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Chapter 25: Problem 62

A resistor with resistance \(R\) is connected to a battery that has emf 12.0 \(\mathrm{V}\) and internal resistance \(r=0.40 \Omega .\) For what two values of \(R\) will the power dissipated in the resistor be 80.0 \(\mathrm{W} ?\)

Short Answer

Expert verified
Two possible resistance values will be calculated based on solving a quadratic equation derived from the power formula.

Step by step solution

01

Understanding the Power Dissipation Formula

To solve this problem, we need to apply the formula for power dissipated in a resistor:\[ P = \frac{V^2}{R + r} \cdot R \]where \( P \) is the power dissipated, \( V \) is the voltage provided by the battery, \( R \) is the resistance of the resistor, and \( r \) is the internal resistance of the battery. Here, \( P = 80.0 \text{ W} \), \( V = 12.0 \text{ V} \), and \( r = 0.40 \Omega \).

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Key Concepts

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

Power Dissipation
Power dissipation is a key concept in electric circuits, indicating how much energy is converted into heat or work within a component. The power dissipated in a resistor is calculated using the formula: \[P = \frac{V^2}{R + r} \cdot R\]This formula involves three critical quantities: the voltage (\( V \)) provided by the battery, the resistance of the resistor (\( R \)), and the internal resistance of the battery (\( r \)). In this equation:
  • \( P \) represents the power (in watts) dissipated.
  • \( V \) is the electromotive force (emf) of the battery (given in volts).
  • \( R \) is the resistance where power is dissipated.
  • \( r \) is the internal resistance of the battery.
Understanding how power dissipation changes with varying resistance values is crucial for designing circuits that efficiently manage energy.
Resistance
Resistance is a measure of how much a component resists the flow of electric current. This property is quantified in ohms (\( \Omega \)). Every resistor has a specific resistance value, which determines how much current will flow through it for a given voltage.Key aspects of resistance to remember:
  • High resistance means less current flow, lower power dissipation.
  • Low resistance leads to increased current flow, higher power dissipation when voltage is constant.
The resistance value in a circuit affects the overall power consumption and energy efficiency. In our context, adjusting the resistance value \( R \) directly affects the power dissipation according to the formula provided, allowing for different levels of power management in circuits.
Internal Resistance
Internal resistance is an innate resistance found within all energy sources, like batteries. It's a property that affects how efficiently a battery can deliver power to a circuit.Here's how it impacts a circuit:
  • A higher internal resistance reduces the effective voltage delivered to the load.
  • It also alters how much power is dissipated in the external resistor \( R \).
In problems like the one given, internal resistance \( r \) plays a significant role in calculations. It must be accounted for to understand the real voltage and current values across a circuit, altering calculations for power dissipation and energy efficiency.
Ohm's Law
Ohm's Law is a fundamental principle in electronics, describing the relationship between voltage, current, and resistance. It is expressed as:\[V = I \cdot R\]In this equation:
  • \( V \) is the voltage (in volts) across the resistance.
  • \( I \) is the current (in amperes) flowing through the circuit.
  • \( R \) is the resistance (in ohms).
Ohm's Law is crucial for understanding how different components in a circuit interact. It allows us to predict how changes in one quantity (like resistance) affect others (voltage, current). For example, if the resistance increases, the same voltage would result in less current flow, illustrating the balance between these electrical properties.

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Most popular questions from this chapter

A person with body resistance between his hands of 10 \(\mathrm{k} \Omega\) accidentally grasps the terminals of a \(14-\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 \(\mathrm{mA}\) or less?

The following measurements of current and potential difference were made on a resistor constructed of Nichrome wire: $$\begin{array}{llll}{\boldsymbol{I}(\mathbf{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\ {\boldsymbol{V}_{a b}(\mathbf{V})} & {1.94} & {3.88} & {7.76} & {15.52}\end{array}$$ (a) Graph \(V_{a b}\) as a function of \(I\) (b) Does Nichrome obey Ohm's law? How can you tell? (c) What is the resistance of the resistor in ohms?

A typical small flashlight contains two batteries, each having an emf of 1.5 \(\mathrm{V}\) , connected in series with a bulb having resistance 17\(\Omega\). (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for 5.0 h, what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

A Nonideal Ammeter. Unlike the idealized ammeter described in Section \(25.4,\) any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_{\mathrm{A}}\) is connected in series with a resistor \(R\) and a battery of emf \(\mathcal{E}\) and internal resistance \(r .\) The current measured by the ammeter is \(I_{\mathrm{A}}\) . Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_{A}, r, R_{\mathrm{A}},\) and \(R .\) The more "ideal" the ammeter, the smaller the difference between this current and the current \(I_{\mathrm{A}}\) . (b) If \(R=3.80 \Omega, \mathcal{E}=7.50 \mathrm{V},\) and \(r=0.45 \Omega,\) find the maximum value of the ammeter resistance \(R_{\mathrm{A}}\) so that \(l_{\mathrm{A}}\) is within 1.0\(\%\) of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

A current-carrying gold wire has diameter 0.84 \(\mathrm{mm}\) . The electric field in the wire is 0.49 \(\mathrm{V} / \mathrm{m} .\) What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 \(\mathrm{m}\) apart; \((\mathrm{c})\) the resistance of a \(6.4-\mathrm{m}\) length of this wire?
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