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An aluminum wire with a cross-sectional area of
4 x 10^{- 6} m ^{2} carries a current of
5 A. Find the drift speed of the electrons in the
wire, assuming each aluminum atom contributes on electron.
The density of aluminum is 2.7 g/cm ^{3}.
Solution:
For this we will use
I = nqv_{d}A.
To find the number of charges per unit volume, n , we find from the periodic table that the atomic mass of aluminum is 26.98 g/mol. Since 1 mol of a substance contains Avagadro's number N_{A} = 6.02 x 10^{23} of particles, we findn = 6.02 x 10^{23} x x x ^{3} = 6.02 x 10^{28}
We then findv_{d} = = = 1.3 x 10^{- 4} m / s .
Thus, the average drift speed is about 0.13 mm/s.
Suppose one wants to make a 0.5 resistor
out of 1 g of Copper. If the resistor is a uniform
cylinder, what is the diameter and length required?
Solution:
We will relate the resistance R to the resistivity by
R = ,
where for Copper = 1.7 x 10^{- 8} m . Since Copper has a density of 8.95 g/cm ^{3}, we know that the volume V of Copper present isV = x ^{3} = 1.12 x 10^{- 7} m ^{3}.
Since V = LA , we then findR = = L = = = = 1.8 m .
We can then also find the diameter d required asA = = d^{ 2} d = = = 2.8 x 10^{- 4} m .
Thus, the wire must have a length of 1.8 m and a diameter of 2.8 x 10^{- 4} m.
A small motor draws a current of 2 A from a 120 V line.
Assuming 100% efficiency, what is the cost of operating
the motor for 10 hours if the power company charges
$0.050/kWh?
Solution:
The motor has a power output of
P = VI = 2 x 120 = 240 W .
Operating it for 10 hours will consume an amount of energyE = Pt = 240 W x 10 h x = 2.4 kWh .
It will thus cost 2.4 x $0.05=$.12 to operate the motor.
Find the current through R_{4} in the circuit in Fig. 17.9
if V = 30 V,
R_{1} = 12 ,
R_{2} = 18 ,
R_{3} = 9 , and
R_{4} = 6 .
Solution:
We first reduce the four resistors to one equivalent resistance.
The resistors R_{1} , R_{2} , and R_{3} are in parallel;
they have an equivalent resistance
= + + = + + R_{p} = 3 .
This equivalent resistance is then in series with R_{1} ; the equivalent resistance of these two resistors in series isR = R_{1} + R_{p} = 12 + 3 = 15 .
Thus, the equivalent resistance of the four resistors is 15 . The current flowing through this equivalent circuit is thusI = = = 2 A ,
from which we deduce the potential difference across the equivalent resistance R_{p} = 3 asV_{p} = IR_{p} = 2 x 3 = 6 V .
This potential difference is the same across R_{2} , R_{3} , and R_{4} , and in particular we have through R_{4}I_{4}R_{4} = 6 I_{4} = = 1 A .
Thus, the current through R_{4} is 1 A.
Find the power lost in the 50 resistor in the
circuit in Fig. 17.10.
Solution:
We begin by labelling the unknown currents as indicated.
Applying Kirchhoff's junction rule, we find
I_{1} = I_{2} + I_{3}.
We next use Kirchhoff's loop rule; for the top loop, going clockwise, we find30I_{3} + 50I_{3} - 40I_{2} = 0 I_{3} = I_{2},
while for the bottom loop, also going clockwise,40I_{2} - 20 + 10I_{1} = 0 I_{1} = 2 - 4I_{2}.
Substituting these last two relations into the first yields2 - 4I_{2} = I_{2} + I_{2} I_{2} = A ,
from which followsI_{3} = I_{2} = A , | |||
I_{1} = 2 - 4I_{2} = A . |
As a check on our algebra, we can verify that by going around the outer loop
30I_{3} + 50I_{3} - 20 + 10I_{1} = 0,
as it should. We then know the current through the 50 resistor to be I_{3} = A, from which we find the power lost asP = I^{ 2}R = ^{2} x 50 = 1.65 W .
Thus, the 50 resistor dissipates 1.65 W of power.
Find the currents through all three resistors in the
circuit in Fig. 17.11.
Solution:
We begin by labeling the unknown currents as indicated.
Applying Kirchhoff's junction rule, we find
I_{1} + I_{3} = I_{2}.
We next use Kirchhoff's loop rule; for the top loop, going clockwise, we find20 - 100I_{2} - 20I_{3} = 0 I_{3} = 1 - 5I_{2},
while for the bottom loop, also going clockwise,100I_{2} - 60 + 10I_{1} = 0 I_{1} = 6 - 10I_{2}.
Substituting these last two relations into the first yields(1 - 5I_{2}) + (6 - 10I_{2}) = I_{2} I_{2} = A ,
from which followsI_{3} = 1 - 5I_{2} = - A , | |||
I_{1} = 6 - 10I_{2} = A . |
As a check on our algebra, we can verify that by going around the outer loop
- 20I_{3} + 20 - 60 + 10I_{1} = 0,
as it should. Note that for this check we must explicitly use the negative value found for I_{3} .www-admin@theory.uwinnipeg.ca