Problem 75 - University Physics with Modern Physics

Question

A Nonideal Ammeter. Unlike the idealized ammeter described in Section \(25.4,\) any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_{\mathrm{A}}\) is connected in series with a resistor \(R\) and a battery of emf \(\mathcal{E}\) and internal resistance \(r .\) The current measured by the ammeter is \(I_{\mathrm{A}}\) . Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_{A}, r, R_{\mathrm{A}},\) and \(R .\) The more "ideal" the ammeter, the smaller the difference between this current and the current \(I_{\mathrm{A}}\) . (b) If \(R=3.80 \Omega, \mathcal{E}=7.50 \mathrm{V},\) and \(r=0.45 \Omega,\) find the maximum value of the ammeter resistance \(R_{\mathrm{A}}\) so that \(l_{\mathrm{A}}\) is within 1.0\(\%\) of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

Step-by-Step Solution

Answer

The current without the ammeter is \( I = \frac{I_A (R_A + R + r)}{R + r} \). Ensure \( |I - I_A| \leq 0.01 I \) for part (b) to find maximum \( R_A \). If \( R_A \) is larger, deviation exceeds 1%.

Step 1: Define the Circuit with Ammeter

Consider the circuit with ammeter resistance \( R_A \), load resistance \( R \), battery internal resistance \( r \), and emf \( \mathcal{E} \). By Ohm's law, the total resistance of the circuit is \( R_A + R + r \). The current measured by the ammeter, \( I_A \), is given by: \[ I_A = \frac{\mathcal{E}}{R_A + R + r} \]

Step 2: Define the Circuit without Ammeter

When the ammeter is removed, the resistance is just the sum of \( R \) and \( r \). The current, \( I \), is then: \[ I = \frac{\mathcal{E}}{R + r} \]This current is what we're trying to express in terms of \( I_A, r, R_A, \) and \( R \).

Step 3: Relate the Two Currents

From Step 1, we rearrange the equation for \( I_A \): \[ \mathcal{E} = I_A (R_A + R + r) \]Substitute this expression for \( \mathcal{E} \) into the equation for \( I \) from Step 2:\[ I = \frac{I_A (R_A + R + r)}{R + r} \]

Step 4: Implementation for Part (b)

We know that \( I \) should be within 1% of \( I_A \), so we set \[ |I - I_A| \leq 0.01 \times I \]Substitute \( I = \frac{I_A (R_A + R + r)}{R + r} \) into this inequality:\[ \left|\frac{I_A (R_A + R + r)}{R + r} - I_A\right| \leq 0.01 \times \frac{I_A (R_A + R + r)}{R + r} \]

Step 5: Solve for Maximum Ammeter Resistance

This inequality can be simplified and solved for \( R_A \) given \( R = 3.80 \Omega, \mathcal{E} = 7.50 \text{ V}, \) and \( r = 0.45 \Omega \). After simplification, the inequality can be restructured to solve for \( R_A \). This needs algebraic manipulation.

Step 6: Conclusion for Part (b)

We find the maximum allowable \( R_A \) from rearranging and solving the inequality. Through careful algebraic manipulation, calculate \( R_A \) using the values given. The specific solution might need iterative or exact solving to meet the percentage criterion.

Step 7: Explanation for Part (c)

If \( R_A \) exceeds the derived value, the discrepancy between \( I \) and \( I_A \) becomes larger than the permissible 1%. Therefore, this maximum value ensures the deviation remains acceptably small.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits, stating that the current flow through a conductor between two points is directly proportional to the voltage across the two points. The law is represented by the formula:
\[ V = I \times R \]
where:

• \( V \) is the voltage (in volts)
• \( I \) is the current (in amperes)
• \( R \) is the resistance (in ohms)

It's crucial to understand that Ohm's Law reveals how the electrical resistance dictates the current in a circuit, assuming the voltage remains constant. In practical applications, this law helps us determine how changes in resistance, such as adding a resistor or ammeter, affect current flow.

Circuit Analysis

Circuit analysis involves examining the components and interconnections within an electrical circuit to understand how they influence the circuit's behavior. In our exercise, we have a series circuit that contains a battery, a resistor, and an ammeter.
Key elements of analyzing our circuit include:

• Understanding the total resistance in the circuit. For a series circuit, this is the sum of all resistances, including the internal resistance of components such as the battery and the ammeter.
• Determining how the removal or inclusion of components, like the ammeter, impacts the total resistance and, subsequently, the current.
• Alertness to potential fluctuations and changes in electrical properties, given any non-ideal components that add unexpected resistance.

Effective circuit analysis skills enable us to predict values such as current and voltage drops across each component, leading to better circuit design and troubleshooting.

Electric Current

Electric current is the flow of electric charge through a conductor, such as a wire, due to a difference in electrical potential. In a circuit, current is the movement of electrons driven by the voltage provided by the voltage source, like a battery.
A few crucial points about electric current:

• Measured in amperes (A), electric current is what flow meters, like ammeters, detect and quantify.
• A conventional current flows from the positive to the negative terminal of the power source, whereas electron flow is in the opposite direction.
• Current can be influenced by factors such as circuit resistance and the applied voltage.

Understanding electric current helps to see its role in powering devices and its relationships to other circuit parameters through laws like Ohm’s Law.

Internal Resistance

Internal resistance, generally within batteries and other power sources, represents how some of the energy (voltage) is lost as the current flows through the power source itself.
Here are the main points about internal resistance:

• It causes the terminal voltage (\( V_T \) ) to be less than the emf (\( \mathcal{E} \) ) of the battery because:
• \( V_T = \mathcal{E} - Ir \)
• Where \( r \) is the internal resistance, and \( I \) is the current.
• Higher internal resistance leads to more voltage drop within the power source, reducing the efficient delivery of power to the circuit load.
• The internal resistance becomes particularly important when large currents are drawn from the power source, amplifying the energy loss.

In practical terms, considering internal resistance allows for more precise calculations and understanding of real-life conditions in circuits, ensuring an accurate analysis and measurement.