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Chapter 25: Problem 86

A source with emf \(\mathcal{E}\) and internal resistance \(r\) is connected to an external circuit. (a) Show that the power output of the source is maximum when the current in the circuit is one-half the short-circuit current of the source. (b) If the external circuit consists of a resistance \(R,\) show that the power output is maximum when \(R=r\) and that the maximum power is \(\mathcal{E}^{2} / 4 r_{1}\)

Short Answer

Expert verified
The power is maximum when the current is half the short-circuit current and \( R = r \), with maximum power \( \mathcal{E}^2 / 4r \).

Step by step solution

01

Circuit Analysis and Short-Circuit Current

Consider the circuit where an EMF source \( \mathcal{E} \) is in series with an internal resistance \( r \) and an external resistance \( R \). The total resistance of the circuit is \( R + r \). Hence, using Ohm's law, the current \( I \) is given by:\[ I = \frac{\mathcal{E}}{R + r} \]In the short-circuit condition, \( R = 0 \), so the short-circuit current \( I_{sc} \) is:\[ I_{sc} = \frac{\mathcal{E}}{r} \]
02

Condition for Maximum Power Output

The power \( P \) delivered to the external resistance \( R \) is given by:\[ P = I^2 R = \left( \frac{\mathcal{E}}{R + r} \right)^2 R \]To find the condition for maximum power, compute the derivative of \( P \) with respect to \( R \) and set it to zero:\[ \frac{dP}{dR} = \frac{d}{dR} \left[ \frac{\mathcal{E}^2 R}{(R + r)^2} \right] = 0 \]
03

Solving the Derivative Equation

Differentiate the power expression:\[ \frac{dP}{dR} = \frac{\mathcal{E}^2 (R + r)^2 - 2\mathcal{E}^2 R (R + r)}{(R + r)^4} = 0 \]Simplifying, we find:\[ (R + r)^2 - 2R(R + r) = R + r - 2R = r - R = 0 \]Thus, \( R = r \).
04

Maximum Power Calculation

Using \( R = r \) to find the maximum power:\[ I = \frac{\mathcal{E}}{2r} \]The power is then:\[ P_{max} = I^2 R = \left( \frac{\mathcal{E}}{2r} \right)^2 r = \frac{\mathcal{E}^2}{4r} \]
05

Relating Current to Half of Short-Circuit Current

From Step 1, the short-circuit current \( I_{sc} = \frac{\mathcal{E}}{r} \). At maximum power, the actual current \( I = \frac{\mathcal{E}}{2r} \), showing:\[ I = \frac{1}{2} I_{sc} \]Hence, the power output is maximum when the current in the circuit is one-half of the short-circuit current.

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Key Concepts

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

EMF and Internal Resistance
When discussing circuits, the terms **emf (electromotive force)** and **internal resistance** are crucial. Electromotive force, denoted by \( \mathcal{E} \), refers to the potential difference generated by a source when no current is flowing. It is essentially the maximum energy per charge a power source can provide. Internal resistance, symbolized as \( r \), is the inherent resistance within the power source itself, which causes some of the energy to be lost as heat before it even leaves the source.
  • **EMF:** The total potential difference produced by the source.
  • **Internal Resistance:** Resistance within the source that affects the total output voltage.
Understanding these concepts is vital for analyzing how energy is distributed in circuits, particularly when aiming to optimize power output.
Ohm's Law
**Ohm's Law** is a fundamental principle in circuit analysis, stating that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor. Mathematically, it is expressed as:\[ V = I imes R \]Where:
  • \( V \) is the voltage across the conductor
  • \( I \) is the current flowing through it
  • \( R \) is the resistance of the conductor
In the context of the given problem, Ohm's Law helps to determine the current in a circuit both under normal and short-circuit conditions. It forms the basis for deriving the short-circuit current and understands energy transfer in the circuit.
Short-Circuit Current
The **short-circuit current**, \( I_{sc} \), is the maximum current that can flow through a circuit when the external resistance (\( R \)) goes to zero, effectively removing any barriers to current flow. It can be calculated using the internal resistance and emf of the source:\[ I_{sc} = \frac{\mathcal{E}}{r} \]This formula indicates that the short-circuit current is inversely proportional to the internal resistance of the source, meaning that less internal resistance allows more current to flow freely when there's no external load. Understanding this concept is crucial for analyzing maximum power transfer conditions, which occur only under certain circuit configurations.
Circuit Analysis
**Circuit analysis** is the process of determining the voltages across, and the current through, every component in an electrical circuit. For our problem, it involves understanding how components like resistances and power sources interact.
  • Calculate total resistance as \( R + r \), where \( R \) is external resistance and \( r \) is internal resistance.
  • Determine current using Ohm鈥檚 Law: \( I = \frac{\mathcal{E}}{R + r} \).
The analysis focuses on how altering external resistance impacts power output. By setting the derivative of power with respect to \( R \) to zero, we can find the configuration that maximizes power output, which is essential in optimizing circuit performance for various applications.
Power Output Optimization
**Power output optimization** in circuits aims to adjust conditions so that energy is most efficiently transferred from the source to the load. Maximizing power output depends on matching the external resistance (\( R \)) to the source's internal resistance (\( r \)).
  • The maximum power transfer theorem states that this is achieved when \( R = r \).
  • At this condition, the current is half of the short-circuit current, i.e., \( I = \frac{1}{2} I_{sc} \).
  • The maximum power is given by \( \frac{\mathcal{E}^2}{4r} \).
By choosing the external resistance to equal the internal resistance, circuits can deliver maximum power to the load, a principle widely applied in electrical and electronic design.

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