# Electric Current problem.

1. Mar 4, 2006

### jrd007

I did all the word but my answers are not matching my books?

1) Can a 2.5 mm-diameter copper wire have the same resistence as a tungsten wire of the same length? Give numerical detials. Answer, yes, 4.6 mm

Here is what I did:
( p = resistivity )

R = p(L/A) so I set R(Cu) = R(tug) and since L is the same it cancels out.

Area = 1/2pie2^2 ---> Cu area = (.5)(3.14)(1.25mm) ---> A = .393 mm

So p(A) = p(A) ---> (1.68 x 10^-8)(.393 mm) = (5.6 x 10^-8)(A)
A = .1179 mm

therefore diameter would be... .1179 mm = 2(pie)r^2
r = .2729 mm which mean diameter = .546 mm... so what did I do incorrect?

2) A small immersion heater can be used in a car to heat a cup of water. If the heater can heat 120 mL of water from 25C to 95 C in 8.0 minutes, (a) How much current does it draw from the car's 12 V battery, (b) What is the resistence? Assume the manufacturer's clain of 60% effciency. Answers: 10 A, 1.2 Olhms

What I did...

Used specific heat equation ---> Q = mcT where I changed 120 mL of water into .120 kg ---> Q = (.120 kg)(4185 J/kg C)(95C-25C)
Q = 35154 J

Then I used the equation PE = 1/2Q(where Q is the charger)V
35151 J = (.5)(Q)(12 V)
Q = 5859 C

Finally I used I(current) = Q/T(time) ---> 5859C/480s
I = 12 A

And for Resistence I used V/I = R ---> 12 V/12 A = 1 Olhm

I am off my two units for some odd reason in my current and off by .2 in my resistence. Did I make a mistake?

2. Mar 4, 2006

### Integral

Staff Emeritus
Last edited by a moderator: Apr 22, 2017
3. Mar 4, 2006

### Integral

Staff Emeritus
After sorting through your work, I see several things wrong. What is the area of a circle? Where did you get the number for the area of the copper? Re examine your work you have make some very basic errors.

It is essential that you make a habit of working with symbols until you have isolated the parameter you need a number for. In this case you have.

$$R = \frac {\rho L} A$$

I will use a T subscript for Tungsten and a C subscript for copper.

$$R_T = R_C$$

so

$$\frac {\rho_T L_T} {A_T} = {\frac {\rho_C L_C} {A_C}$$

further you are given that
$$L_T = L_C$$

so now we have:
$$\frac {\rho_T } {A_T} = \frac {\rho_C } {A_C}$$

$$A_T = \frac {\rho_C } { \rho_T A_C }$$

Now can you complete the problem in this same manner until you have isolated the radius of the tungsten?

I have found that it is often necessary to repeat calculations multiple times. Do it until you can get the same answer several times in a row.

Last edited: Mar 4, 2006
4. Mar 5, 2006

### jrd007

That does stil not work. Area of copper will be 4.91 mm^2. If you then use.

$$A_T = \frac {\rho_C } { \rho_T A_C }$$

You get Area of T = .0611

and then if we use the equation of a circle for this area we get d = .27 mm.

5. Mar 5, 2006

### jrd007

Now I see. You actually manipulated the equation wrong. The Area of T is equal to AcPt/Pc. That gives the correct answer. Thanks for a good try though.

Does number two look okay?

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