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Chapter 26: Problem 80

A certain galvanometer has a resistance of 65.0\(\Omega\) and deflects full scale with a current of 1.50 \(\mathrm{mA}\) in its coil. This is to be replaced with a second galvanometer that has a resistance of 38.0\(\Omega\) and deflects full scale with a current of 3.60\(\mu A\) in its coil. Devise a circuit incorporating the second galvanometer such that the equivalent resistance of the circuit equals the resistance of the first galvanometer, and the second galvanometer deflects full scale when the current through the circuit equals the full-scale current of the first galvanometer.

Short Answer

Expert verified
Use a shunt resistor to adjust for full-scale current and add a series resistor to match the total resistance.

Step by step solution

01

Understanding the Problem

We have two galvanometers: the first with a resistance of 65.0 \(\Omega\) and a full-scale current of 1.50 mA. The second has a resistance of 38.0 \(\Omega\) and a full-scale current of 3.60 \(\mu\mathrm{A}\). We need a circuit with the second galvanometer that has the same total resistance as the first and reaches full scale with the first's full-scale current.
02

Calculating Shunt Resistor for Current Adjustment

To make the second galvanometer deflect full scale with 1.50 mA, we use a shunt resistor. The current needed by the galvanometer for full-scale deflection is 3.60 \(\mu\mathrm{A}\), so the remaining current \((1.50\, \mathrm{mA} - 3.60\,\mu\mathrm{A})\) must pass through the shunt resistor.
03

Applying Current Division Formula

Using the current division rule: \( I_s = I - I_g \), where \(I\) is 1.50 mA and \(I_g\) is 3.60 \(\mu\mathrm{A}\), calculate \( I_s \). Then, determine the value of the shunt resistor \(R_s\) that will carry \( I_s \) such that the voltage across it is the same as that across the galvanometer: \( R_s = \frac{V}{I_s} = \frac{I_g \cdot R_g}{I_s} \).
04

Calculating Series Resistor for Resistance Adjustment

Next, to match the total resistance to 65.0 \(\Omega\), add a series resistor \(R_x\). Calculate \( R_x = R_{total} - (R_g \parallel R_s)\), where \( R_{total} \) is 65.0 \(\Omega\) and \( R_g \parallel R_s \) is the equivalent resistance of the second galvanometer and the shunt resistor in parallel.
05

Combining Results

Add the calculated series resistor \( R_x \) to the galvanometer-shunt parallel combination. The sum of these resistances will match the desired total resistance of 65.0 \(\Omega\). Now, the circuit is completed, ensuring both the correct equivalent resistance and full-scale current deflection.

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Key Concepts

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

Circuit Resistance
The concept of circuit resistance is crucial in understanding how electrical circuits function. Resistance measures how much a component opposes the flow of electric current. The unit for resistance is the ohm (惟). When designing a circuit, we often aim for a specific total resistance, which influences how the circuit behaves in response to electrical current and potential difference.

For the galvanometer circuit in question, achieving the correct circuit resistance鈥攅quivalent to the first galvanometer's 65.0 惟鈥攊s essential. This ensures that the overall electrical characteristics remain unchanged despite using a different galvanometer. To achieve this, both shunt and series resistors may be used to fine-tune the total resistance of the circuit. This combination allows for flexibility in circuit design, accommodating different components while maintaining desired electrical properties.
Full-scale Deflection
Full-scale deflection in a galvanometer corresponds to the maximum current that can pass through before the display or needle reaches its limits. It's essentially the maximum capacity that the device can measure accurately. For accurate readings, the galvanometer should deflect fully at the specific current intended for the measurement.

In this exercise, the first galvanometer is designed to reach full-scale deflection with a current of 1.50 mA. For the second galvanometer, which has a lower full-scale deflection current of 3.60 碌A, adjustments need to be made using additional resistors (shunt and series) to ensure it matches the first galvanometer鈥檚 requirements. By understanding full-scale deflection, one can ensure that measurements remain consistent even when replacing or adjusting components in a circuit.
Shunt Resistor
A shunt resistor is crucial for redirecting excess current away from sensitive components of a circuit, like a galvanometer, to prevent damage and maintain accurate readings.
  • It allows only a fraction of the total current to flow through the galvanometer.
  • It must be carefully calculated to handle the extra current without introducing significant error.
In the context of our exercise, a shunt resistor is used to adjust the total current flowing through the second galvanometer, down to its specified full-scale current of 3.60 碌A. The remaining current, calculated as 1.50 mA minus 3.60 碌A, will flow through the shunt resistor. The correct value of this resistor ensures that the voltage across it is the same as that across the galvanometer, allowing the circuit to function as intended.
Series Resistor
A series resistor is used to adjust the overall resistance of a circuit by adding it directly in line with other components. This adjustment is vital when modifying circuits to maintain the desired total resistance.
  • It helps increase the circuit's resistance to match specified needs.
  • It's placed in series to ensure all current flows through it, thus adding its resistance to the circuit.
In our galvanometer circuit, the series resistor helps achieve the overall required resistance of 65.0 惟. By calculating and adding this resistor, the equivalent resistance, considering the shunt resistor and galvanometer in parallel, matches the desired resistance value. This ensures that the new galvanometer functions under the same conditions as the original, making it a seamless replacement.
Current Division Rule
The current division rule helps distribute current in a parallel circuit. It's especially useful when using shunt resistors, as it explains how the current divides between different paths.
  • Used to calculate how much current should travel through each component.
  • Allows precise control over current flow in circuits with multiple branches.
When you have a galvanometer with a shunt resistor, the current division rule calculates the current through each. By knowing the full-scale deflection current (3.60 碌A for the second galvanometer), and the total input current (1.50 mA), we can find the current through the shunt using the relation: \[ I_s = I - I_g \]This ensures the galvanometer receives only its maximum allowable current, preventing overload and ensuring accuracy. Understanding this rule ensures precise measurements and efficient circuit design.

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Most popular questions from this chapter

An emf source with \(\mathcal{E}=120 \mathrm{V},\) a resistor with \(R=80.0 \Omega\) and a capacitor with \(C=4.00 \mu F\) are connected in series. As the capacitor charges, when the current in the resistor is 0.900 \(\mathrm{A}\) , what is the magnitude of the charge on each plate of the capacitor?

A. \(2.36-\mu F\) capacitor that is initially uncharged is connected in series with a \(4.26-\Omega\) resistor and an emf source with \(\mathcal{E}=120 \mathrm{V}\) and negligible intermal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

Light Bulbs in Series. A \(60-\mathrm{W}, 120-\mathrm{V}\) light bulb and a \(200-\mathrm{W}, 120-\mathrm{V}\) light bulb are connected in series across a \(240-\mathrm{V}\) line. Assume that the resistance of each bulb does not vary with current. (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a \(25-\mathrm{W}, 120-\mathrm{V}\) light bulb dissipates 25 \(\mathrm{W}\) when connected to a \(120-\mathrm{V}\) line. (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

A capacitor that is initially uncharged is connected in series with a resistor and an emf source with \(\mathcal{E}=110 \mathrm{V}\) and negligible internal resistance. Just after the circuit is completed, the current through the resistor is \(6.5 \times 10^{-5} \mathrm{A}\) . The time constant for the circuit is 6.2 \(\mathrm{s}\) . What are the resistance of the resistor and the capacitance of the capacitor?

A resistor \(R_{1}\) consumes electrical power \(P_{1}\) when connected to an emf \(\mathcal{E}\) . When resistor \(R_{2}\) is connected to the same emf, it consumes electrical power \(P_{2}\) . In terms of \(P_{1}\) and \(P_{2}\) , what is the total electrical power consumed when they are both connected to this emf source (a) in parallel and \((b)\) in series?
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