# G25_hw by ajizai

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```									Current

G25.1. A current of 1.50 A flos in a wire. How many electrons are flowing past any point
in the wire per second? The charge on one electron is 1.6       C

Solution:

G25.4. What is the resistance of a toaster if 110 V produces a current of 4.2 A?

Solution:

G25.10. A bird stands on a dc electric transmission line carrying 2500 A. The line has
resistance per meter and the bird's feet are 4.0 cm apart. What potential
difference does the bird feel?

Solution:
The resistance of 1 m wire is                .
The resistance of 0.04 m wire is then
V

G25.15. A certain copper wire has a resistance of 10.0 . At what point along the length
must the wire be cut so that the resistance of one piece is 5.0 times the resistance of the
other? What is the resistance of each piece?

Solution:
total resistance
resistance of piece 1
resistance of piece 2

The wire must be cut so that one piece is 5 times longer than the other

G25.21. A length of wire is cut in half and the two lengths are wrapped together side by
side to make a thicker wire. How does the resistance of this new combination compare to
the resistance of the original wire?

Solution
Initially we have :
After the cutting:

The resistance is less by a factor of 4.

G25.27. What is the maximum power consumption of a 9.0 V portable cassette player
that draws a maximum of 350 mA of current?

Solution:
! "#       ! "# ! "#                A     V           W

G25.31. (a) What is the resistance and current through a 60-W light bulb if it is
connected to its proper source voltage of 120 V? (b) Repeat for a 150-W bulb.

Solution:
(a) "Proper" means that the light bulb is connected to the maximum voltage possible.
That is the voltage at which the power consumed will be equal to the power on the label
60 W.
A
(b)
A

G25.34. At \$0.110 per KWh, what does it cost to leave a 60-W porch light on day and
night for a year?

Solution:
1 KWh is the energy consumed if we run an appliance with power 1 KW                 W for 1
hour. Note, usually, we calculate the energy in J, but this time we need to calculate it in
other units - the units that the electric company uses!

Power: 60 W is equal to 60 10 KW      KW
Time: 1 year        days)    hours)    hours
Energy (in KWh) (Power in KW) (number of hours in year)
KWh
Cost:                    \$   a year

G25.47. Calculate the peak voltage across, and peak current through, and 1800-W arc
welder connected to a 450-V ac line.

Solution:
\$! % \$! %

\$! %         V                    \$! %            V
\$! %      \$! %
A
\$! %      A

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