϵ_r (frac{ε_0 A}{d}))
= 2.5 × 0.005 × 10^-6
∴ C_NEW = 12.5 × 10^-9 F
Now, q = CV = 0.005 × 10^-6 × 500 = 2.5 × 10^-6
V_NEW = (frac{q} {C_{NEW}})
= (frac{2.5 × 10^{-6}}{12.5 × 10^{-9}}) V_NEW = 200
E = (frac{1}{2}) CV²
= (frac{1}{2}) × 12.5 × 10^-9 × (200)²
= 250 μJ.

8. The resistance of copper motor winding at t=20°C is 3.42 Ω. After extended operation at full load, the motor windings measures 4.22 Ω. If the temperature coefficient is 0.0426, what is the rise in temperature?
A. 60°C
B. 45.2°C
C. 72.9°C
D. 10.16°C

Answer: D
Clarification: Given that, R₁ = 3.42 Ω
T₁ = 20° and α = 0.0426
R₂ = 4.22 Ω
Now, (frac{R_1}{1 + αT_1} = frac{R_2}{1 + αT_2})
Or, (frac{3.42}{1 + 0.0426 × 20} = frac{4.22}{1 + 0.0426 T_2})
∴ Rise in temperature = T₂ – T₁
= 30.16 – 20
= 10.16°C.

9. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 50 μF is?
A. 1.57 A
B. 1.87 A
C. 1.67 A
D. 2.83 A

Answer: C
Clarification: As the capacitors are in parallel, then the voltage V is given by,
V = (frac{1}{C_1} int I_1 ,dt )
∴ I₁ = C₁ (frac{dV}{dt})
That is, (frac{I_1}{I_2} = frac{C_1}{C_2} = frac{50}{100} = frac{1}{2}) ………………….. (1)
Also, I₁ + I₂ = 5 A ………………………….. (2)
Solving (1) and (2), we get, I₁ = 1.67 A.

10. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 100 μF is?
A. 2.33 A
B. 3.33 A
C. 1.33 A
D. 4.33 A

Answer: B
Clarification: As the capacitors are in parallel, then the voltage V is given by,
V = (frac{1}{C_2} int I_2 ,dt )
∴ I₂ = C₂ (frac{dV}{dt})
That is, (frac{I_1}{I_2} = frac{C_1}{C_2} = frac{50}{100} = frac{1}{2}) ………………….. (1)
Also, I₁ + I₂ = 5 A ………………………….. (2)
Solving (1) and (2), we get, I₂ = 3.33 A.

11. In the circuit given below, the resonant frequency is _______________

A. (frac{1}{2sqrt{2} π}) Hz
B. (frac{1}{2π}) Hz
C. (frac{1}{4π}) Hz
D. (frac{1}{sqrt{2}2π}) Hz

Answer: C
Clarification: I_EQ = L₁ + L₂ + 2M
L_EQ = 1 + 2 + 2 × (frac{1}{2}) = 4 H
∴ F_O = (frac{1}{2πsqrt{LC}} )
= (frac{1}{2πsqrt{4 × 1}} )
= (frac{1}{4π}) Hz.

12. In a series resonant circuit, V_C = 300 V, V_L = 300 V and V_R = 100 V. What is the value of the source voltage?
A. Zero
B. 100 V
C. 350 V
D. 200 V

Answer: B
Clarification: As V_C and V_L are equal, then X_C is equal to X_L and both the voltages are then cancelled out.
That is V_S = V_R
∴ V_S = 100 V.

13. For the circuit given below, what is the value of the Q factor for the inductor?

A. 4.74
B. 4.472
C. 4.358
D. 4.853

Answer: C
Clarification: Q_IN = (sqrt{frac{L}{CR^2} – 1})
= (sqrt{frac{1}{2 × 5^2 × 10^{-3} – 1}})
= (sqrt{frac{1}{50 × 10^{-3} – 1}})
= (sqrt{19}) = 4.358.

14. In the circuit given below, the value of the voltage source E is _______________

A. -65 V
B. 40 V
C. -60 V
D. 65 V

15. In the circuit given below, bulb X uses 48 W when lit, bulb Y uses 22 W when lit and bulb Z uses 14.4 W when lit. The number of additional bulbs in parallel to this circuit, that would be required to below the fuse is _______________

A. 4
B. 5
C. 6
D. 7

Answer: A
Clarification: I_X = (frac{48}{12}) = 4 A
I_Y = (frac{22}{12}) = 1.8 A
I_Z = (frac{14.4}{12}) = 1.2 A
Current required to below the fuse = 20 A
∴ Additional bulbs must draw current = 20 – (4 + 18 + 1.2)
= 20 – 7 = 13
∴ Number of additional bulbs required = (frac{13}{3}) = 4.33
So, 4 additional bulbs are required.