Question

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.

Answer 1

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:

        $I_{1} = \frac{V}{R_{int} +r_{L} }$

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

       $I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

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

       $I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

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

        $\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

• 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.