Question

Consider the HPPC test depicted on slide 3 of this lesson for ΔT=10 s and Ichg =Idis =10 A. If the cell voltages at times {0,10+,20−,30−,30+,40−, and 50} seconds are {3.8,3.42,3.4,3.78,4.15, 4.19 , and 3.81}, what is the HPPC discharge resistance Rdis,ΔT of this cell (in mΩ )? Round your answer to the nearest mΩ.

Answer 1

The HPPC discharge resistance (Rdis,ΔT) of the cell is calculated using Ohm's law with the given voltages before and after the discharge. The resistance is found to be 41 mΩ.

Step-by-step explanation:

To calculate the HPPC discharge resistance (Rdis,ΔT) of a cell, we can use the provided voltages and the current. The resistance during discharge can be found using Ohm's law, which states that Voltage (V) = Current (I) × Resistance (R). Since we are given the cell voltages immediately before and after a discharge event, we can calculate the resistance.

We can calculate the drop in voltage (Voltage final immediately after discharge - Voltage initial immediately before discharge), which is 3.78V - 4.19V = -0.41V. We are also given that the current during discharge (Idis) is 10A. Using Ohm's law (V=IR), we can find the discharge resistance. The negative sign indicates the direction of voltage change (discharging).

Rdis,ΔT = ΔV / Idis = -0.41V / 10A = -0.041 Ω or -41 mΩ. Since resistance cannot be negative, we take the absolute value, resulting in Rdis,ΔT = 41 mΩ as the HPPC discharge resistance.