Question

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.207 A 0.207 A from it, the potential difference between its terminals is 5.03 V 5.03 V . What is the potential difference when you draw 0.523 A 0.523 A ?

Answer 1

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:

         $V_{rint} = I* r_{int}$

• 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:

        $V = V_{b} - V_{rint} = 6.00 V - 0.207A* r_{int}$

• We can solve for rint, as follows:

        $r_{int} = \frac{V_{b} - V}{I} = \frac{6.00 V - 5.03V}{0.207A} = 4.7 \Omega$

• 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:

         $V = V_{b} - V_{rint} = 6.00 V - 0.523A* 4.7 \Omega = 3.55 V$

• 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.