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Alexeev081 [22]
3 years ago
13

A car battery has a rating of 250 ampere-hours. This rating is one indication of the total charge that the battery can provide t

o a circuit before failing. (a) What is the total charge (in coulombs) that this battery can provide? (b) Determine the maximum current that the battery can provide for 37 minutes.
Physics
1 answer:
Serga [27]3 years ago
4 0

Answer:

Total charge provided by the battery could be 900000 C.

Maximum current provided by the battery for 37 minutes could be 405.405 A

Explanation:

Rating= 250 A-h

a. Total charge:

Rating=Q/t\\Q=Rating.t\\

Suppose t=1h

Q=250 A(1h)\\Q=250 A(3600 sec)\\Q= 900000 A.sec

We konw that Ampere=\frac{Coulomb}{sec}, replacing:

Q=900000(\frac{Coulomb}{sec})(sec)\\Q=900000 Coulomb

Total charge provided by the battery could be 900000 C.

b. Maximum current for 37 minutes

I=\frac{Q}{t} \\I=\frac{900000 C}{37*60 sec}\\I=405.405 A

Maximum current provided by the battery for 37 minutes could be 405.405 A

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Answer:

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From the question we are told that

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Using one term approximation

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                    = 36.24

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        \lambda = 3.06632

        A = 1.9942

Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

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Taking natural log of both sides

           -1.993 =  -9.383\  \tau

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substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

         t = 40007 sec  

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