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Chapter 20: Problem 56

Two identical resistors are connected in parallel across a 25-V battery, which supplies them with a total power of 9.6 W. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged. Find (a) the initial resistance of each resistor and (b) the total power delivered to the resistors after one resistor has been heated.

Short Answer

Expert verified
(a) Initial resistance: \( 130.2 \, \Omega \); (b) Power after heating: 7.2 W.

Step by step solution

01

Determine Total Initial Resistance

Let the initial resistance of each resistor be \( R \). When resistors are in parallel, the total resistance \( R_{parallel} \) is given by \( \frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \). Therefore, \( R_{parallel} = \frac{R}{2} \).
02

Use Power Formula to Find Initial Resistance

The total power \( P \) supplied by the battery is 9.6 W, and the voltage \( V \) is 25 V. Using the power formula \( P = \frac{V^2}{R_{parallel}} \), we solve for \( R_{parallel} \):\[ 9.6 = \frac{25^2}{R_{parallel}} \]Solving for \( R_{parallel} \), we find:\[ R_{parallel} = \frac{625}{9.6} \approx 65.1042 \, \Omega \]Given \( R_{parallel} = \frac{R}{2} \), therefore \( R = 2 \times 65.1042 \approx 130.2084 \, \Omega \).
03

Determine New Total Resistance After Heating

After heating, one resistor's resistance doubles, becoming \( 2R \). The other resistor remains \( R \). Now, the total resistance \( R'_{parallel} \) when they are in parallel is given by:\[ \frac{1}{R'_{parallel}} = \frac{1}{R} + \frac{1}{2R} = \frac{2}{2R} + \frac{1}{2R} = \frac{3}{2R} \]Thus, \( R'_{parallel} = \frac{2R}{3} \).
04

Calculate New Total Power Delivery

With the new total resistance \( R'_{parallel} = \frac{2R}{3} \) and using \( R = 130.2084 \, \Omega \), we find:\[ R'_{parallel} = \frac{2 \times 130.2084}{3} \approx 86.8056 \, \Omega \]The power supplied by the battery now is given by:\[ P' = \frac{V^2}{R'_{parallel}} = \frac{25^2}{86.8056} \approx 7.2 \, \text{W} \]
05

Conclusion

The initial resistance of each resistor is approximately \( 130.2 \, \Omega \) and the total power delivered to the resistors after one resistor has been heated is approximately 7.2 W.

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Key Concepts

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

Electric Circuits
Electric circuits are pathways that allow electric current to flow through them. They consist of various electrical components, like batteries, resistors, and wires, all connected to form a closed loop.
Resistors are elements that restrict the flow of electric current, and they can be connected in different configurations, such as series or parallel.
In a parallel circuit, like the one described in the exercise, multiple pathways allow the current to flow through each resistor independently. This means that the voltage across each resistor in a parallel configuration is the same.
  • The benefit of connecting resistors in parallel is that it decreases the total resistance of the circuit.
  • This gives multiple paths for the current, which affects the total power used by the circuit.
Understanding the structure of electric circuits helps in solving problems related to current flow and resistance distribution.
Power Formula
The power formula is a crucial concept in understanding electric circuits. Power in electrical terms is defined as the rate at which energy is used or transferred. It is measured in watts (W). In our exercise, the power delivered by the battery is crucial for determining the resistance of the resistors. The power formula is given by:\[ P = \frac{V^2}{R} \]Where:
  • \( P \) is the power in watts.
  • \( V \) is the voltage across the component in volts.
  • \( R \) is the resistance in ohms (\( \Omega \)).
This formula shows how power, voltage, and resistance are interconnected. By having the voltage and total power, you can calculate the resistance, which is a key step in the exercise as seen when determining the initial resistance of each resistor.
Ohm's Law
Ohm's Law is a fundamental principle in the study of electric circuits. It describes the relationship between voltage, current, and resistance in an electrical circuit. Ohm's Law is stated as:\[ V = IR \]Where:
  • \( V \) is the voltage across the resistor in volts.
  • \( I \) is the current flowing through the resistor in amperes.
  • \( R \) is the resistance in ohms (\( \Omega \)).
By knowing any two of these values, you can calculate the third. This is vital in electric circuits, as it helps to analyze how the circuit operates under different conditions. In the problem at hand, knowing the voltage and being able to calculate resistance helps to understand how much current flows and power is consumed after one resistor's resistance changes.
Resistance Calculation
Resistance calculation is an essential part of solving problems involving electric circuits. Resistance determines how much a resistor will oppose the flow of current. When resistors are configured in parallel, the calculation differs from when they are in series. For resistors in parallel, the total or equivalent resistance \( R_{parallel} \) is calculated with the formula:\[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \]In our exercise, two identical resistors are initially in parallel, so their equivalent resistance is reduced.
After one resistor's resistance doubles, the resistance calculation changes, impacting the total resistance and power delivered to the circuit.
  • Utilizing the resistance formula enables precise calculations of individual and total resistances in parallel circuits.
  • Knowing these calculations helps determine the efficiency and functionality of electric circuits under various conditions.
Understanding how to calculate resistance in different configurations is crucial for solving complex electric circuit problems.

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Most popular questions from this chapter

A circuit contains a resistor in series with a capacitor, the series combination being connected across the terminals of a battery, as in Figure 20.37\(a .\) The time constant for charging the capacitor is 1.5 \(\mathrm{s}\) when the resistor has a resistance of \(2.0 \times 10^{4} \Omega .\) What would the time constant be if the resistance had a value of \(5.2 \times 10^{4} \Omega ?\)

A coffee cup heater and a lamp are connected in parallel to the same \(120-\mathrm{V}\) outlet. Together, they use a total of 111 \(\mathrm{W}\) of power. The resistance of the heater is \(4.0 \times 10^{2} \Omega .\) Find the resistance of the lamp.

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In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7 A of current at 120 V. With the energy used by the clothes dryer, how long (in hours) could you use this computer to 鈥渟urf鈥� the Internet?

A battery delivering a current of 55.0 \(\mathrm{A}\) to a circuit has a terminal voltage of 23.4 \(\mathrm{V}\) . The electric power being dissipated by the internal resistance of the battery is 34.0 \(\mathrm{W}\) . Find the emf of the battery.
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