**Quiz: Electrical EngineeringExam: DFCCILTopic: Miscellaneous**

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

**Q1.** A transformer having a turn’s ratio of 1:5 and a resistance of 500 Ω is connected across the secondary terminals. What is the equivalent resistance for the current flowing in the primary?

(a) 10 Ω

(b) 50 Ω

(c) 20 Ω

(d) 100 Ω

**Q2. **The number of secondary turns of a coil in a transformer is 4 times the number of turns in primary coil. If the voltage in primary coil is 200 V, the secondary voltage will be?

(a) 200 V

(b) 800 V

(c) 50 V

(d) 2000 V

**Q3.** The basic function of a transformer is to change

(a) The level of the voltage

(b) The power level

(c) The power factor

(d) The frequency

**Q4. **Shunt conductance in the power transmission is due to

(a) Leakage over the insulator

(b) Leakage over the conductors

(c) Leakage over the poles

(d) Leakage between ground and conductor

**Q5.** The gang capacitor is a variable capacitor in which capacitance is varied by changing the

(a) Dielectric

(b) Number of plates

(c) Plate area

(d) Distance between the points

**Q6. **The type of wattmeter commonly used for the measurement if Power in an AC circuit is

(a) Moving Iron type

(b) Thermocouple Type

(c) Rectifier Type

(d) Dynamometer Type

**Q7.** In the given circuit with the shown ideal 5V DC source, the magnitude of the total current drawn from the source at a steady-state is

(a) 2.5 A

(b) 7.5 A

(c) 5 A

(d) 10 A

Q8. Consider the function F(s) = ω/(s^2+ω²)

Where F(s) = Laplace transform of f(t). The final value of f(t) is equal to

(a) infinite

(b) zero

(c) finite constant

(d) a value in between – 1 and + 1

**Q9. **The power gain in decibels can be expressed as

(a) 20 log10(PO/Pi )

(b) 10 log10(VO/Vi )

(c) 10 log10(PO/Pi )

(d) 20 log10(PO/Vi )

**Q10.** Negative feedback used in amplifier results in-

(a) More gain, more bandwidth

(b) More gain, less bandwidth

(c) Less gain, more bandwidth

(d) Less gain, less bandwidth

**SOLUTIONS**

SOLUTIONS

**S1. Ans.(c)Sol.** ACQ: N_1/N_2 =1/5

∴K=5

So, equivalent resistance refereed to primary side=R2/K^2 =500/(5)^2 =20 Ω.

**S2. Ans.(b)Sol. **V2/V1 =N2/N1

ACQ: N2=4N1

∴V2/200=(4N1)/N1

⇒V2=200×4=800 V.

**S3. Ans.(a)Sol.** The basic function of a transformer is to change the level of the voltage.

**S4. Ans.(a)Sol. **Shunt conductance in the power transmission is due to Leakage over the insulator.

Shunt conductance is due to leakage currents flowing across insulators and air.

For transmission line modelling, shunt conductance is not considered due to smaller value.

**S5. Ans.(c)Sol. **The gang capacitor is a variable capacitor in which capacitance is varied by changing the plate area.

```
Gang capacitors are variable capacitors.
The ganged capacitor shown above is a combination of two capacitors connected together.
```

**S6. Ans.(d)Sol.** Dynamometer Type

**S7. Ans.(d)Sol.** Under steady-state conditions, the inductor is replaced by short-circuited. Therefore, a 1 Ω resistor that is connected parallel to the inductor is neglected.

Hence total resistance of the circuit will be:

R_T = 1Ω || 1Ω =0.5 Ω

Therefore, current drawn by the 5 V source is

∴I = V/R = 5/0.5

⇒I = 10 A

**S8. Ans. (d)Sol.** F(s) = ω/(s^2+ω²)

Taking inverse Laplace:

f(t) = sin ωt

Therefore, the final value for sinωt is equals to a value between -1 and +1.

**S9. Ans.(c)Sol.**

`Power gain =Po/Pi`

in decibel (dB):

dB = 10 log10(Po/Pi )

=10 log10(Vo/Vi )^2

=20 log10(Vo/Vi )

**S10. Ans.(c)Sol.** Effects of Negative feedback:

In negative feedback, the feedback energy (voltage or current), is out of phase with the input signal and thus opposes it.

Negative feedback reduces gain of the amplifier. It also reduces distortion, noise and instability.

This feedback increases bandwidth and improves input and output impedances.