Estimation of battery charging from definite integral

Brief note on definite integrals

Definite integrals are used to find the area under the given limits ‘a’ and ‘b’ for a differential derivative function. Though many rules are there to determine definite integrals for every specific function, basic algebraic function is given. 

For example, if ‘y’ as a function of ‘x’ or f(x) is given as,

where ‘a’ is lower limit, ‘b’ is upper limit and n ≠ –1.

 

Find the definite integral between the limits 1 to 4 for the function:

y = 2x² + 3x.

On solving this, the value is 64.5.

 

Now applying this concept to charging a battery... a battery is charged under constant voltage by varying the current (C) in Amperes. The current passed with reference to time (t) in hours is governed by the equation, C = 0.5t² + 0.5t. Estimate the charge (Ah) given from 2.5 hours to 4.5 hours.

(1 Ah is One Ampere given in one hour.)

 

The problem is represented by the following figure shown below. This figure shows the current passed with reference to time (x, in hours) during the charging of the battery and is governed by the equation, C = 0.5t² + 0.5t.

 

 

 

 

 

 

 

 

 

 

 

Charge (Coulomb) = Current (Ampere) × Time (second)

 

The net charge as well as charging rate between 2.5 to 4.5 hours can be estimated by definite integration method.

 

The integral formula for the function, C = 0.5t² + 0.5t can be given as,

 

 

 

= [(0.1667 × 4.5³) + (0.25 × 4.5²)] – [(0.1667 × 2.5³) + (0.25 × 2.5²)]

 

= [(15.1905) + (5.0625)] – [(2.6047) + (1.5625)]

 

= [20.2530 – 4.1672] = 16.0858 Ah.

 

So, the quantity of the charge passed within the time duration of two hours (between 2.5 to 4.5 hours) is 16.0858 Ah.

 

This estimated quantity of charge is for two hours. And hence the charging rate for an hour is equals to (16.0858 Ah/2) = 8.0429 Ah.

  

It can be estimated using the given equation, C = 0.5t² + 0.5t.

The total current (I) in Amperes passed between these two hours (2.5 to 4.5 hours) of time duration (t, in hours) is calculated as follows:

Substitute the ‘t’ value as 4.5 hours and 2.5 hours.

I = [(0.5 × 4.5²) + (0.5 × 4.5)] – [(0.5 × 2.5²) + (0.5 × 2.5)]

 

= [(10.125) + (2.25)] – [(3.125) + (1.25)]

 

= [12.375 – 4.375] = 8 A.

So, the quantity of current passed between the two hours is 8 A. Hence, the charge given at two hours is (8 A × 2 hours) = 16 Ah.