Question E7D09:
How is battery operating time calculated?  

Choices:  
A: Average current divided by capacity in amp-hours  
B: Average current divided by internal resistance  
C: Capacity in amp-hours divided by average current  
D: Internal resistance divided by average current    

Correct Answer is  C: Capacity in amp-hours divided by average current  

To calculate battery operating time:  
$$t\ (\text{hours}) = \frac{C\ (\text{amp-hours})}{I_{\text{avg}}\ (\text{amps})}$$

Why C is correct (Choice C):
- Capacity is given in amp-hours (Ah), which means the battery can ideally supply $C$ amps for 1 hour, or $C/2$ amps for 2 hours, etc.  
- Dividing capacity by average current gives time:  
  $$\frac{\text{Ah}}{\text{A}} = \text{hours}$$

Why the other choices are incorrect:
- A: Average current divided by capacity $\rightarrow \frac{\text{A}}{\text{Ah}} = \frac{1}{\text{hours}}$, not time.
- B: Average current divided by internal resistance has units $\frac{\text{A}}{\Omega}$, which is not time and has no direct relation to operating duration.
- D: Internal resistance divided by average current $\rightarrow \frac{\Omega}{\text{A}} = \frac{\text{V}}{\text{A}^2}$, not time; internal resistance affects voltage drop and heating, not time directly.

Useful formulas and tips:
- Average current for intermittent loads (e.g., TX/RX duty cycle):
  $$I_{\text{avg}} = p_{\text{TX}} I_{\text{TX}} + (1 - p_{\text{TX}}) I_{\text{RX}}$$
- C-rate relation: if a pack is discharged at $n$C (i.e., $I = nC$), then
  $$t \approx \frac{1}{n}\ \text{hours}$$
- Real-world adjustments: actual time is often less due to Peukert effect (especially lead-acid), temperature, cutoff voltage, and aging. As a rule of thumb, derate estimated time by 20–30% unless specified otherwise.