Basics Electrical Engineering MCQs Part 20

Contents

- 1 What is value of current if one billion electrons path through a cross sectional area in 10^-3 s:
- 2 An energy source provides 50 mJ of energy for every microcoulomb of charge that flows. What is value of potential enery of source:
- 3 Which of following is the best conductor
- 4 A nichrome wire used as heater coil has resistance of 2 Ω/m. For a heater of 2 kW at 200 V, the length of wire required is
- 5 An electron is moving in cirlce of radius 4.5*10^-11 m at a frequency of 5.28*10^15 revolutions per second. The equivalent current is approximately:

## What is value of current if one billion electrons path through a cross sectional area in 10^-3 s:

What is value of current if one billion electrons path through a cross sectional area in 10^-3 s:

- 1.2 * 10^-6 A
- 1.6 * 10^-6 A
- 1.2 * 10^-7 A
- 1.6 * 10^-7 A

Correct answer: 4. 1.6 * 10^-7 A

Solution:

Current (I) = ne/t = 10^9 * 1.6 * 10^-19/10^-3 = 1.6 * 10-7 A

Where n = number of electrons; and e = charge on one electron

## An energy source provides 50 mJ of energy for every microcoulomb of charge that flows. What is value of potential enery of source:

An energy source provides 50 mJ of energy for every microcoulomb of charge that flows. What is value of potential enery of source:

- 5 V
- 15 V
- 5 kV
- 50 kV

Correct answer: 4. 50 kV

Solution: V = W/Q = 50 * 10^-3 / 1 * 10^-6 = 50 * 10^3 = 50 kV

## Which of following is the best conductor

Which of following is the best conductor:

- Copper
- Silver
- Gold
- Zinc

Correct answer: 2. Silver

## A nichrome wire used as heater coil has resistance of 2 Ω/m. For a heater of 2 kW at 200 V, the length of wire required is

A nichrome wire used as heater coil has resistance of 2 Ω/m. For a heater of 2 kW at 200 V, the length of wire required is:

- 1 m
- 2 m
- 5 m
- 10 m

Correct answer: 4. 10 m

Solution:

Resistance of heater wire (R) = V²/P = (200)²/2 * 1000 = 20 Ω

Length of heater wire = 20/2 = 10 m

## An electron is moving in cirlce of radius 4.5*10^-11 m at a frequency of 5.28*10^15 revolutions per second. The equivalent current is approximately:

An electron is moving in cirlce of radius 4.5*10^-11 m at a frequency of 5.28*10^15 revolutions per second. The equivalent current is approximately:

- 4.2 * 10^-4 A
- 2.3 * 10^-4 A
- 5.86 * 10^-4 A
- 8.45 * 10^-4 A

Correct answer: 4. 8.45 * 10^-4 A

Solution:

Current = Charge/Time period = Charge * Frequency

=1.6810^-19 * 5.28*10^15 = 8.45 * 10^-4 A