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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 second 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

Find Current for R1

First, find the current flowing through the first resistor using the power formula. Given the power \( P_1 = 36.0 \, \text{W} \) for \( R_1 = 25.0 \, \Omega \), use the formula for power: \[ P_1 = I^2 R_1 \]Solving for \( I \), we have:\[ I^2 = \frac{P_1}{R_1} = \frac{36.0}{25.0} = 1.44 \]Thus, \( I = \sqrt{1.44} = 1.2 \, \text{A} \).
02

Calculate Total Resistance

In a series circuit, the total resistance \( R_{\text{total}} \) is the sum of the individual resistances. Therefore, calculate it as follows:\[ R_{\text{total}} = R_1 + R_2 = 25.0 \, \Omega + 15.0 \, \Omega = 40.0 \, \Omega \]
03

Find Total Power Dissipated

Using the current \( I = 1.2 \, \text{A} \) found earlier, calculate the total power dissipated using:\[ P_{\text{total}} = I^2 \times R_{\text{total}} \]Substitute the known values:\[ P_{\text{total}} = (1.2)^2 \times 40.0 = 1.44 \times 40.0 = 57.6 \, \text{W} \]

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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 used to describe the relationship between voltage (\(V\)), current (\(I\)), and resistance (\(R\)) in an electrical circuit. This law states 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. The formula is expressed as:
\[V = I \, R\]
This equation is integral in solving many electric circuit problems, as it allows you to calculate one of the three elements if the other two are known. In the exercise, once the current through the first resistor was calculated using another form of power equation, Ohm's Law can further analyze the circuit.
  • Known **Relevance**: It relates voltage, current, and resistance, which are core properties of an electric circuit.
  • **Uses**: Calculate any missing circuit quantity when two quantities are known.
By applying Ohm’s Law in conjunction with the power concept, you can understand how components in an electrical circuit interact and influence energy dissipation.
Power Dissipation
Power dissipation in electrical circuits refers to the process in which electrical energy is converted into heat energy in resistive components. In resistors, power dissipation is expressed using the formula:
\[P = I^2 R = IV = \frac{V^2}{R}\]
Each form of the power equation highlights different circuit relationships. In our example problem, the first step was to determine the power dissipation using \(P = I^2 R\). With the power dissipated by resistor \(R_1\) provided, it was used to find the circuit current. This current then applies to all components since they are in series:
  • **Practical Application**: Ensures components operate safely and aren't overloaded with excess heat.
  • **Connection**: Ties in with energy conservation principles and overall circuit efficiency.
Understanding power dissipation is essential because it helps assess a circuit's energy use efficiently, guiding decisions for energy distribution or component choices to improve energy efficiency.
Electrical Resistance
Electrical resistance is the opposition that a material offers to the flow of electric current. It is quantified in ohms (\(\Omega\)) and depends on various factors, including the material type, cross-sectional area, and temperature. In any circuit, resistance affects both current and power. In series circuits specifically, the total resistance is the sum of individual resistances:
\[R_{\text{total}} = R_1 + R_2 + \ldots + R_n\]
In the provided problem, this leads to calculating the overall effect of two resistors in series, enabling the correct computation of total power dissipation:
  • **Impact**: Determines current flow and voltage distribution within a circuit.
  • **Significance in Series Circuits**: Each resistor adds to the total resistance, affecting current flow similarly throughout components.
Recognizing how resistance behaves in series configurations helps diagnose circuit behaviors and guides the design for specific voltage or current requirements.

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Most popular questions from this chapter

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (\(Note:\) This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer's heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

A 12.0-mF capacitor is charged to a potential of 50.0 V and then discharged through a 225-\(\Omega\) resistor. How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy?

Three resistors having resistances of 1.60 \(\Omega\), 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 \(2.00-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(6.00-\mathrm{k} \Omega\) resistor and an emf source with \(\mathcal{E}=90.0 \mathrm{~V}\) and negligible internal resistance. The circuit is completed at \(t=0 .\) (a) Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? (b) At what value of \(t\) is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor? (c) At the time calculated in part (b), what is the rate at which electrical energy is being dissipated in the resistor?

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