Circuits Question -- Charging a car battery in 4 hours...

  • #1

Homework Statement


A constant current of 3A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + 1/2 t V, where t is in hours,

(a) how much charge is transported as a result of the charging?
(b) how much energy is expended?
(c) how much does the charging cost? Assume electricity costs 9 cents/kWh.

Homework Equations


P = IV
q = dI/dt
P = dW/dt

The Attempt at a Solution


I originally thought I would need to convert the time into seconds to figure out the charge transported when solving for charge using an integral, the answer turned out to be right. However my problem comes to part b, and indirectly, c. When I was solving for work, I took the integral of power. I thought I needed to convert the voltage equation in terms of seconds so that's what I did. My answer was quite a ways off. Any help is appreciated.
 

Answers and Replies

  • #2
gneill
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Can you show your calculation details?
 
  • #3
Can you show your calculation details?
I'll try as best as I can using a computer.
(A) q(t) = int(3 dt) from 0 - 14400 s.
q(t) = 3t| from 0 - 14400 s
q(t) = 43.2 kC

(B) V = (10 + 1/2t) x (60 min/hr) x (60 sec/min) = 36000 + 1800t
P = IV = 3(36000 + 1800t) = 5400t + 108000
W = int(5400t + 108000 dt) from 0 - 14400s
W = 5.614e11 J.

The correct answer to part B should be 475.2 kJ
 
  • #4
gneill
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Okay, your answer for part (a) looks good.

(B) V = (10 + 1/2t) x (60 min/hr) x (60 sec/min) = 36000 + 1800t
Note that the time conversion should not apply to the constant 10V term!. t is associated with only one term in the terminal voltage expression.
 
  • #5
Okay, your answer for part (a) looks good.


Note that the time conversion should not apply to the constant 10V term!. t is associated with only one term in the terminal voltage expression.

Ah I see, but where would I go from there? I feel like I'm still missing something since when using a similar form to the integral I have set up above. I still get answers that are to the 11th power of 10 versus where they should be around the 5th power of 10.
 
  • #6
gneill
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Do the integration using "4" as the upper limit for time. The result will be in V*A*hr (Volt Amp Hours):

##\int_0^4 (10 + \frac{1}{2} t) dt =~ ... VAhr##

You can do this because you're told that in the expression for the terminal voltage the time is specified in hours.

You can convert the resulting V*A*hr to V*A*sec afterwards.
 
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  • #7
Do the integration using "4" as the upper limit for time. The result will be in V*A*hr (Volt Amp Hours):

##\int_0^4 (10 + \frac{1}{2} t) dt =~ ... VAhr##

You can do this because you're told that in the expression for the terminal voltage the time is specified in hours.

You can convert the resulting V*A*hr to V*A*sec afterwards.

Oh okay! I never thought of work as Volt*Amps*Second. That makes a lot of sense. Thanks!
 
  • #9
gneill
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I think you've noticed this isn't right?
Gaaah! I should have spotted that! Thanks for catching that. :smile:
 
  • #10
I think you've noticed this isn't right?

That was a mistake haha. I know it should be I = dq/dt. My brain was just moving too quick for my fingers to comprehend and type the information!
 

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