**RVUNL Recruitment 2021:** The RVUNL exam date is released for a total of 1075 vacancies.

There are two parts for the RVUNL test, Part A and Part B.

Part A will have 60% weightage while Part B has 40% weightage out of the total marks in the online written exam.

**RVUNL’21 EE: Daily Practices Quiz 24-Aug-2021**

**Each question carries 2 marks.**

**Negative marking: 0.5 mark**

**Total Questions: 06**

**Time: 08 min.**

Q1. A 6-pole, 50 Hz, 1-phase induction motor runs at a speed of 900 r.p.m. The frequency of currents in the cage rotor will be

(a) 5 Hz, 50 Hz

(b) 5 Hz, 55 Hz

(c) 5 Hz, 95 Hz

(d) 55 Hz, 95 Hz

(e) None of the above

Q2. A parallel RLC circuit resonates at a frequency of 2 MHz. At a frequency 2.2 MHz, the nature of circuit impedance will be

(a) Resistive

(b) Inductive

(c) Capacitive

(d) Can’t say

(e) None of the above

Q3. If the Op-Amp in the figure is ideal, the output voltage Vo will be equal to

(a) 1 V

(b) 6 V

(c) 14 V

(d) 17 V

(e) 13 V

Q4. A 40 KVA transformer has a core loss of 400 W and full-load copper loss of 800 W. The fraction of rated load at maximum efficiency is

(a) 50 %

(b) 62.3 %

(c) 70.7 %

(d) 100 %

(e) 86.6 %

Q5. The inductance of a high Q inductor can be measured using a

(a) Schering bridge

(b) Wein bridge

(c) Maxwell bridge

(d) Hay’s bridge

(e) All of the above

Q6. A series R-C circuit having R = 5 Ω and C = 100 μF has a time constant of

(a) 20 ×10^(-6) sec.

(b) 5 × 10^(-4) sec.

(c) 0.005 sec.

(d) 500 sec.

(e) 250 sec.

SOLUTIONS

S1. Ans.(c)

Sol. Synchronous speed for forward rotating flux, Ns = 120 f/P = 120 × 50/6 = 1000 r.p.m.

Therefore, slip for forward rotating flux is

s_f=s=(N_s-N)/N_s =(1000-900)/1000=0.1

∴ Rotor current frequency for sf is f’ = sf × f = 0.1 × 50 = 5 Hz.

Slip for backward rotating flux s_b =(2 – s) =(2 – 0.1) = 1.9. Therefore, rotor current frequency, f” = (2 – s) × f = 1.9 × 50 = 95 Hz.

S2. Ans.(c)

Sol. As given frequency is greater than resonant frequency for parallel RLC Circuit. So, nature will be capacitive and vice-versa for series RLC circuit.

S3. Ans.(b)

Sol. Given:

Op-amp is ideal.

feedback resistor (R_f) = 5 kΩ

R_1 = 1 kΩ

Inverting terminal voltage (V_1) = 2 V

non-Inverting terminal voltage (V_2) = 3 V

As input voltage is applied to both the inverting and non-inverting terminal, so the output voltage will be due to both.

We will calculate V_x using voltage divider rule,

V_x = 8/(1 + 8) × V_2 = 8/9 × 3

∴ V_x = 8/3 V

So, V_o= voltage due to inverting+ voltage due to non-inverting terminals

Now, V_o = -R_f/R_1 × V_1 + (1 + R_f/R_1 ) × V_x

⇒ V_o = -5/1 × 2 + (1 + 5/1) × 8/3

⇒ V_o = -10 + 6 × 8/3

⇒ V_o = -10 + 16

So, V_o = 6 V

S4. Ans.(c)

Sol. η=√(P_C/P_cufl )=√(400/800) ×100=70.7 %

S5. Ans.(d)

Sol. Bridges used for measurement of inductance:

Anderson bridge (Q<1).

Maxwell bridge (1<Q<10).

Hay bridge (Q>10).

S6. Ans.(b)

Sol. Time constant is

τ = RC = 5 × 100 × 10^(-6)= 5 × 10^(-4) sec.