Answer:
1) L = 3.465 × 10^2; 31250000 S/m
2) terminal pd = 11V
3) ₦2
Step-by-step 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
