Question

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference between its terminals is 5.03 V 5.03 V . What is the potential difference when you draw 0.469 A 0.469 A ?

Answer 1

Answer:

V = 4.81 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:

       $V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

• We can solve for rint, as follows:

         $r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

• When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

       $V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

• 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.