Quiz: Electrical Engineering

Exam: RVUNL-JE

Topic: Miscellaneous

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

Q1. In a series RLC circuit, the phase difference between the voltage across the inductor and the current in the circuit is?

(a) 0⁰

(b) 90⁰

(c) 180⁰

(d) 360⁰

Q2. The current in the inductor lags the voltage in a series RLC circuit **_** resonant frequency.

(a) Above

(b) Below

(c) Equal to

(d) Depends on the circuit

Q3. The current in the capacitor leads the voltage in a series RLC circuit **_** resonant frequency.

(a) Above

(b) Below

(c) Equal to

(d) Depends on the circuit

Q4. A PN junction diode’s dynamic conductance is directly proportional to

(a) the applied voltage

(b) the temperature

(c) the thermal voltage

(d) its current

Q5. Insulators have **__** temperature coefficient of resistance.

(a) negative

(b) positive

(c) zero

(d) none of these

Q6. A 200 W, 200 V bulb and a 100 W, 200V bulb are connected in series and the voltage of 400 V is applied across the series connected bulbs. Under this condition

(a) 100 W bulb will be brighter than 200 W bulb

(b) 200 W bulb will be brighter than 100 W bulb

(c) Both the bulbs will have equal brightness

(d) Both the bulbs will be darker than when they are connected across rated voltage

Q7. An ac voltage source of 250 V(rms) supplies active power of 300 watts and reactive power of 400 VAR. Find the value of current(rms) drawn by the source.

(a) 10 A

(b) 6 A

(c) 4 A

(d) 2 A

Q8. Which one of the following correct?

(a) Synchronous motor is supplied with D.C. voltage in the armature winding

(b) Synchronous motor is supplied with ac voltage in the field winding

(c) Synchronous motor is supplied with rectified voltage in the armature winding

(d) Synchronous motor is supplied with dc voltage in the filed winding

Q9. In order to eliminate the fifth harmonic voltage from the phase voltage of an alternator, the coil should be pitched by:

(a) 72°

(b) 36°

(c) 15°

(d) 18°

Q10. Differential relays are used for protection of

(a) Transformers

(b) Feeders

(c) Alternators

(d) All of the above

SOLUTIONS

S1. Ans.(b)

Sol. In a series RLC circuit, voltage across inductor lead the current by 90⁰. so, the phase difference between the voltage across the inductor and the current is 90⁰.

NOTE: For Series RLC circuit……………….

When XL > XC, the phase angle ϕ is positive. The circuit behaves as RL series circuit in which the current lags behind the applied voltage and the power factor is lagging.

When XL < XC, the phase angle ϕ is negative, and the circuit acts as a series RC circuit in which the current leads the voltage by 90 degrees. i.e., power factor is leading.

When XL = XC, the phase angle ϕ is zero, as a result, the circuit behaves like a purely resistive circuit. In this type of circuit, the current and voltage are in phase with each other. The value of the power factor is unity.

S2. Ans.(a)

Sol. The current in the inductor lags the voltage in a series RLC circuit if circuit is inductive dominant i.e., ω > ω_0. So, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency.

S3. Ans.(b)

Sol. The current in the capacitor leads the voltage in a series RLC circuit if circuit is capacitive dominant i.e., ω < ω0. So, the current in the capacitor leads the voltage in a series RLC circuit below the resonant frequency.

S4. Ans.(d)

Sol. In a PN junction diode,

r = (ηV_T)/I

Transconductance = 1/resistance

So, g_m = I/(ηV_T )

Therefore, in case of a PN junction diode’s dynamic conductance is directly proportional to its current.

S5. Ans.(a)

Sol. Insulators have Negative Temperature Coefficient of resistance because its resistance decreases with increment of temperature. On increasing temperature insulator may

start conducting that means their resistance is decreasing.

S6. Ans.(a)

Sol. For series connection, lower power rating bulb will glow more.

S7. Ans.(d)

Sol. tan ϕ =400/300=4/3

∴cos ϕ =3/5

P= V_rms I_rms cosϕ

⇒300= 250 ×I_rms×3/5

⇒I_rms= 2 A.

S8. Ans.(d)

Sol. in synchronous motor

>3-ϕ AC supply on → Stator

>DC supply on → Rotor

S9. Ans.(b)

Sol. To eliminate n^th harmonic, the displacement angle (β) = (180°)/n

According to Question,

To eliminate 5th harmonic, β = (180°)/5=36°

S10. Ans.(d)

Sol. Differential relays:

>Used against internal faults

>Used for protection of transformers, feeders, alternators, bus-bars etc.