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Chapter 20: Problem 80

The coil of wire in a galvanometer has a resistance of \(R_{C}=60.0 \Omega\). The galvanometer exhibits a full-scale deflection when the current through it is \(0.400 \mathrm{~mA}\). A resistor is connected in series with this combination so as to produce a voltmeter. The voltmeter is to have a full- scale deflection when it measures a potential difference of \(10.0 \mathrm{~V}\). What is the resistance of this resistor?

Short Answer

Expert verified
The resistance required is 24,940 \(\Omega\).

Step by step solution

01

Understanding the Problem

We need to determine the resistance required in series with a galvanometer coil to function as a voltmeter with a full-scale deflection at 10 V. We know the coil resistance and full-scale current of the galvanometer.
02

Relationship between Voltage, Current, and Resistance

Using Ohm's Law, we know that the full-scale voltage \( V \) is related to the current \( I \) and the total resistance \( R_{total} \) by: \[ V = I \times R_{total} \]
03

Expressing the Total Resistance

The total resistance \( R_{total} \) is the sum of the coil resistance \( R_C \) and the resistance \( R \) we need to find:\[ R_{total} = R_C + R \]
04

Applying Ohm's Law to the Given Values

Substituting known values into Ohm's Law gives:\[ 10.0 = 0.0004 imes (60 + R) \]
05

Solving for the Unknown Resistance

Rearrange and solve the equation:\[ 10.0 = 0.0004 imes R + 0.0004 imes 60 \]\[ 10.0 = 0.0004R + 0.024 \]Subtract 0.024 from both sides:\[ 9.976 = 0.0004R \]Solve for \( R \) by dividing both sides by 0.0004:\[ R = \frac{9.976}{0.0004} = 24940 \Omega \]
06

Conclusion

The required resistance to make the galvanometer act as a voltmeter with a full-scale deflection at 10 V is \( 24,940 \Omega \).

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Key Concepts

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

Galvanometer
A galvanometer is a sensitive instrument designed to detect and measure small electric currents. It works on the principle that a current-carrying coil placed in a magnetic field experiences a torque proportional to the current. This causes a needle or pointer to move across a scale, indicating the current's magnitude. A galvanometer is quite similar to a basic moving coil meter which makes it suitable for detecting minute currents.

Essential features of a galvanometer include:
  • High sensitivity: Capable of detecting current as small as microamperes.
  • Coil resistance: The coil offers some inherent resistance, denoted as \( R_{C} \), usually measured in ohms.
  • Full-scale deflection: This indicates the maximum current a galvanometer can measure without getting damaged or misreading, measured in milliamps (mA).
Series Resistance
In electrical measurements, series resistance is essential when converting a galvanometer to a voltmeter. This added resistance prolongs the range of measurable voltages without altering the galvanometer's sensitivity to the current. When a resistance is added in series to a galvanometer, the overall resistance of the circuit increases, enhancing the device's potential handling capacity.

The significance of series resistance lies in:
  • Protecting the galvanometer: By limiting how much current can flow through the galvanometer, preventing potential damage.
  • Adjusting sensitivity: Allows the measurement of higher potential differences by increasing total resistance.
  • Enabling voltmeter range extension based on required scales, such as from 0 V to 10 V in this exercise.
Voltmeter Conversion
Converting a galvanometer into a voltmeter is a common procedure due to the galvanometer's sensitivity. This conversion involves adding a precise value of series resistance so that the combined setup can measure higher voltages. The prime tool responsible for this is Ohm's Law, which relates voltage (\( V \)), current (\( I \)), and resistance (\( R \)) with the formula: \( V = I \times R \).

For voltmeter conversion:
  • Determine the necessary full-scale voltage the new voltmeter needs to measure.
  • Use the coil's known resistance and full-scale deflection current to calculate the total desired resistance.
  • The series resistance \( R \) is found by subtracting the galvanometer's coil resistance \( R_{C} \) from the total resistance, derived by rearranging and solving the equation \( V = I \times (R_{C} + R) \).
This process allows us to convert delicate galvanometers into robust voltmeters with specific voltage ranges suitable for various applications.

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Most popular questions from this chapter

A resistor is connected across the terminals of a \(9.0\) -V battery, which delivers \(1.1 \times 10^{5} \mathrm{~J}\) of energy to the resistor in six hours. What is the resistance of the resistor?

Three identical resistors are connected in parallel. The equivalent resistance increases by \(700-\Omega\) when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor.

Two wires have the same length and the same resistance. One is made from aluminum and the other from copper. Obtain the ratio of the cross-sectional area of the aluminum wire to that of the copper wire.

Concept Question Two capacitors, \(C_{1}\) and \(C_{2}\), are connected to a battery whose voltage is \(V\). Recall from Section \(19.5\) that the electrical energy stored by each capacitor is \(\frac{1}{2} C_{1} V_{1}^{2}\) and \(\frac{1}{2} C_{2} V_{2}^{2}\), where \(V_{1}\) and \(V_{2}\) are, respectively, the voltages across \(C_{1}\) and \(C_{2}\). If the capacitors are connected in series, is the total energy stored by them greater than, less than, or equal to the total energy stored when they are connected in parallel? Justify your answer. Problem The battery voltage is \(V=60.0 \mathrm{~V}\), and the capacitances are \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=4.00 \mu \mathrm{F}\). Determine the total energy stored by the two capacitors when they are wired (a) in parallel and (b) in series. Check to make sure that your answer is consistent with your answer to the Concept Question.

In Section \(12.3\) it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is \(125 \Omega\) when its temperature is \(20.0^{\circ} \mathrm{C}\). The wire is then immersed in boiling chlorine, and the resistance drops to \(99.6 \Omega\). The temperature coefficient of resistivity of platinum is \(\alpha=3.72 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\). What is the temnerature of the boiling chlorine?
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