Analyzing % Change in Power of a Battery

[Brimley]

Hello everyone,

I'm trying to do a simple percentage change formula to analyze the difference that a material change can have on a battery and would just like some advice as to whether or not I proceeded correctly.

Essentially, my reference text gives an example of a Battery whose cast lead electrode grid would be replaced with a pierced-and-expanded metal grid. It notes the following:

"In a nominally 12 V battery, 6 cells are connected in series and within each cell there are two lead electrodes."

It is noted that the substitution of the expanded metal grid will increase the resistance of each electrode from 0.17 mOhm (or 0.17 * 10^-3 Ohm) to 0.3 mOhm (or 0.3 *10^-3 Ohm). It says the the battery initially had an overall resistance of 6mOhm (or 6*10^-3 Ohm) with the cast lead electrodes.

Objective: Estimate the % Change in Available Power

Here is the formula I used:

(P_new - P_old)/(P_old) * 100% = %Change (where P is power)

Here are my procedure:

I treat this as a circuit problem:

V_Battery = 12 V.
R_total-Old = 6 mOhm
R_{terminal-A-Old} = 0.17 mOhm
R_{terminal-B-Old} = 0.17 mOhm
R_internal = R_total-Old - R_{terminal-A-Old} - R_{terminal-B-Old} = 6-(0.17+0.17) = 5.66 mOhm

I_old = V_Battery / R_total-Old = 12 V / 6 mOhm = 2000 A
P_old = (I_old²)*R_total-Old = (2000² A²) (6 mOhm) = 24000 W

R_{terminal-A-New} = 0.3 mOhm
R_{terminal-B-New} = 0.3 mOhm
R_total-New = R_internal + R_{terminal-A-New} + R_{terminal-B-New} = 5.66+0.3+0.3 = 6.26 mOhm
I_new = V_Battery / R_total-New = 12 V/ 6.26 mOhm = 1916.93 A
P_new = (I_new²)*R_total-New = (1916.93² A²)(6.26 mOhm) = 23003.2 W

%Change = 100%*(P_new - P_old)/(P_old) = 100 * (23003.2 - 24000)/(24000) = -4.15335 %

Final Answer: %Change = -4.15335 %

Does this procedure seem correct? Can anyone verify this answer?

Thank you!

 

[Valkarie]

</code>Yes, your procedure seems correct. The answer you have obtained is also correct as per the given data.