/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 (a) What is the internal resista... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by \(5.00 \mathrm{A} ?\) (b) Can the emf of the voltage source be found with the information supplied?

Short Answer

Expert verified
The internal resistance of the voltage source is 0.4 ohms. The emf of the voltage source cannot be found with the information supplied.

Step by step solution

01

Understanding the Concept of Internal Resistance

Internal resistance is a characteristic of a voltage source that causes a drop in the terminal voltage as the current supplied by the source increases. This phenomenon is described by the formula \(V = \text{emf} - Ir\), where \(V\) is the terminal voltage, \(\text{emf}\) is the electromotive force (or simply emf), \(I\) is the current, and \(r\) is the internal resistance.
02

Applying Ohm's Law to Find Internal Resistance

According to Ohm's law for a voltage source, \(\text{emf} = V + Ir\). Considering that the terminal voltage drops by 2.00 V when the current increases by 5.00 A, we can rearrange this formula to solve for \(r\). The change in voltage (denoted as \(\triangle V\)) is related to the change in current (\(\triangle I\)) by the equation \(\triangle V = -r \triangle I\). By substituting the given values, the internal resistance can be calculated.
03

Calculating the Internal Resistance

Substitute the change in voltage and the change in current into the equation \(r = -\frac{\triangle V}{\triangle I}\) to find the internal resistance. Thus \(r = -\frac{2.00\text{ V}}{5.00\text{ A}}\).
04

Simplifying the Calculation

Simplify the fraction to find the internal resistance \(r = -\frac{2.00}{5.00} = -0.4\text{ ohms}\). The negative sign indicates that we are considering a drop in voltage. However, resistance is always positive, so we take the absolute value: \(r = 0.4\text{ ohms}\).
05

Discussing the Possibility of Finding emf

Without the initial terminal voltage or the emf of the voltage source before and after the change in current, we cannot determine the exact emf value. We only have the information about the change in voltage due to a change in current caused by the internal resistance.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Terminal Voltage
Terminal voltage is a crucial concept in the study of circuits and electronics. It refers to the voltage measured across the terminals of a power source when a load is connected. This voltage is what is effectively available for use by the external circuit. It is essential to understand that the terminal voltage of a battery or other voltage source will differ from its emf when there is a current flowing due to the presence of internal resistance.

The equation that describes this relationship is given by:
\( V = \text{emf} - I \cdot r \)
where \( V \) represents the terminal voltage, \( \text{emf} \) is the electromotive force, \( I \) is the current, and \( r \) is the internal resistance of the source.
In the context of the exercise, a drop in terminal voltage by 2.00 V when the current increases suggests that internal resistance is at play. By understanding how terminal voltage is influenced by current and internal resistance, it becomes possible to calculate the unknown internal resistance using these measured changes.
Electromotive Force (emf)
The term electromotive force, or emf, represents the energy provided by a power source per unit charge that passes through it. Although referred to as a 'force,' it is actually a potential difference measured in volts. Emf is the maximum voltage a power source can provide when no current is flowing, which means there is no voltage drop due to internal resistance. The emf can be thought of as the 'full' or 'ideal' voltage output of the source, before any internal factors, such as resistance, come into play.

To connect it to our original exercise, even though we can determine the internal resistance with the given information, we cannot ascertain the exact emf without knowing the initial terminal voltage or emf and the change in current. The equation involving emf, terminal voltage, and internal resistance is:
\( \text{emf} = V + I \cdot r \)
Understanding emf allows us to see the big picture of how power sources behave and influence the functioning of circuits.
Ohm's Law
Ohm's law is a fundamental principle in the field of electrical engineering and physics that relates current (\(I\)), voltage (\(V\)), and resistance (\(R\)) in a simple and direct manner. The law is mathematically represented as:
\( V = I \cdot R \)
This equation tells us that the voltage across a conductor is directly proportional to the current flowing through it, with the resistance being the factor of proportionality.

When it comes to internal resistance, Ohm’s law is modified to account for the voltage source itself, leading us to the formula:
\( \text{emf} = V + I \cdot r \)
Through this modified version, we see how the internal resistance of a voltage source affects the terminal voltage. In our original exercise, Ohm’s law helps us understand why there is a drop in terminal voltage with an increase in current and stands as the basis for calculating the internal resistance by relating the change in voltage to the change in current through the resistor.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An ECG monitor must have an \(R C\) time constant less than \(1.00 \times 10^{2}\) \mus to be able to measure variations in voltage over small time intervals. (a) If the resistance of the circuit (due mostly to that of the patient's chest) is \(1.00 \mathrm{k} \Omega\) what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)?

Why is a conventional fission nuclear reactor not able to explode as a bomb?

Another set of reactions that result in the fusing of hydrogen into helium in the Sun and especially in hotter stars is called the carbon cycle. It is $$ \begin{aligned} { }^{12} \mathrm{C}+{ }^{1} \mathrm{H} & \rightarrow{ }^{13} \mathrm{~N}+\gamma \\ { }^{13} \mathrm{~N} & \rightarrow{ }^{13} \mathrm{C}+e^{+}+v_{e}, \\ { }^{13} \mathrm{C}+{ }^{1} \mathrm{H} & \rightarrow{ }^{14} \mathrm{~N}+\gamma, \\ { }^{14} \mathrm{~N}+{ }^{1} \mathrm{H} & \rightarrow{ }^{15} \mathrm{O}+\gamma, \\ { }^{15} \mathrm{O} & \rightarrow{ }^{15} \mathrm{~N}+e^{+}+v_{e}, \\ { }^{15} \mathrm{~N}+{ }^{1} \mathrm{H} & \rightarrow{ }^{12} \mathrm{C}+{ }^{4} \mathrm{He} . \end{aligned} $$ Write down the overall effect of the carbon cycle (as was done for the proton- proton cycle in \(2 e^{-}+4^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 v_{e}+6 \gamma\) ). Note the number of protons \(\left({ }^{1} \mathrm{H}\right)\) required and assume that the positrons \(\left(e^{+}\right)\) annihilate electrons to form more \(\gamma\) rays.

In considering potential fusion reactions, what is the advantage of the reaction \({ }^{2} \mathrm{H}+{ }^{3} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+n\) over the reaction \({ }^{2} \mathrm{H}+{ }^{2} \mathrm{H} \rightarrow{ }^{3} \mathrm{He}+n ?\)

The timing device in an automobile's intermittent wiper system is based on an \(R C\) time constant and utilizes a \(0.500-\mu \mathrm{F}\) capacitor and a variable resistor. Over what range must \(R\) be made to vary to achieve time constants from 2.00 to \(15.0 \mathrm{s} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.