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Chapter 19: Problem 85

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 27 \(\mathrm{W}\) . What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

Short Answer

Expert verified
The power dissipated in parallel is 243 W.

Step by step solution

01

Understand the Problem

We have three identical resistors connected in series, and we know that when a certain voltage is applied, the total power dissipated is 27 W. The question asks us to find the power dissipated when these resistors are connected in parallel with the same voltage applied.
02

Recall the Formula for Power Dissipated

The power dissipated by a resistor can be calculated using the formula: \[ P = \frac{V^2}{R} \]where \( V \) is the voltage across the resistor and \( R \) is the resistance.
03

Power for Series Connection

In series, the total resistance is the sum of individual resistances, i.e., \( R_s = 3R \). Thus, the power dissipated is:\[ P = \frac{V^2}{3R} = 27 \] This allows us to find \( R \) in terms of \( V \): \[ R = \frac{V^2}{81} \].
04

Find Total Resistance for Parallel Connection

In parallel, the total resistance is given by: \[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \]So, \( R_p = \frac{R}{3} \).
05

Calculate Power for Parallel Connection

Using the formula for power, the power dissipated in the parallel arrangement is:\[ P_p = \frac{V^2}{R_p} = \frac{V^2}{\frac{R}{3}} = 3 \cdot \frac{V^2}{R} \]We already know \( \frac{V^2}{R} = 81 \) (from Step 3), so:\[ P_p = 3 \times 81 = 243 \text{ W} \].
06

Conclusion

When the three identical resistors are connected in parallel across the same potential difference, the power dissipated is 243 W.

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Key Concepts

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

Series and parallel circuits
Imagine resistors as a chain of hurdles for electric current. They can be arranged in sequences called circuits. In a **series circuit**, all resistors are lined up one after another:
  • Current has one pathway to flow.
  • The total resistance increases as you add more resistors.
Think of it as water flowing through a pipe. If you make the pipe longer (adding more resistors), it becomes tougher for water to flow, thus more effort (or voltage) is needed. On the other hand, in a **parallel circuit**, resistors branch out:
  • Current has multiple pathways to flow.
  • The total resistance decreases as more resistors are added.
Imagine several pipes converging into one. Water (current) can select which route to take, facing less resistance overall.
Electric power calculation
To unravel electric power, think about the work done by electricity in a circuit. We determine it with the formula:
  • \[ P = \frac{V^2}{R} \]
Where:
  • \( P \) is the power in watts, reflecting energy consumption or dissipation.
  • \( V \) represents voltage across the resistor.
  • \( R \) is the resistance, opposing current flow.
These elements interplay just like pushing a box over a surface:
  • **Voltage (\( V \))**: Like the push force on the box.
  • **Resistance (\( R \))**: Like friction hindering movement.
  • **Power (\( P \))**: The effort exerted to keep the box moving.
By controlling these factors, we balance the energy used and lost to keep devices running smoothly.
Resistor combinations
Resistors, often called the guardians of the circuit, can be combined in multiple ways to tailor circuit behavior. Understanding the impact of different combinations helps optimize electric circuits effectively.**Series Combination**:
  • Resistors add up, resulting in: \( R_s = R_1 + R_2 + R_3 + ... \)
  • The same current flows through each resistor, but the voltage varies across each one.
  • Overall, uses more voltage to get the current moving.
**Parallel Combination**:
  • Total resistance is less than any single resistor: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... \)
  • Voltage remains consistent across all resistors, but the current divides.
  • Benefits in allowing more current to flow through the circuit.
These combinations impact how resistors handle electricity, determining energy usage and efficiency in circuits. Understanding them is like mastering a puzzle, ensuring all pieces fit precisely for optimal function.

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Most popular questions from this chapter

A carbon resistor having a temperature coefficient of resistivity of \(-0.00050\left(\mathrm{C}^{\circ}\right)^{-1},\) is to be used as a thermometer. On a winter day when the temperature is \(4.0^{\circ} \mathrm{C},\) the resistance of the carbon resistor is 217.3\(\Omega .\) What is the temperature on a spring day when the resistance is 215.8\(\Omega ?\) (Take the reference temperature \(T_{0}\) to be \(4.0^{\circ} \mathrm{C.}\) .

Typical household currents are on the order of a few amperes. If a 1.50 A current flows through the leads of an electrical appliance, (a) how many electrons per second pass through it, (b) how many coulombs pass through it in 5.0 min, and (c) how long does it take for 7.50 \(\mathrm{C}\) of charge to pass through?

A \(540-\mathrm{W}\) electric heater is designed to operate from 120 \(\mathrm{V}\) lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

A fully charged 6.0\(\mu\) F capacitor is connected in series with a \(1.5 \times 10^{5} \Omega\) resistor. What percentage of the original charge is left on the capacitor after 1.8 s of discharging?

Electricity through the body, II. The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0\(\Omega \cdot \mathrm{m} .\) The conducting path between the hands can be represented approximately as a cylinder 1.6 \(\mathrm{m}\) long and 0.10 \(\mathrm{m}\) in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of 100 \(\mathrm{mA}\) ? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?
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