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.

Answer 2

The potential difference is $4.81V$.

Ohm's law :

It is given as,  $V=IR$

Where ,

• $V$ is potential difference

• $I$ is current

• $R$ is resistance

Given that, When you draw a current of $0.383 A$ from it, the potential difference between its terminals is $5.03 V$.

So that,  $R=\frac{6-5.03}{0.383}=2.53$

The potential difference when current is $0.469 A$ ,

                $V=6-(0.469*2.53)\\\\V=6-1.19=4.81V$

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