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Chapter 25: Problem 75

A Nonideal Ammeter. Unlike the idealized ammeter described in Section 25.4, any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_{A}\) 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 \(I_{A}\) . 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 \(I_{A}, r, R_{A},\) and \(R\) . The more "ideal" the ammeter, the smaller the difference between this current and the current \(I_{A}\) . (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 \(R_{A}\) so that \(I_{A}\) 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.

Short Answer

Expert verified
\( R_A = 0.0395 \Omega \). This is the maximum acceptable ammeter resistance for a 1% difference.

Step by step solution

01

Understand Initial Circuit

When the ammeter is connected, the total resistance in the circuit is the sum of the ammeter resistance \( R_A \), the resistance \( R \), and the internal resistance \( r \). The current through the circuit is \( I_A = \frac{\mathcal{E}}{R_A + R + r} \). This is the current measured by the ammeter.
02

Current Without Ammeter

With the ammeter removed, the total resistance in the circuit is just \( R + r \). The new current \( I \) without the ammeter is given by \( I = \frac{\mathcal{E}}{R + r} \).
03

Relate the Currents

To relate \( I_A \) and \( I \), use the expressions derived: \( I_A = \frac{\mathcal{E}}{R_A + R + r} \) and \( I = \frac{\mathcal{E}}{R + r} \). Notice that \( I_A < I \) due to the additional resistance from the ammeter.
04

Set Up the Condition for Part (b)

Given \( I_A \) is within 1% of \( I \): \( |I - I_A| \leq 0.01I \). Substitute the expression for \( I_A \) and \( I \) to solve for \( R_A \), i.e., \( \left|\frac{\mathcal{E}}{R + r} - \frac{\mathcal{E}}{R_A + R + r}\right| \leq 0.01 \times \frac{\mathcal{E}}{R + r} \).
05

Calculate Maximum Ammeter Resistance \( R_A \)

Substitute the given values: \( R = 3.80 \Omega \), \( \mathcal{E} = 7.50 \mathrm{V} \), \( r = 0.45 \Omega \). Solve \( \left(\frac{1}{R + r} - \frac{1}{R_A + R + r}\right) \leq 0.01 \times \frac{1}{R + r} \) to find \( R_A \). Simplify and solve for \( R_A \).
06

Interpretation of Maximum \( R_A \)

The result indicates the highest allowable value of \( R_A \) ensuring that the current measured by the ammeter \( I_A \) does not differ from the actual current \( I \) by more than 1%. This is because higher resistance in the ammeter would reduce \( I_A \) further, worsening the precision of the measurement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circuit Resistance
Resistance is the hindrance that slows down the flow of electric current in a circuit. In our exercise, the circuit consists of an ammeter, a resistor, and a battery with an internal resistance.

The total resistance in the circuit when the ammeter is included is the sum of all these resistances: the ammeter resistance ( R_A ), the given resistor ( R ), and the battery's internal resistance ( r ). When the ammeter is removed, the circuit only has the resistor and the internal battery resistance.

- **Ammeter Resistance:** This adds to the total circuit resistance when present. - **Total Circuit Resistance:** With the ammeter, it's R_A + R + r . Without, it's R + r .

This difference affects the current flowing through the circuit, influencing how accurately the ammeter measures the electric current.
Current Measurement
Current measurement is key in understanding how much electricity is flowing through a circuit. When we use an ammeter, it helps us determine the current. However, if the ammeter isn't ideal, it might slightly affect the measurement due to its inherent internal resistance.

- **Identifying Current:** The principal task of an ammeter is to measure the current in amperes (A). - **Impact of Ammeter:** As it introduces additional resistance, the measured current ( I_A ) can differ from the true current ( I ) when the ammeter is absent.

To ensure accuracy, we want an ammeter that interferes as little as possible with the actual current, i.e., by having negligible resistance. This allows us to achieve precise current measurement in the circuit.
Ammeter Resistance
The resistance of the ammeter ( R_A ) has a critical influence on its performance. Ideal ammeters have zero resistance, but real ones do not, meaning they can introduce errors in circuit measurements.

- **Effect of Non-Ideal Ammeter:** Because it adds resistance, the total resistance becomes R_A + R + r when it's connected. - **Importance of Low Resistance:** The lower the R_A , the closer the measured current ( I_A ) will be to the actual current ( I ) without the ammeter.

If the ammeter resistance is too high, it might significantly alter the current measurement, reducing accuracy and reliability of the data.
Electrical Circuits
Electrical circuits function as pathways for current to flow through using various components like batteries, resistors, and in this case, an ammeter. Understanding the right combination of these components is essential for accurate functionality and measurement.

- **Components:** - **Batteries** provide energy that sets electrons in motion. - **Resistors** control the flow and manage the current levels. - **Ammeter in Circuits:** When placed in series, it measures the current by becoming part of the established path. - **Total Resistance:** It's vital to understand how each element contributes to the overall circuit resistance.

A well-designed circuit ensures that each component functions optimally to provide accurate measurements and desired outcomes in electronic applications.
Ohm's Law
Ohm's Law is a foundational principle in the study of electrical circuits. It's expressed as V = I imes R , where V is the voltage, I is the current, and R is the resistance.

This law helps us comprehend how changes in voltage or resistance can impact current flow in a circuit, making it essential for calculations involving electrical circuits.

- **Application:** Ohm's Law allows us to determine currents and predict outcomes in a circuit when variables change. - **For Calculation:** Using Ohm's Law with total resistance, we can calculate the current in circuits with or without additional components like an ammeter. - **Practical Understanding:** Helps in designing circuits with specific characteristics by determining required resistance to achieve desired current flow.

By applying this law, we can solve for different parameters within the circuit to ensure everything functions smoothly and up to expectation.

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Most popular questions from this chapter

You apply a potential difference of 4.50 \(\mathrm{V}\) between the ends of a wire that is 2.50 \(\mathrm{m}\) in length and 0.654 \(\mathrm{mm}\) in radius. The resulting current through the wire is 17.6 \(\mathrm{A}\) . What is the resistivity of the wire?

A \(12.6-\mathrm{V}\) car battery with negligible internal resistance is connected to a series combination of a \(3.2-\Omega\) resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law but instead has a current- voltage relationship \(V=\alpha I+\beta I^{2},\) with \(\alpha=3.8 \Omega\) and \(\beta=1.3 \Omega / \mathrm{A}\) . What is the current through the \(3.2-\Omega\) resistor?

An 18 -gauge wire (diameter 1.02 \(\mathrm{mm} )\) carries a current with a current density of \(1.50 \times 10^{6} \mathrm{A} / \mathrm{m}^{2}\) . Calculate (a) the current in the wire and \((\mathrm{b})\) the drift velocity of electrons in the wire.

Light Bulbs. The power rating of a light bulb (such as a \(100-W\) bulb) is the power it dissipates when connected across a \(120-V\) potential difference. What is the resistance of (a) a \(100-W\) bulb and (b) a \(60-\mathrm{W}\) bulb? (c) How much current does each bulb draw in normal use?

The resistivity of a semiconductor can be modified by adding different amounts of impuritics. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L\) . The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\) \(\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L\) (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in therod as a function of \(x\) (c) Find the electric potential \(V(x)\) in the rod as a function of \(x\) . (d) Graph the functions \(\rho(x) . E(x)\) . and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\) .
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