Answer:
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:
[tex]I_{1} = \frac{V}{R_{int} +r_{L} }[/tex]
- where Rint = r and RL = r
- Replacing these values in I₁, we have:
[tex]I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)[/tex]
- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:
[tex]I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)[/tex]
- We can find the relationship between I₂, and I₁, dividing both sides, as follows:
[tex]\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}[/tex]
- 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.
