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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
Upon solving the quadratic equation, we find the two possible values for the resistance \(R\) that will result in a power dissipation of 80W.

Step by step solution

01

Use the Power Formula

The power dissipated by a resistor (in Watts) is given by \(P= I^2 * R\), where \(I\) is the current (in Amps), and \(R\) is the resistance (in Ohms). However, in this case, we also have to consider the influence of the internal resistance of the battery. Recall that the current provided by the battery can be given by Ohm's law as \(I= \frac{{\varepsilon}}{{R + r}}\) where \(\varepsilon\) is the electromotive force or voltage provided by the battery, \(R\) is the resistance, and \(r\) is the internal resistance of the battery.
02

Substitute the Current Equation into the Power Equation

Plug the above equation of \(I\) into the equation for \(P\) to get a new equation for \(P\) in terms of \(R\), \(\varepsilon\), and \(r\). \(P = \frac{{\varepsilon^2 * R}}{{(R + r)^2}}\)
03

Set Up the Quadratic Equation

We know that \(P\) is 80W and \(\varepsilon\) is 12V. Substituting these values, along with the given \(r\) value of 0.40 \(\Omega\), into the equation we then get: \(80 = \frac{{12^2 * R}}{{(R + 0.40)^2}}\). This equation can be simplified to obtain a quadratic equation.
04

Solve the Quadratic Equation

The quadratic equation can be solved using the quadratic formula \(R = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\) or by factoring. However, this is a real world problem and only a positive value for R will make sense in this context. By solving the quadratic equation, you will obtain two values for \(R\).

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the world of electricity and electronics. It relates three key quantities: voltage (\(V\)), current (\(I\)) and resistance (\(R\)). The law is simply expressed as:
\[ V = I \times R \]This equation tells us that the voltage across a resistor is the product of the current flowing through it and its resistance. To find any one of these values, you can rearrange the equation:
  • To find current: \( I = \frac{V}{R} \)

  • To find resistance: \( R = \frac{V}{I} \)
Ohm's Law helps us understand the relationship between these variables. When voltage increases and resistance remains constant, the current also increases. It’s an essential concept for circuit analysis, as it allows us to calculate how much current flows through a resistor when a known voltage is applied.
Power Dissipation
Power dissipation in electrical circuits refers to the conversion of electrical energy into heat energy. This is particularly relevant for resistors, which consume electrical power and dissipate it as heat. The power dissipated by a resistor with known current can be calculated using the equation:
\[ P = I^2 \times R \]Here, \( P \) represents the power in watts, \( I \) is the current in amps, and \( R \) is the resistance in ohms.If we know the voltage across the resistor, we can also use:
\[ P = \frac{V^2}{R} \]Both formulas underscore the dependency between power, voltage, current, and resistance. When analyzing a circuit, it is crucial to consider power dissipation to prevent potential overheating, which could result in damage to components or reduced efficiency. Knowing how to calculate power dissipation helps maintain a safe and effective electronic or electrical system.
Resistor
Resistors are fundamental components in electronic circuits. Their primary function is to resist or restrict the flow of electrical current, providing precise control over the circuit. Resistance is measured in ohms (\( \Omega \)), and each resistor has a specific resistance value that determines how it limits current.Resistors are categorized based on material, size, and power rating. Here's how they work:
  • They help control voltage levels within a circuit.

  • They protect sensitive components by limiting current that could otherwise cause overheating or damage.

  • They play critical roles in timing circuits, voltage dividers, and signal conditioning.
In combination with Ohm’s Law and power equations, understanding how resistors affect circuits enables engineers and technicians to design and troubleshoot various electrical systems.
Internal Resistance
Internal resistance is a characteristic of batteries and power sources that indicates resistance within the power source itself. Although not always desirable, it's inherent in real-world power supplies due to the materials and chemical processes involved.This small resistance affects the total output voltage and current of a battery. When current flows, internal resistance causes a voltage drop inside the battery, reducing the effective voltage available to the external circuit components. The effective voltage can be calculated as:
\[ V_{\text{effective}} = \varepsilon - I \times r \]where \( \varepsilon \) is the electromotive force, \( I \) is the current, and \( r \) is the internal resistance.Considering internal resistance is essential for calculating precise voltage and current distributions in circuits, particularly when high accuracy is necessary. It helps in understanding the limitations of battery-supplied circuits and the importance of selecting appropriate power sources for different applications.

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Most popular questions from this chapter

A metal ring \(4.50 \mathrm{~cm}\) in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of \(1.12 \mathrm{~T}\) between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at \(0.250 \mathrm{~T} / \mathrm{s}\). (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

Current passes through a solution of sodium chloride. In \(1.00 \mathrm{~s}, 2.68 \times 10^{16} \mathrm{Na}^{+}\) ions arrive at the negative electrode and \(3.92 \times 10^{16} \mathrm{Cl}^{-}\) ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

A battery has emf \(30.0 \mathrm{~V}\) and internal resistance \(r .\) A \(9.00 \Omega\) resistor is connected to the terminals of the battery, and the voltage drop across the resistor is \(27.0 \mathrm{~V}\). What is the internal resistance of the battery?

A circular loop of wire with a radius of \(12.0 \mathrm{~cm}\) and oriented in the horizontal \(x y\) -plane is located in a region of uniform magnetic field. A field of \(1.5 \mathrm{~T}\) is directed along the positive \(z\) -direction, which is upward. (a) If the loop is removed from the field region in a time interval of \(2.0 \mathrm{~ms}\), find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

In Fig. 29.23 the capacitor plates have area \(5.00 \mathrm{~cm}^{2}\) and separation \(2.00 \mathrm{~mm} .\) The plates are in vacuum. The charging current \(i_{\mathrm{C}}\) has a constant value of \(1.80 \mathrm{~mA}\). At \(t=0\) the charge on the plates is zero. (a) Calculate the charge on the plates, the electric field between the plates, and the potential difference between the plates when \(t=0.500 \mu \mathrm{s}\). (b) Calculate \(d E / d t,\) the time rate of change of the electric field between the plates. Does \(d E / d t\) vary in time? (c) Calculate the displacement current density \(j_{\mathrm{D}}\) between the plates, and from this the total displacement current \(i_{\mathrm{D}}\). How do \(i_{\mathrm{C}}\) and \(i_{\mathrm{D}}\) compare?

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