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 ?

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:

         

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:

       

We can solve for rint, as follows:

       

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:

         

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.  

Since the pully is a simple machine so the mechanical advantage is a 1

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