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Ch. 26 - DC Circuits
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 26 - DC CircuitsProblem 58a
Chapter 25, Problem 58a

A galvanometer has an internal resistance of 32 Ω and deflects full scale for a 48-μA current. Describe how to use this galvanometer to make an ammeter to read currents up to 25 A.

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Understand the problem: A galvanometer is a sensitive device that measures small currents. To convert it into an ammeter capable of measuring larger currents (up to 25 A), we need to add a shunt resistor in parallel with the galvanometer. This shunt resistor will bypass most of the current, allowing only a small fraction to pass through the galvanometer.
Step 1: Calculate the voltage across the galvanometer when it is at full-scale deflection. Use Ohm's Law: \( V_g = I_g \cdot R_g \), where \( I_g = 48 \mu A \) (full-scale current) and \( R_g = 32 \Omega \) (internal resistance of the galvanometer).
Step 2: Determine the current that needs to bypass the galvanometer. The total current to be measured is \( I = 25 \, A \), so the current through the shunt resistor is \( I_s = I - I_g \).
Step 3: Use the fact that the voltage across the galvanometer and the shunt resistor must be the same (since they are in parallel). The voltage across the shunt resistor is \( V_s = I_s \cdot R_s \), where \( R_s \) is the resistance of the shunt resistor. Set \( V_g = V_s \) and solve for \( R_s \): \( R_s = \frac{V_g}{I_s} \).
Step 4: Substitute the values of \( V_g \), \( I_g \), and \( I_s \) into the equation for \( R_s \) to calculate the required shunt resistance. This will give you the value of the resistor needed to convert the galvanometer into an ammeter capable of measuring up to 25 A.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Galvanometer

A galvanometer is an electromechanical device used to detect and measure small electric currents. It operates by using a coil of wire that experiences a magnetic field, causing it to rotate and indicate the current level on a scale. The full-scale deflection of a galvanometer occurs at a specific current, which is crucial for its calibration and use in other applications.

Shunt Resistor

A shunt resistor is a low-resistance component connected in parallel with a galvanometer to allow most of the current to bypass the galvanometer. This enables the galvanometer to measure larger currents without being damaged. The value of the shunt resistor is calculated based on the maximum current desired and the full-scale deflection current of the galvanometer.
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Current Division

Current division is a principle used in electrical circuits to determine how current is distributed among parallel branches. In the context of using a galvanometer as an ammeter, the total current entering the circuit is divided between the galvanometer and the shunt resistor. Understanding current division is essential for calculating the appropriate shunt resistance needed to ensure accurate readings at higher currents.
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Related Practice
Textbook Question

(III) Determine the net resistance in Fig. 26–61 (a) between points a and c, and (b) between points a and b. Assume R' ≠ R. [Hint: Apply an emf between the two points in each case and determine currents; use symmetry at junctions.]

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Textbook Question

A galvanometer has a sensitivity of 45kΩ/V and internal resistance 20.0 Ω. How could you make this into an ammeter that reads 1.0 A full scale?

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Textbook Question

(II) Suppose two batteries, with unequal emfs of 2.00 V and 3.00 V, are connected as shown in Fig. 26–63. If each internal resistance is r = 0.350Ω and R = 4.00Ω, what is the voltage across the resistor R?

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Textbook Question

A galvanometer has an internal resistance of 32 Ω and deflects full scale for a 48-μA current. Describe how to use this galvanometer to make a voltmeter to give a full scale deflection of 250 V.

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Textbook Question

In the circuit shown in Fig. 26–75, the 33-Ω resistor dissipates 0.80 W. What is the battery voltage?

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Textbook Question

The performance of the starter circuit in a car can be significantly degraded by a small amount of corrosion on a battery terminal. Figure 26–88a depicts a properly functioning circuit with a battery (12.5-V emf, 0.02-Ω internal resistance) attached via corrosion-free cables to a starter motor of resistance Rs = 0.15Ω. Sometime later, corrosion between a battery terminal and a starter cable introduces an extra series resistance of only RC = 0.10Ω into the circuit as suggested in Fig. 26–88b. Let P0 be the power delivered to the starter in the circuit free of corrosion, and let P be the power delivered to the starter with corrosion. Determine the ratio P/P0.

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