Question

A battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?

Short answer

Answer: The approximate internal resistance of the battery is 1.67 Ω.

Step-by-step solution

Understand the problem

First, we need to understand the concept involved in this problem. A battery has an internal resistance, which causes the potential difference (voltage) across the battery to change depending on the electrical load connected to it. In this case, we are given the open-circuit voltage (no load connected) and the voltage when a load (resistor) is connected across it. We need to find the internal resistance of the battery using these values.

Apply Ohm's law

To solve this problem, we will apply Ohm's law, which states that the potential difference across a resistor is equal to the product of the current passing through it and its resistance, i.e., \(V=IR\). We will write an equation for the potential difference across the external resistor and the potential difference across the internal resistance of the battery when the load is connected.

Write the equations

When the load is connected, the potential difference across the battery is 12.68 V. Therefore, \(V_\text{external} + V_r = 12.68\) Where: - \(V_\text{external}\) is the potential difference across the internal resistance of the battery, - \(V_r\) is the potential difference across the external resistor. We can rewrite this equation using Ohm's law as: \(I r_\text{int} + IR = 12.68\) Where: - \(I\) is the current passing through the circuit, - \(r_\text{int}\) is the internal resistance of the battery, - \(R\) is the resistance of the external resistor (17.91 Ohms).

Calculate the current

The potential difference across the external resistor is 12.68 V, and its resistance is 17.91 Ohms. We can find the current passing through the external resistor using Ohm's law: \(I = \frac{V_r}{R} = \frac{12.68}{17.91} = 0.7076 \mathrm{~A}\)

Find the internal resistance

Now, we can substitute the current value in the equation: \(0.7076 r_\text{int} + 0.7076 \times 17.91 = 12.68\) Let's solve for \(r_\text{int}\): \(r_\text{int} = \frac{12.68 - 0.7076 \times 17.91}{0.7076} = 1.67 \mathrm{~\Omega}\) (approximately) The internal resistance of the battery is approximately \(1.67 \mathrm{~\Omega}\).