Question

Consider a cell having open-circuit voltage of 3.8V at its present state of charge, operational voltage limits of Umin = 2.5V and Vmax = 4.2V, HPPC resistances of Rchg, AT 0.012 and Rdis, At = 0.00812. What amount of (absolute) charge power (in W) does the HPPC method predict? Round your answer to the nearest W, and report a positive number.

Answer 1

Using the HPPC method and Ohm's Law, the predicted absolute charge power for a cell with an open-circuit voltage of 3.8V and a charging resistance of 0.012 ohms is 1197 W after rounding to the nearest whole watt.

Step-by-step explanation:

To calculate the charge power using the Hybrid Pulse Power Characterization (HPPC) method for the cell described in the question, we use the cell's open-circuit voltage and HPPC resistances. Since the open-circuit voltage is 3.8V, we can find the charge power by applying Ohm's Law, P = V^2 / R, where P is the power, V is the voltage, and R is the resistance. The resistance for charging, Rchg, is given as 0.012 ohms. Therefore, the charge power (Pchg) is Pchg = (3.8^2) / 0.012 = 1196.67 W. Rounding to the nearest whole number, the predicted charge power is 1197 W.