A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the internal resistance triples. How much is thecurrent trough the load resistor reduced? Explain.

The current is reduced to half of its original value.

Explanation:

Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

       

where Rint = r and RL = rReplacing these values in I₁, we have:

       

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

       

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  

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