# Basic Electrical Engineering MCQs Part 28

Basic Electrical Engineering MCQs Part 28

## In gases the flow of current is due to:

In gases the flow of current is due to:

1. Positive ions only
2. Electrons only
3. Electrons and positive ions
4. Electrons, positive ions and negative ions

Correct answer: 4. Electrons, positive ions and negative ions

## The temperature coefficient of resistance of certain wire is 0.00150 per °C. At 300 K, its resistance is 1 Ω. The resistance of wire will be 2 Ω at what temp value

The temperature coefficient of resistance of certain wire is 0.00150 per °C. At 300 K, its resistance is 1 Ω. The resistance of wire will be 2 Ω at what temp value:

1. 585.53 K
2. 966.66 K
3. 1052.33 K
4. 2200 K

Explanation:

The resistance of wire at 300 K or 27° C is 1 Ω. (Since Celsius = Kelvin – 273.15)

Let’s assume R1 = 1 Ω

Let’s assume at t° C the resistance of wire is 2 Ω

From formula:

R2 = R1 [ 1 + α (t – 27) ]

Where α = Temperature coefficient of resistance

OR

2 = 1 [ 1 + 0.0015 (t – 27)]

Solving above:

t = 693.66° C

Temp in Kelvin

t = 693.66 + 273 = 966.66 K

## The total charge entering terminal of an element is given by equation

The total charge entering terminal of an element is given by equation:

q = (9t2 – 15t)mC

The current at t = 3s is:

1. 3 mA
2. 6 mA
3. 12 mA
4. 18 mA

Explanation:

i = dq/dt = d/dt (q) = d/dt (9t2 – 15t) mA = (9t – 15) mA

At t = 3s

i = 9(3) – 15 = 12 mA

## No current flows between two charged bodies when they have same

No current flows between two charged bodies when they have same

1. Charge
2. Capacity
3. Potential
4. None of the above

## The rating of the fuse wire is always expressed in

The rating of the fuse wire is always expressed in:

1. Ampere
2. Ampere-hours
3. Volts
4. kWH