1)

Since the current is equal to 6.0 x 10^-5 A, this means that every second, 6.0 x 10^-5 C of charge is striking the screen. This corresponds to

6.0 x 10^-5 C/(1.6 x 10^-19 C/1 e) = 3.75 x 10 ¹⁴ e.

2)

(a) From Ohm's law, we know that V = IR or I = V/R. For this problem, V = 9.0 V. The resistance is given by R = r L/A, where L = 34.5 m and A = p (2.5 x 10 ^-4 m)² = 2.0 x 10^-7 m². At 20 ^o C, the resistivity of copper is 1.7 x 10^-8 W m. Thus, the resistance is

R = (1.7 x 10^-8W m)(34.5 m)/(2.0 x 10^-7 m²) = 2.9 W .

This means that the current is given by

I = (9.0 V)/(2.9 W) = 3.1 A.

(b) If the wire is heated to 30 ^o, then the resistivity changes to

r = 1.7 x 10^-8 W m [1 + (3.9 x 10^-3/C)(30 - 20)^oC] = 1.7 x 10^-8W m[1 + .039] = 1.8 x 10 ^-8 W m.

This gives a resistance

R = (1.8 x 10^-8W m)(34.5 m)/(2.0 x 10^-7 m²) = 3.1 W .

Thus, the current is given by

I = 9.0 V/3.1 W = 2.9 A.

3)

The power dissipated in a resistor is given by P = I² R. For the line, we have

R = (.31 W/km)(160 km) = 49.6 W.

With a 1000 A current, we get

P = (1000 A)²(49.6 W ) = 4.96 x 10⁷ W.

4)

Since the television is rated at 90 W and is left on for 21 hours, we know that the amount of energy consumed is

E = (90 W)(21 hours) = (.09 kW)(21 hr) = 1.89 kWh.

If energy costs $.07/kWh, then the cost is

(1.89 kWh)($.07/kWh) = $.13,

a far cry from the hundreds of dollars you would have to pay to sit in the stands.

If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu