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Chapter 10: Problem 99

Find the resistance that must be placed in parallel with a \(60.0-\Omega\) galvanometer having a 1.00 -mA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 25.0 - A full-scale reading. Include a circuit diagram with your solution.

Short Answer

Expert verified
The resistance that must be placed in parallel with a 60.0 惟 galvanometer having a 1.00 mA sensitivity to allow it to be used as an ammeter with a 25.0 A full-scale reading can be determined as follows: First, calculate the current in the parallel resistor at full scale: I_resistor = I_total - I_galvanometer = 25.0 A - (1.00 x 10^{-3} A) = 24.999 A Next, find the equivalent resistance in the parallel combination: R_eq = (R_1 * I_galvanometer) / I_total = (60.0 惟 * 1.00 x 10^{-3} A) / 25.0 A = 0.0024 惟 Finally, determine the unknown parallel resistor value R_2: \( \frac{1}{R_{2}} = \frac{1}{R_{eq}} - \frac{1}{R_{1}} \Rightarrow R_2 = \frac{1}{(\frac{1}{0.0024} - \frac{1}{60.0})} \) R_2 鈮� 0.00241379 惟 The required parallel resistance is approximately 0.0024 惟. The circuit diagram would show the 60.0 惟 galvanometer resistor in parallel with the determined 0.0024 惟 resistor, with the ammeter reading a 25.0 A full-scale.

Step by step solution

01

1. Calculate the current in the parallel resistor at full scale

: The current passing through the parallel resistor when the ammeter reads full scale (25.0 A) is given as follows: I_total = I_galvanometer + I_resistor Where, I_total is the total current, I_galvanometer is the current passing through the galvanometer, and I_resistor is the current passing through the parallel resistor. Given that galvanometer鈥檚 sensitivity is 1.00 mA, we have I_galvanometer = 1.00 x 10^{-3} A. We can now solve for I_resistor: I_resistor = I_total - I_galvanometer
02

2. Find the equivalent resistance in the parallel combination

: Use the fact that the equivalent resistance of parallel resistors is given by the formula: \( \frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \) Where, R_eq is the equivalent resistance, R_1 is the resistance of the galvanometer, and R_2 is the resistance of the parallel resistor. We are given that the resistance of the galvanometer, R_1, is 60.0 惟. We need to determine the equivalent resistance, R_eq, which allows the full-scale reading to be 25.0 A. To find R_eq, use Ohm's law, V = IR: R_eq = V / I_total However, we do not have the voltage (V). But we can relate the voltage to the galvanometer's resistance and current: V = R_1 * I_galvanometer Now we can substitute the expression for voltage into our equivalent resistance formula to solve for R_eq: R_eq = (R_1 * I_galvanometer) / I_total
03

3. Determine the unknown parallel resistor value

: Now we will use the formula for equivalent resistance of parallel resistors to solve for the unknown parallel resistor value R_2: \( \frac{1}{R_{2}} = \frac{1}{R_{eq}} - \frac{1}{R_{1}} \)
04

4. Circuit diagram

: Draw a circuit diagram with the galvanometer (represented as a resistor with resistance 60.0 惟) connected in parallel with the calculated unknown resistor, to visualize the solution. Label the diagram appropriately, showing the ammeter reading 25.0 A full-scale. With these steps completed, you would have determined the resistance of the unknown parallel resistor that must be placed in parallel with the given galvanometer to allow it to be used as an ammeter with a 25.0 - A full-scale reading.

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Key Concepts

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

Parallel Circuits
Parallel circuits are a common way to configure electrical components in a circuit. In a parallel configuration, components like resistors are connected across the same two points in a circuit. This set-up is beneficial as it ensures that the full voltage of the source is applied to each component. Furthermore, it allows the total current to be divided among the parallel branches, thus enabling parallel circuits to handle high current loads efficiently.

This characteristic of sharing the current is highly utilized in electronics to ensure that components such as resistors, batteries, or lights do not burn out due to an excessive current flow. In the exercise given, a galvanometer is connected in parallel with another resistor, making them both subject to the same voltage. However, each part of the circuit will carry its own current, contributing to the total current experienced by the circuit.
  • In parallel circuits, the voltage is the same across all components.
  • The total current is the sum of the currents through each component.
  • The inverse of the total resistance ( (R_{eq} ) in a parallel circuit is the sum of the inverses of each resistance.
Ohm's Law
Ohm's Law is one of the fundamental principles of electronics that relates the voltage, current, and resistance in a circuit using the formula: \( V = IR \). Here, \( V \) stands for voltage, \( I \) for current, and \( R \) for resistance. This law helps in understanding how current behaves within a circuit when voltage and resistance are known.

When dealing with a galvanometer and the need to transform it into an ammeter, Ohm鈥檚 Law is crucial. We can calculate how the total current is split between the original galvanometer and any additional parallel resistor. This law allows for the calculation of the equivalent resistance \( (R_{eq}) \), which is needed to obtain a specific current reading, like the 25.0 A in the exercise.
  • Ohm鈥檚 Law is fundamental to understanding basic electrical circuits.
  • It is used to determine how voltage, current, and resistance interact in any part of a circuit.
  • The formula can be rearranged to solve for any one of the three variables if the other two are known.
Galvanometer Sensitivity
A galvanometer is a device used to detect and measure small electric currents. Galvanometer sensitivity refers to the smallest current that it can detect, usually measured in milliampere (mA). In the provided exercise, the galvanometer has a sensitivity of 1.00 mA, meaning it can sense very low currents, which is critical for precise measurements.

When converting a galvanometer to function as an ammeter, which measures larger currents, a shunt resistor with very low resistance is placed in parallel. This configuration directs most of the circuit's current through the shunt, while maintaining a small, quantifiable current through the galvanometer. Understanding the sensitivity helps in calculating the exact specifications of the shunt resistor necessary to reach the desired full-scale reading without saturating the galvanometer.
  • Galvanometer sensitivity is vital for accurate low-current measurements.
  • Increasing the current measuring capacity requires incorporating additional circuitry, such as parallel resistors.
  • Sensitivity is key to choosing the right configurations and components for adapted functionality, like transforming a galvanometer to an ammeter.

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Most popular questions from this chapter

What effect will the internal resistance of a rechargeable battery have on the energy being used to recharge the battery?

A 12.0-V emf automobile battery has a terminal voltage of \(16.0 \mathrm{V}\) when being charged by a current of 10.0 A. (a) What is the battery's internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in \(^{\circ} \mathrm{C} / \mathrm{min}\) ) will its temperature increase if its mass is \(20.0 \mathrm{kg}\) and it has a specific heat of \(0.300 \mathrm{kcal} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) assuming no heat escapes?

A car battery with a 12-V emf and an internal resistance of \(0.050 \Omega\) is being charged with a current of 60 A. Note that in this process, the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted into chemical energy?

A heart defibrillator passes 10.0 A through a patient's torso for \(5.00 \mathrm{ms}\) in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if \(500 \mathrm{J}\) of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the \(8.00 \mathrm{kg}\) of affected tissue.

(a) Given a 48.0-V battery and \(4.0-\Omega\) and \(96.0-\Omega\) resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.
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