# Basic Electrical Engineering MCQ Part 2

Basic Electrical Engineering MCQ Part 2

## Point Source Example

The figure below displays an electric circuit drawn using point source:

The voltmeter connected across 15 kohm resistor will read:

1. 3 V
2. 5 V
3. 10 V
4. 15 V

Explanation:

The circuit redrawn to normal circuit as follows:

Total input: Vin = 25 V

Using voltage divider rule:

V (15k) = (15 kohm)/(10 kohm + 15 kohm) * 25 V = 15 V

## Resistance calculation for 5 ohms wire whose length is increased four times and area by two times

A certain length of wire has resistance 5 ohms. What is the resistance of wire that is made of the same material and is four times longer and has twice the cross-sectional area:

1. 2.5 ohms
2. 5 ohms
3. 7.5 ohms
4. 10 ohms

Explanation: Suppose R1 and R2 are values of resistances and R1 = 5 ohms.

From Formula R1 = ρ*l1/a1 and R2 = ρ*l2/a2

R2/R1 = (l2/l1) * (a1/a2)

R2/R1 = (4.l/l) * (a/2a) = (4.l/l) * (a/2a) = 4/2 = 2

R2 = 2.R1 = 2. 5 ohms = 10 ohms

## For constant resistor by increasing the voltage the current will

For constant resistor, by increasing the voltage, the current will:

1. Increase
2. Decrease
3. Remain the same

## 3 mA current flowing through 1 kohm resistance will produce a voltage drop of

3 mA current flowing through 1 kohm resistance will produce a voltage drop of:

1. 0.33 V
2. 1 V
3. 3 V
4. 3 mV

## Norton theorem is ___________ Thevenin theorem

Norton’s theorem is ___________ Thevenin’s theorem:

1. Identical to
2. Converse of