Problem 68 - Physics

Question

A battery has an internal resistance of \(0.012 \Omega\) and an emf of \(9.00 \mathrm{~V}\). What is the maximum current that can be drawn from the battery without the terminal voltage dropping below \(8.90 \mathrm{~V}\) ?

Step-by-Step Solution

Answer

The maximum current is approximately 8.33 A.

Step 1: Understand the Problem

We have a battery with an emf (electromotive force) of 9.00 V and an internal resistance of 0.012 Ω. We need to find the maximum current that can be drawn without dropping the terminal voltage below 8.90 V.

Step 2: Apply Ohm's Law for Internal Resistance

Use the relation for terminal voltage: \( V_t = \varepsilon - I \cdot r \), where \( \varepsilon \) is the emf, \( I \) is the current, and \( r \) is the internal resistance. We need \( V_t \geq 8.90 \). Substitute the given values: \( 8.90 = 9.00 - I \cdot 0.012 \).

Step 3: Solve the Equation

Rearrange the equation: \( I \cdot 0.012 = 9.00 - 8.90 \). This simplifies to \( I \cdot 0.012 = 0.10 \).

Step 4: Calculate the Maximum Current

Divide both sides by 0.012 to solve for \( I \): \( I = \frac{0.10}{0.012} \). Calculate this to find \( I \approx 8.33 \) A.

Step 5: Verify the Solution

Check that the terminal voltage is indeed 8.90 V when \( I = 8.33 \) A. Substitute back into \( V_t = \varepsilon - I \cdot r \): \( V_t = 9.00 - 8.33 \times 0.012 = 8.90 \) V. This indicates our calculation is correct.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in the field of electronics and physics. It establishes the relationship between voltage, current, and resistance in an electrical circuit. The formula is usually given by \( V = I \cdot R \), where:

• \( V \) represents voltage
• \( I \) stands for current
• \( R \) is resistance

In this specific exercise, Ohm's Law helps determine the current in circuits with internal resistance. When dealing with internal resistance, Ohm's Law aids in calculating how much current will allow a certain terminal voltage and emf to be maintained. This relationship is crucial for understanding how different components of a circuit work together.

Terminal Voltage

Terminal voltage is the potential difference measured across the terminals of a battery. This voltage differs from the battery's emf due to the internal resistance. In essence, the greater the current passing through a battery, the more significant the voltage drop across its internal resistance, which reduces the terminal voltage.For any battery, the terminal voltage \( V_t \) is given by the equation \( V_t = \varepsilon - I \cdot r \). Here, \( \varepsilon \) is the electromotive force (emf), \( I \) is the current, and \( r \) is the internal resistance. Therefore, to maintain a certain terminal voltage when the battery is under load, the current must be adjusted accordingly. Understanding terminal voltage is key for ensuring that devices powered by the battery function properly without risk of overloading.

emf

Electromotive force (emf) is a key concept in understanding how batteries work. It is the total voltage a battery can provide when no current is flowing through it. The symbol used to denote emf is \( \varepsilon \).

• Emf is considered the battery's ideal voltage.
• It is a critical factor in determining the maximum potential difference the battery can provide.

In an ideal scenario, requiring no current is drawn, a battery would provide its emf across its terminals. However, when a battery is in use, internal resistance causes a reduction in the actual voltage available at the terminals compared to the emf. Recognizing this difference is vital for accurately calculating and predicting the behavior of circuits using the battery.

Current Calculation

Calculating current is essential for predicting the performance of circuits and ensuring they function within desired parameters. The exercise involves determining the maximum current a battery can provide without the terminal voltage dropping significantly.To find the current \( I \) when the terminal voltage \( V_t \) is already known, use the rearranged terminal voltage formula: \( I = \frac{\varepsilon - V_t}{r} \). In this exercise, plug in the values for \( \varepsilon = 9.00 \, V\), \( V_t = 8.90 \, V\), and \( r = 0.012 \, \Omega\). Solving \( I = \frac{9.00 - 8.90}{0.012} \) gives \( I \approx 8.33 \, A\).This calculation ensures that the terminal voltage remains within acceptable limits while drawing the maximum possible current from the battery. Such computations are vital for designing circuits that safeguard power sources and make efficient use of them.