Answer:
V= 3.55 V
Explanation:
- As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
- We can calculate this loss, applying Ohm's law to the internal resistance, as follows:
[tex]V_{rint} = I* r_{int}[/tex]
- The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:
[tex]V = V_{b} - V_{rint} = 6.00 V - 0.207A* r_{int}[/tex]
- We can solve for rint, as follows:
[tex]r_{int} = \frac{V_{b} - V}{I} = \frac{6.00 V - 5.03V}{0.207A} = 4.7 \Omega[/tex]
- When the circuit draws from battery a current I of 0.523A, we can find the potential difference between the terminals of the battery, as follows:
[tex]V = V_{b} - V_{rint} = 6.00 V - 0.523A* 4.7 \Omega = 3.55 V[/tex]
- As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.
