1. Applied Electricity Two
Formula for Resistance
R =ρl
A
ρ = resistivity of the material
l = is the length
A=is the cross sectional area

2. Applied Electricity Two
Resistance would be proportional to the length
and an increase in the length will increase the resistance by
a similar amount
and is
Inversely proportional to the Cross sectional area (A)

3. Applied Electricity Two
An increase in cross sectional area would decrease the
resistance but by a factor calculated by
multiplying the increase by itself
i.e.
by doubling the diameter (or radius) we have increased the
area
by 4 (2 x 2 = 4)
if we had increased the diameter (or radius) by 3
we would increase the area by 9
(3 x 3 = 9)

4. Applied Electricity Two
Rh = Rc[ 1 + α(t2-t1)]
t1 = temp of Rc
t2 = temp of Rh
α = temperature coefficent
Rc = starting resistance
Rh = final resistance

5. Applied Electricity Two
Rh = Rc[ 1 + α(t2-t1)]
t1 = temp of Rc
t2 = temp of Rh
α = temperature coefficent
Rc = starting resistance
Rh = final resistance

6. Applied Electricity Two
The field windings of generator have a resistance of
125 ohms
at a temperature of 20 deg C.
What will be the resistance of the windings when the
machine temperature rises on full load to 60 deg C.
Rh = Rc[ 1 + α(t2-t1)
t1 = temp of 20c
t2 = temp of 60c
α = .004 (copper windings)
Rc = 125 ohms
Rh = final resistance

7. Applied Electricity Two
R (at 60C) = 125 x ( 1 + (.004(60 - 20)))
= 145 ohms
Remember to use brackets and do the
addition separate to the multiplication.

8. Applied Electricity Two
R2 = R1 ( 234.5 + t2)
( 234.5 + t1)
t1 = temp of R1
t2 = temp of R2
R1 = starting resistance
R2 = final resistance

9. Applied Electricity Two
The resistance of a coil of copper wire is 34 ohms
at 15 deg C. What would be its resistance at 70
deg C?
R2 = R1 ( 234.5 + t2)
( 234.5 + t1)
t1 = temp of 15
t2 = temp of 70
34 = starting resistance
R2 = final resistance

10. Applied Electricity Two
R2 = 34 ( 234.5 + 70)
( 234.5 + 15)
= 41.49 ohms