Question

Determine the length and conductivity of a wire with diameter 4.2 mm constructed from an alloy of resistivity 3.2 x 10^-8 ohms-meter, if its resistance is 0.8 0hms. [Take pie = 22/7]. (b) A battery of emf 12.0 V and internal resistance 0.5 is connected to 1.5 and 4.0 series resistors. Calculate the terminal voltage of the battery. (c) A lamp marked 100 W, 250 V is lit for 10 hours. if it operates normally and 1 kWh of electrical energy cost ₦2, what is the cost of lighting the lamp

Answer 1

1) L = 3.465 × 10^2; 31250000 S/m

2) terminal pd = 11V

3) ₦2

Explanation:

1)

diameter of wire (d) = 4.2mm = 4.2 × 10^-3m

resistivity 3.2 x 10^-8 ohms-meter

Resistance(R) = 0.8 ohms

Area of wire (A)= πd^2 / 4

A =[ ( 22/7) * (4.2×10^-3)^2] / 4

A = [22/7 * 17.64×10^-6] / 4

A = 13.86×10^-6m^2

Recall:

Resistivity = (R ×A) / Length

Length = (R × A) / resistivity

Length(L) = [(0.8)(13.86×10^-6)] / 3.2 x 10^-8

L = (11.088 × 10^-6) / 3.2 x 10^-8

L = 3.465 × 10^2

Conductivity = 1/resistivity

Conductivity = 1/3.2 x 10^-8

Conductivity = 31250000 S/m

2)

Series resistor = 1.5 and 4.0 ohms

e.m.f (E) = 12V, internal resistance 'r' = 0.5

Recall :

I = E / (R + r)

Where I = current

R = circuit resistance

R = Rs = 1.5 + 4.0 (series connection) = 5.5

Recall : I = V/R

Where V = voltage ( terminal pd)

Therefore,

V/R = E/(R+r)

V/5.5 = 12/(5.5 + 0.5)

V/5.5 = 12/6

V/5.5 = 2

V = 5.5 × 2

V = 11V

3) power = 100W, t =10 hours

Energy consumed = power × time

Energy consumed = 100W × 10 hrs = 1000Whr

Recall: 1 KW = 1000 W

Therefore, energy consumed = 1KW

Cost of energy per KW = ₦2

Therefore, cost of lightning the lamp = ₦2 × 1KW = $2