In this case, the extra voltage (terminal voltage - emf) is due to the internal resistance of the battery. Therefore, we can write:
\[ r = \frac{\Delta V}{I} \]
where \( \Delta V = V_{terminal} - V_{emf} = 16.0V - 12.0V = 4.0V \) is the extra voltage, \( I = 10.0A
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