11.3 Part 1 Solution
11.3 PART 1 Solution.
Equivalent Circuits
Equations
R
Variables

R = resistance (ohms, Ω)
A
  length of resistor (m)
Rs   Ri
A = cross sectional area of
resistor (m2)
i
1
1

Rp
i Ri
ρ = material resistivity
(Ω·m)
Resistors
1. The instructions for an electric device suggest that a 20-gauge extension cord can be sued for a distance p
to 35 meters, but a 16-guage cord should be used for longer distances, to keep the resistance of the wire as
small as possible. The cross-sectional area of 20-gauge wire is 5.2 x 10-7 m2, while that of 16-gauge wire
is 13 x 10-7 m2. Determine the resistance of 35 m of 20-gauge wire and the resistance of 75 meters of 16gauge wire.
Resistance of 35 m of 20-guage Wire
L = 35 m
A = 5.2 x 10-7 m2
ρ = 1.72 x 10-8 (Ω·m) (assume copper since most extension cords are made of copper)
R=?

(1.72  10 8 m)(35m)
R

 1.15
A
5.2  10 7 m 2
Resistance of 75 m of 16-guage Wire
L = 75 m
A = 5.2 x 10-7 m2
ρ = 1.72 x 10-8 (Ω·m) (assume copper since most extension cords are made of copper)
R=?

(1.72  108 m)(75m)
R

 .99
A
13  107 m 2
2. High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some
of these lines has a cross-sectional area of 4.9 x 10-4 m2. What is the resistance of ten kilometers of this
wire?
L = 10,000 m
A = 4.9 x 10-4 m2 (
ρ = 2.82 x 10-8 (Ω·m)
R=?

(2.82  10 7 m)(10,000m)
R

 5.75
A
4.9  10 4 m 2
3. Two wires have the same length and same resistance. One is made from aluminum and the other from
copper. Obtain the ratio of the cross-sectional area of the aluminum wire to the copper wire.
LAl = LCu
RAl = RCu
R
A

A

R
solve for Area
create the ratio
 Al LAl
 Al LAl
AAl
RAl
RAl



 Al
ACu Cu LCu Cu LCu Cu
RCu
RCu
4. A wire connecting two points is made of copper with a cross-sectional area A. (The wire takes direct line
between the points) By what factor must the area be increased in order to reduce the resistance in the wire
by a factor of 2? 4?
R1 

A
if R is reduce to R/2 what must A be.
R1= 2R2 create a ratio


1
R
A
A
A
A
2  1  1  1  1  2 so 2A1 = A2 so A2 is twice as large as A1.
1
R2  
A1
A2
A2
A2
By the same logic, to reduce the resistance by a factor of 4 A2 must be 4 times greater.
5. A wire connecting two points is made of copper with a cross-sectional area A. (The wire takes direct line
between the points) Assuming the cross-section is a circle by what factor must the radius be increased in
order to reduce the resistance in the wire by a factor of 2? 4?
See previous question. Substitute the equation for the area of a circle in for A.


1
R
 r
 r
r2 r2
2 1 

 1  22


1 r1
R2
2
2
  r2   r2
r22
2
1
2
1
2r12  r22 take the square root of both sides to get
2r1  r2
To reduce the resistance by a factor of two the radius must be increased by a factor of √2. The radius must
be increased by the square root of the reduction factor that is desired. To reduce the factor by 4 the radius
must be increased by a factor of 2.
6. A wire has a resistance of 21.0 ohms. It is melted down, and from the same volume of metal a new wire is
made that is three times longer than the original wire. What is the resistance of the new wire?
R1 = 21 ohms
V = constant = (L1) x (A1)
L2 = 3L1
Find A2
(L2) x (A2)= (L1) x (A1) replace L2 with 3L1
(3L1) x (A2) = (L1) x (A1) solve for A2
A2 
A1
3
plug in the values for A2 and L2
R2 
 3L1
A1
3
or
R2 L1

 R1
9
A1
Adding Resistors in Series and Parallel
7. Calculate the equivalent resistor for the two resistors in series.
Rs   Ri  4  8  12
i
8. Calculate the equivalent resistor for the two resistors in parallel.
( R1  R2 ) PARALLEL 
6585  36.8
R1R2

R1  R2 65  85
9. Calculate the equivalent resistor for resistors in parallel and series.
A
B
Add the three resistors on the
end of the arrangement.
These are in series
6Ω + 5Ω + 3 Ω = 14Ω
Add the two in parallel and
replace it with the sum,
which is 5.1 Ω
Add the two in series and
replace it with the sum,
which is 9.1 Ω
Add the two in parallel and
replace it with the sum,
which is 4.76 Ω
Add the two in series and
replace it with the sum,
which is 6.76 Ω
The Total Resistance for this
arrangement of resistors is
6.76 Ω
4.2 Ω
10. Create a diagram for the toaster, iron, microwave and circuit breaker shown.
Some websites worth checking out for further study or reference.
http://engr.calvin.edu/courses/engr204/examples/equRes/index.html - Equivalent Resistance
www.physics.brocku.ca/.../1p93/ElectricCircuits/ - Question *6