Question

A battery with an internal resistance of r and an emf of 10.00 V is connected to a load resistor R=r. As the battery ages, the internal resistance triples. How much is the current through the load resistor reduced?

Answer 1

Answer

Internal resistance

$i=\frac{E-V}{r+R}$

In first case

$i=\frac{10-V}{r+r}=\frac{10-V}{2r}$

Now in second case

$i'=\frac{10-V}{3r+r}=\frac{10-V}{4r}=\frac{i}{2}$

So current is halves.