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A resistor with \(R_{1}=25.0 \Omega\) is connected to a battery that has negligible internal resistance and electrical energy is dissipated by \(R_{1}\) at a rate of 36.0 W. If a scond resistor with \(R_{2}=15.0 \Omega\) is connected in series with \(R_{1},\) what is the total rate at which electrical energy is dissipated by the two resistors?

Short Answer

Expert verified
The total power dissipated is 57.6 W.

Step by step solution

01

Calculate the Current through the First Resistor

First, let's determine the current flowing through the original circuit when only \(R_1\) is connected. We know that the power \(P\) dissipated in a resistor is given by \(P = I^2 R\). Therefore, \(I^2 = \frac{P}{R}\). For \(R_1 = 25.0 \Omega\) and \(P = 36.0\) W, we can calculate: \[I^2 = \frac{36.0}{25.0}\]Calculating this gives:\[I^2 = 1.44\]Then, the current \(I\) is:\[I = \sqrt{1.44} = 1.2\text{ A}\]
02

Determine Total Resistance in Series

When a second resistor \(R_2 = 15.0 \Omega\) is added in series to \(R_1\), the total resistance \(R_{\text{total}}\) becomes the sum of the two resistances:\[R_{\text{total}} = R_1 + R_2 = 25.0\, \Omega + 15.0\, \Omega = 40.0\, \Omega\]
03

Calculate Total Power Dissipated in Series Circuit

Now that we know both the current \(I = 1.2\, \text{A}\) and the total resistance in the series, we can calculate the total power dissipated using the formula:\[P_{\text{total}} = I^2 R_{\text{total}}\]Substituting the known values:\[P_{\text{total}} = (1.2)^2 \times 40.0\]\[P_{\text{total}} = 1.44 \times 40.0 = 57.6\, \text{W}\]
04

Conclusion

The total rate at which electrical energy is dissipated by the two resistors when they are connected in series is \(57.6\, \text{W}\).

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Key Concepts

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

Power Dissipation
When we discuss power dissipation in an electric circuit, we refer to the rate at which electrical energy is converted into heat energy in a resistor. Power in the context of resistors is typically calculated using the formula:
  • \( P = I^2 R \)
Here, \( P \) is the power dissipated in watts, \( I \) is the current through the resistor in amperes, and \( R \) is the resistance in ohms.
This formula highlights the relationship between the current flowing through a resistor and the energy that is transformed into heat. This power dissipation is an essential aspect of electrical circuits because it influences how resistors manage energy and heat.
Resistors are designed to handle a specific power rating. Exceeding this rating can lead to overheating or damage to the component. By understanding power dissipation, we can select appropriate resistors that function safely within the circuit, preventing failures and ensuring efficiency.
Series Resistors
In electrical circuits, resistors are often arranged in series to achieve a desired total resistance. When connected in series, the resistors share the same current, but the voltage across each may vary depending on their resistance values.
The total resistance in a series circuit is simply the sum of each individual resistance:
  • \( R_{\text{total}} = R_1 + R_2 + \ldots + R_n \)

This concept is straightforward but crucial, as it affects how the circuit behaves overall. With increased total resistance from multiple resistors in series, the circuit will draw less current for a given voltage than if fewer resistors were present.
Understanding series resistors is vital for designing circuits with specific resistance requirements. For situations like our exercise, knowing how to calculate the total resistance helps in determining other factors such as overall power dissipation.
Ohm's Law
Ohm's Law is a fundamental principle in the study of electricity, forming the basis of understanding how currents, voltages, and resistances interrelate. It is expressed through the equation:
  • \( V = I R \)
Here, \( V \) represents voltage in volts, \( I \) is the current in amperes, and \( R \) is resistance in ohms. This relationship shows that the voltage across a resistor is directly proportional to the current flowing through it and its resistance.
Ohm's Law is key to solving many electrical problems, as it allows you to determine any one of the three variables if the other two are known.
In practice, this law helps engineers and users design circuits to achieve desired electrical behavior. By selecting appropriate resistances, modifying currents, or adjusting voltages based on Ohm's Law, better and safer electrical installations can be achieved. Learning and applying this law ensures a deeper understanding of electric circuits and their operation.

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Most popular questions from this chapter

Working Late! You are working late in your electronics shop and find that you need various resistors for a project. But alas, all you have is a big box of 10.0 .0 resistors. Show how you can make each of the following equivalent resistances by a combination of your \(10.0-\Omega\) resistors: (a) \(35 \Omega,(\mathrm{b}) 1.0 \Omega,(\mathrm{c}) 3.33 \Omega,\) (d) 7.5\(\Omega\)

A \(12.0-\mu \mathrm{F}\) capacitor is charged to a potential of 50.0 \(\mathrm{V}\) and then discharged through a 175 - \(\Omega\) resistor. How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy?

A resistor with \(R=850 \Omega\) is connected to the plates of a charged capacitor with capacitance \(C=4.62 \mu \mathrm{F}\) . Just before the connection is made, the charge on the capacitor is 6.90 \(\mathrm{mC.}\) (a) What is the energy initially stored in the capacitor? (b) What is the electrical power dissipated in the resistor just after the connection is made? (c) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part (a)?

Three resistors having resistances of 1.60 ? \(2,2.40 \Omega,\) and 4.80\(\Omega\) are connected in parallel to a 28.0 -V battery that has negligible internal resistance. Find (a) the equivalent resistance of the combination; (b) the current in each resistor; (c) the total current through the battery; (d) the voltage across each resistor; (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power. the one with the greatest resistance or the least resistance? Explain why this should be.

A \(12.4-\mu \mathrm{F}\) capacitor is connected through a \(0.895-\mathrm{M\Omega}\) resistor to a constant potential difference of 60.0 \(\mathrm{V}\) . (a) Compute the charge on the capacitor at the following times after the connections are made: \(0,5.0 \mathrm{s}, 10.0 \mathrm{s}, 20.0 \mathrm{s},\) and 100.0 \(\mathrm{s}\) . (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and 20 \(\mathrm{s}\) .

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