Problem 29 - College Physics

Question

With a 1500 \(\mathrm{M\Omega}\) resistor across its terminals, the terminal voltage of a certain battery is 2.50 \(\mathrm{V}\) . With only a 5.00\(\Omega\) ? resistor across its terminals, the terminal voltage is 1.75 \(\mathrm{V}\) . (a) Find the internal emf and the internal resistance of this battery. (b) What would be the terminal voltage if the 5.00\(\Omega\) resistor were replaced by a 7.00 \(\Omega\) resistor?

Step-by-Step Solution

Answer

(a) \(\mathcal{E} = 2.50 \, \mathrm{V}\), \(r = 2.14 \, \Omega\). (b) Terminal voltage \(= 1.92 \, \mathrm{V}\).

Step 1: Identify Known Values

We have two scenarios with different resistors connected to the battery's terminals:1. With a 1500 \(\mathrm{M\Omega}\) resistor, the terminal voltage is 2.50 \(\mathrm{V}\).2. With a 5.00 \(\Omega\) resistor, the terminal voltage is 1.75 \(\mathrm{V}\).Our goal is to find the battery's internal emf (\(\mathcal{E}\)) and the internal resistance (\(r\)).

Step 2: Use Ohm's Law for Each Scenario

For each scenario, Ohm's law is used: \(V = \mathcal{E} - Ir\). Here, \(V\) is the terminal voltage, \(\mathcal{E}\) is the emf, \(I\) is the current through the resistor, and \(r\) is the internal resistance.**Scenario 1:**\[V_1 = 2.50 \, \mathrm{V}, \, R_1 = 1500 \, \mathrm{M\Omega}\]\(I_1 = \frac{V_1}{R_1} = \frac{2.50}{1500 \times 10^6}\) (which is negligible)**Scenario 2:**\[V_2 = 1.75 \, \mathrm{V}, \, R_2 = 5.00 \, \Omega\]\(I_2 = \frac{V_2}{R_2} = \frac{1.75}{5}\)

Step 3: Apply Equations to Find \(\mathcal{E}\) and \(r\) in the 5.00 \(\Omega\) Scenario

From **Scenario 2**, we can write:\[\mathcal{E} - I_2r = V_2 = 1.75 \, \mathrm{V} \]Substitute \(I_2 = \frac{1.75}{5} = 0.35 \, \mathrm{A}\):\[\mathcal{E} - 0.35r = 1.75\]

Step 4: Use 1500 \(\mathrm{M\Omega}\) Scenario to Approximate \(\mathcal{E}\)

With a massive resistor like 1500 \(\mathrm{M\Omega}\), the current through the battery is nearly zero, which implies:\[\mathcal{E} \approx 2.50 \, \mathrm{V}\]

Step 5: Use Values to Solve for \(r\)

Substitute \(\mathcal{E} = 2.50 \mathrm{V}\) into the equation from Step 3:\[2.50 - 0.35r = 1.75\]Solve for \(r\):\[0.35r = 2.50 - 1.75 = 0.75\]\[r = \frac{0.75}{0.35} \approx 2.14 \, \Omega\]

Step 6: Calculate Terminal Voltage with 7.00 \(\Omega\) Resistor

Now replace the resistor with 7.00 \(\Omega\):Calculate the new current, \(I_3 = \frac{\mathcal{E}}{R_3 + r} = \frac{2.50}{7 + 2.14} \approx 0.2731 \, \mathrm{A}\).Use the new current to find the terminal voltage:\[V_3 = \mathcal{E} - I_3r = 2.50 - 0.2731 \cdot 2.14 \approx 1.92 \, \mathrm{V}\]

Key Concepts

Internal Resistance

Internal resistance of a battery is a critical factor that determines how much voltage the battery can deliver when connected to a load. It refers to the resistance within the battery itself caused by the chemical components and construction of the battery.
The internal resistance (\( r \) ) is significant because it affects the efficiency of power delivery.
When current flows through a battery, the internal resistance causes a reduction in the output voltage, which is the terminal voltage seen by an external circuit.

• Internal resistance is typically measured in ohms (\(\Omega\)).
• High internal resistance can lead to power losses and inefficient battery performance.

To understand this in practical terms, consider the second scenario in the original exercise where a 5 \(\Omega\) resistor caused a terminal voltage drop from the initial case. Here, internal resistance was a key factor in reducing the actual voltage that reached the external resistor. This is why calculating internal resistance is vital for accurate determination of a battery's capability.

Terminal Voltage

Terminal voltage is the voltage output of a battery that you measure across its terminals when it is connected to a load. It differs from the electromotive force (emf) due to the voltage drop across the internal resistance of the battery.
This concept can be easily understood using Ohm’s law, which tells us that the voltage across any component in a circuit is the product of its resistance and the current flowing through it (\(V = IR\)). For a battery, however, the terminal voltage (\(V)\) can be calculated as:\[V = \mathcal{E} - Ir\]Where \( \mathcal{E} \) is the emf of the battery, \( I \) is the current, and \( r \) is the internal resistance.

• In a perfect battery with no internal resistance, terminal voltage would equal the emf.
• As the internal resistance increases, the difference between the emf and terminal voltage becomes more pronounced.

In our specific scenarios, the terminal voltage changed when different resistors were used, demonstrating how varying loads affect the voltage available to them.

Electromotive Force (emf)

Electromotive force (abbreviated emf) is the energy provided by a battery or cell per coulomb of charge passing through it. It is often mistaken for the actual voltage of a battery, but there is a subtle but essential distinction.
In circuit terms, emf (\( \mathcal{E} \)) is the total voltage across a cell in the absence of any current. Thus, it is the inherent voltage the battery generates, assuming no internal resistance affects its voltage output.

• Emf is measured in volts (\(\mathrm{V}\)).
• It defines the open-circuit voltage.
• It positions as a constant potential difference under no-load conditions.

In practice, learning the difference between emf and terminal voltage helps in understanding battery performance in a circuit. For example, in our problem, the calculated terminal voltage when using different resistors reflected the impact internal resistance had on converting emf into usable energy.