25.29 • when switch s in fig. e25.29 is open, the voltmeter v reads 3.08 v. when the switch is closed, the…

25.29 • when switch s in fig. e25.29 is open, the voltmeter v reads 3.08 v. when the switch is closed, the voltmeter reading drops to 2.97 v, and the ammeter a reads 1.65 a. find the emf, the internal resistance of the battery, and the circuit resistance r. assume that the two meters are ideal, so they don’t affect the circuit.

25.29 • when switch s in fig. e25.29 is open, the voltmeter v reads 3.08 v. when the switch is closed, the voltmeter reading drops to 2.97 v, and the ammeter a reads 1.65 a. find the emf, the internal resistance of the battery, and the circuit resistance r. assume that the two meters are ideal, so they don’t affect the circuit.

Answer

Explanation:

Step1: Determine the emf

When the switch is open, the volt - meter reads the emf of the battery since no current is flowing through the circuit. So, $\mathcal{E}=3.08\ V$.

Step2: Use the formula for terminal voltage when switch is closed

The terminal voltage $V$ when the switch is closed is given by $V = \mathcal{E}-Ir$, where $V = 2.97\ V$, $\mathcal{E}=3.08\ V$ and $I = 1.65\ A$. We can solve for the internal resistance $r$. $r=\frac{\mathcal{E}-V}{I}=\frac{3.08 - 2.97}{1.65}=\frac{0.11}{1.65}\approx0.067\ \Omega$.

Step3: Use Ohm's law to find the circuit resistance $R$

When the switch is closed, $V = IR$. We know $V = 2.97\ V$ and $I = 1.65\ A$. So, $R=\frac{V}{I}=\frac{2.97}{1.65}=1.8\ \Omega$.

Answer:

The emf $\mathcal{E}=3.08\ V$, the internal resistance $r = 0.067\ \Omega$ and the circuit resistance $R = 1.8\ \Omega$.