Why Are My Calculations for Electrical Resistance and Current Incorrect?

In summary, the 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, 10 A, and has a 1.2 OHL resistance, then the current it draws from the car's 12 V battery is 5859 C.
  • #1
jrd007
159
0
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?
 
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  • #3
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.

[tex] R = \frac {\rho L} A [/tex]

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

[tex] R_T = R_C [/tex]

so

[tex] \frac {\rho_T L_T} {A_T} = {\frac {\rho_C L_C} {A_C} [/tex]

further you are given that
[tex] L_T = L_C [/tex]

so now we have:
[tex] \frac {\rho_T } {A_T} = \frac {\rho_C } {A_C} [/tex]

[tex] A_T = \frac {\rho_C } { \rho_T A_C } [/tex]

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.
 
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  • #4
That does stil not work. Area of copper will be 4.91 mm^2. If you then use.

[tex] A_T = \frac {\rho_C } { \rho_T A_C } [/tex]

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
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?
 

FAQ: Why Are My Calculations for Electrical Resistance and Current Incorrect?

What is an electric current?

An electric current is the flow of electric charge through a conductive material. It is typically measured in units of amperes (A) and is caused by the movement of electrons.

How is electric current measured?

Electric current is measured using a device called an ammeter, which is connected in series with the circuit. The ammeter measures the flow of charge through the circuit and displays it in units of amperes (A).

What factors affect the strength of an electric current?

The strength of an electric current is affected by the voltage applied to the circuit, the resistance of the material, and the length and cross-sectional area of the material. These factors are related by Ohm's Law, which states that current is equal to voltage divided by resistance.

What are the dangers of electric current?

Electric current can be dangerous to humans if it passes through the body. It can cause burns, muscle contractions, and even stop the heart. It is important to always handle electricity safely and never touch live wires or outlets.

How can electric current be used?

Electric current has a wide range of applications, including powering devices, heating materials, and creating magnetic fields. It is used in everyday objects such as light bulbs, refrigerators, and computers, as well as in more complex systems like power plants and electric motors.

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