**Losses at High and Low Charging Rates.** A 100-Ah, 12-V battery with a rest voltage of 12.5 V (at its current SOC) is charged at a C/5 rate, during which time the applied voltage is 13.2 V. Using a simple Thevenin equivalent:

a. Estimate the internal resistance of the battery.

b. What fraction of the input power is lost in the internal resistance of the battery?

c. If the charging is done at a C/20 rate, what fraction of the input power would be lost due to the internal resistance?

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a. At C/5, the current is 100 Ah/5 h = 20 A. The internal resistance must have been

R_{i} \ = \ \frac{V_{in} \ – \ V_{B}}{I} \ = \ \frac{13.2 \ − \ 12.5}{20} \ = \ 0.035 \ \Omegab. So the I² R losses as a fraction of the input power would be

\frac{\text{Power lost in} \ R_{i}}{\text{Input power}} \ = \ \frac{I^{2} \ R}{V_{in} I} \ = \ \frac{\left(20\right)^{2} \ \times \ 0.035}{13.2 \ \times \ 20} \ = \ 0.053 \ = \ 5.3 \%c. At C/20, the current is 100 Ah/20 h = 5 A. The input voltage needed to drive 5 A through the 0.035-Ω resistance is

V_{in} \ = \ V_{B} \ + \ I R \ = \ 12.5 \ + \ 5 \ \times \ 0.035 \ = \ 12.68 \ VSo at the C/20 = 5-A rate, the losses are now only

\frac{\text{Power lost in} \ R_{i}}{\text{Input power}} \ = \ \frac{I^{2} \ R}{V_{in} I} \ = \ \frac{\left(5\right)^{2} \ \times \ 0.035}{12.68 \ \times \ 5} \ = \ 0.014 \ = \ 1.4 \%Question: 9.22

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