Basic Electrical Engineering MCQs Part 4

Contents

- 1 If a piece of wire is melted and recast to half of its original length, the new resistance is
- 2 The resistance of human body is about:
- 3 A current flow of 10 A for 1 hour will transfer how much charge
- 4 The thermal speed of electrons is of the order of
- 5 Due to drop of current by 3%, the percentage by which incandescence of lamp decreases is

## If a piece of wire is melted and recast to half of its original length, the new resistance is

If a piece of wire is melted and recast to half of its original length, the new resistance is:

- R
- R/2
- R/4
- 2R

Correct answer: 3. R/4

Explanation:

Suppose R = ρl/A is original resistance

And R’ = ρl’/A’ is new resistance

Since volume of wire remains same

lA = l’A’

A/A’ = l’/l

l’ = 0.5l

Now A/A’ = l’/l = 0.5 = ½

R’R = l’/l * A/A’ = ½ * ½ = ¼

R’ = R/4

## The resistance of human body is about:

The resistance of human body is about:

- 1 Ω
- 10 Ω
- 100 Ω
- 1000 Ω

Correct answer: 4. 1000 Ω

## A current flow of 10 A for 1 hour will transfer how much charge

A current flow of 10 A for 1 hour will transfer how much charge:

- 3.8 * 10^3 C
- 2.2 * 10^3 C
- 1.5 * 10^2 C
- 3.6 * 10^4 C

Correct answer: 4. 3.6 * 10^4 C

Solution: q = It = 10 A * 3600 s = 36000 As = 3.6 * 10^4 C

## The thermal speed of electrons is of the order of

The thermal speed of electrons is of the order of

- 10^-3 m/s
- 10^6 m/s
- 10^8 m/s
- None of these

Correct answer: 2. 10^6 m/s

## Due to drop of current by 3%, the percentage by which incandescence of lamp decreases is

Due to drop of current by 3%, the percentage by which incandescence of lamp decreases is:

- 3%
- 6%
- 9%
- 12%

Correct answer: 2. 6%

Explanation:

P = I²R or ΔP = 2I(ΔI)R

ΔP/P = 2ΔI/I = 2 * 3% = 6%