Resistivity & Electrical Energy

[ScoutFCM]

Just wanted to see if anyone could check my work. I only have 1 opportunity to do this. Thanks yall!

1.) Suppose that you wish to fabricate a uniform wire out of 1.00g of copper. If the wire is to have a resistance of R=.5Ω and if all of the copper is to be used, what will be (a) the length and (b) diameter of this wire?

a) L = [(1 x 10^-3)(.5)/(8.92 x 10^3 x 1.7 x 10^-8)]^1/2 = 1.82m
b) D = 2 (6.19 x 10^-8 / ∏ ) ^1/2 = 2.8 x 10^-4 m

2) The power supplied to a typical black-and-white television set is 90W when the set is connected to 120V. (a) How much electric energy does this set consume in one hour? (b) A color television set draws about 2.5A when connected to 120V. How much time is reuiqred for it to consume the same energy as the black-and-white model consumes in one hour?

a) E = 90x3600= 3.24 x 10^5 J = .09kWh
b) = R = 6.25 x 48 = 300 W 3.24 x 10^5 = 300t
t = 1080sec = 18min

 

[Chen]

I'm just going to show you how to solve the problems because I do have better things to do than trying to figure out what all those numbers in your final answers mean.

1) You have two equations here with two uknowns, the length of the wire and its cross section area (which we will later use to find the diamater).

The first equation is for the resistance of the wire, which is required to be 0.5Ω:
$$R = \rho _r \frac{L}{A^2}$$
Where $$\rho _r$$ is the resistivity of copper.

The second equation is for the mass of copper to be used, which is required to be 1g:
$$m = \rho _m V = \rho _m LA$$
Where $$\rho _m$$ is the mass density of copper.

Solving the two equations for L and A you should get:
$$A^3 = \frac{\rho _rm}{\rho _m R} = \pi (\frac{D}{2})^2$$
$$L^3 = \frac{m^2 R}{\rho _m^2\rho _r}$$

 

[Chen]

2) $$E = Pt$$ indeed.

$$E_1 = E_2$$
$$P_1t_1 = P_2t_2$$
$$t_2 = \frac{P_1}{P_2}t_1$$
You have P1 and t1, and P2 is just VI (2.5A times 120v).