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A resistor with R1= 25.0 Ω is connected to a battery that has negligible internal resistance and electrical energy is dissipated by R1 at a rate of 36.0 W. If a second resistor with R2= 15.0 Ω is connected in series with R1, 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 P1=36.0W for R1=25.0Ω, use the formula for power: P1=I2R1Solving for I, we have:I2=P1R1=36.025.0=1.44Thus, I=1.44=1.2A.
02

Calculate Total Resistance

In a series circuit, the total resistance Rtotal is the sum of the individual resistances. Therefore, calculate it as follows:Rtotal=R1+R2=25.0Ω+15.0Ω=40.0Ω
03

Find Total Power Dissipated

Using the current I=1.2A found earlier, calculate the total power dissipated using:Ptotal=I2×RtotalSubstitute the known values:Ptotal=(1.2)2×40.0=1.44×40.0=57.6W

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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=IR
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=I2R=IV=V2R
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=I2R. With the power dissipated by resistor R1 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 (Ω) 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:
Rtotal=R1+R2++Rn
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

Two light bulbs have constant resistances of 400 Ω and 800 Ω. If the two light bulbs are connected in series across a 120-V line, find (a) the current through each bulb; (b) the power dissipated in each bulb; (c) the total power dissipated in both bulbs. The two light bulbs are now connected in parallel across the 120-V line. Find (d) the current through each bulb; (e) the power dissipated in each bulb; (f) the total power dissipated in both bulbs. (g) In each situation, which of the two bulbs glows the brightest? (h) In which situation is there a greater total light output from both bulbs combined?

A 224-Ω resistor and a 589-Ω resistor are connected in series across a 90.0-V line. (a) What is the voltage across each resistor? (b) A voltmeter connected across the 224-Ω resistor reads 23.8 V. Find the voltmeter resistance. (c) Find the reading of the same voltmeter if it is connected across the 589-Ω resistor. (d) The readings on this voltmeter are lower than the "true" voltages (that is, without the voltmeter present). Would it be possible to design a voltmeter that gave readings higher than the "true" voltages? Explain.

A 12.4-μF capacitor is connected through a 0.895-MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 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 s.

A 42-Ω resistor and a 20Ω resistor are connected in parallel, and the combination is connected across a 240-V dc line. (a) What is the resistance of the parallel combination? (b) What is the total current through the parallel combination? (c) What is the current through each resistor?

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 45.0 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

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