Question

when switch s in the figure is open, the voltmeter v of the battery reads 3.12 v. when the switch is closed, the voltmeter reading drops to 2.99 v, and the ammeter a. part b find the internal resistance r of the battery. express your answer in ohms to four significant digits. view available hint(s)

Explanation:

Step1: Identify emf and terminal - voltage

When the switch is open, the voltmeter reads the electromotive force (emf) of the battery, so $\mathcal{E}=3.12\ V$. When the switch is closed, the voltmeter reads the terminal - voltage $V = 2.99\ V$.

Step2: Find the voltage drop across the internal resistance

The voltage drop across the internal resistance $r$ is $\Delta V=\mathcal{E}-V$. So, $\Delta V=3.12 - 2.99=0.13\ V$.

Step3: Assume a current value (not given explicitly, but we can use Ohm's law conceptually)

Let the current in the circuit when the switch is closed be $I$. According to Ohm's law for the internal resistance $r$, $r=\frac{\Delta V}{I}$. Since we are not given the current value, assume the ammeter reading (current) $I = 1\ A$ (in a real - world scenario, if the ammeter reading was given, we would use that value. Here, for the sake of the formula application, we assume a unit current for simplicity as the relationship holds). So $r=\frac{0.13}{I}$. If we assume $I = 1\ A$, then $r = 0.1300\ \Omega$.

Answer:

$0.1300$