Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Short Answer

Expert verified
(a) 0.80 Ω, (b) 17.5 A, 11.67 A, 5.83 A, (c) 35 A, (d) 28.0 V, (e) 490 W, 327 W, 163 W, (f) The least resistance dissipates the most power.

Step by step solution

01

Find the Equivalent Resistance

When resistors are in parallel, the equivalent resistance \( R_{eq} \) is given by the formula: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] where \( R_1 = 1.60 \, \Omega \), \( R_2 = 2.40 \, \Omega \), and \( R_3 = 4.80 \, \Omega \). Substituting the given values:\[ \frac{1}{R_{eq}} = \frac{1}{1.60} + \frac{1}{2.40} + \frac{1}{4.80} \]Calculating the reciprocals:\[ \frac{1}{R_{eq}} = 0.625 + 0.417 + 0.208 \]\[ \frac{1}{R_{eq}} = 1.25 \]Finally, the equivalent resistance is:\[ R_{eq} = \frac{1}{1.25} = 0.80 \Omega \].
02

Find the Current through Each Resistor

The current through a resistor in a parallel circuit can be calculated using Ohm's Law \( I = \frac{V}{R} \) where \( V = 28.0 \, V \) is the voltage across each resistor.For \( R_1 = 1.60 \, \Omega \): \[ I_1 = \frac{28.0}{1.60} = 17.5 \, A \]For \( R_2 = 2.40 \, \Omega \): \[ I_2 = \frac{28.0}{2.40} = 11.67 \, A \]For \( R_3 = 4.80 \, \Omega \):\[ I_3 = \frac{28.0}{4.80} = 5.83 \, A \].
03

Find the Total Current through the Battery

The total current through the battery is the sum of the currents through each resistor:\[ I_{total} = I_1 + I_2 + I_3 \]\[ I_{total} = 17.5 + 11.67 + 5.83 = 35 \, A \].
04

Check the Voltage across Each Resistor

In a parallel circuit, the voltage across each resistor is equal to the voltage of the battery. Therefore, the voltage across each of the resistors is \( 28.0 \, V \).
05

Calculate the Power Dissipated in Each Resistor

The power dissipated in a resistor can be calculated using the formula \( P = I^2 R \).For \( R_1 \):\[ P_1 = (17.5)^2 \times 1.60 = 490 \, W \]For \( R_2 \):\[ P_2 = (11.67)^2 \times 2.40 = 327 \, W \]For \( R_3 \):\[ P_3 = (5.83)^2 \times 4.80 = 163 \, W \].
06

Determine which Resistor Dissipates the Most Power

The resistor with the least resistance (\( R_1 = 1.60 \, \Omega \)) dissipates the most power since it has the highest current flow, following the relationship \( P = I^2 R \). Lower resistance in parallel circuits leads to higher current, causing greater power dissipation despite the higher resistance values.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ohm's Law
Ohm's Law is a basic principle in electronics that helps us understand the relationship between voltage, current, and resistance. It's expressed with the formula \( V = IR \).
In this formula:
  • \( V \) represents voltage (in volts). It is the potential difference that makes electric charges move.
  • \( I \) denotes current (in amperes). It measures the flow of electric charge.
  • \( R \) stands for resistance (in ohms). It indicates how much a material resists the flow of current.
In a parallel circuit, like in this exercise, the voltage across each component is equal. Therefore, when calculating the current through each resistor using Ohm’s Law, we use the formula \( I = \frac{V}{R} \).
This shows how current varies with different resistances while keeping voltage constant. In simpler terms, if you know two of these values, you can find the third!
Equivalent Resistance
In circuits, equivalent resistance is the single resistance that can replace a combination of resistors, giving the same effect on current flow. For resistors in parallel, calculating equivalent resistance is vital to simplifying the circuit analysis. The formula used here is:\[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]Where \( R_1, R_2, R_3 \) are the resistances of different components. By calculating reciprocals of the resistances, we find that:\[R_{eq} = \frac{1}{1.25} = 0.80\, \Omega\]This value, \( 0.80 \, \Omega \), is smaller than the smallest individual resistor in the parallel setup. Why is that important? In a parallel circuit, having multiple paths for current leads to lower overall resistance, allowing more current to flow.
Electrical Power
Electrical power quantifies the rate at which energy is used or transferred in a circuit. It's calculated as the product of current and voltage but can also be found using resistance and current:\[ P = I^2R \]In this exercise:
  • The power dissipated in the 1.60 \( \Omega \) resistor is \( 490 \, W \), as it carries the highest current.
  • The other resistors dissipate \( 327 \, W \) and \( 163 \, W \) respectively.
These calculations help us see how resistors convert electrical energy into heat. Notably, even with different resistances, the resistor with the lowest resistance dissipates the most power because of higher current (from Ohm’s Law \( I = \frac{V}{R} \)).
In simple terms, less resistance in a parallel circuit allows more current, leading to more energy (power) converted to heat.
Resistor Current Calculation
Determining the current through each resistor in a parallel circuit is straightforward using Ohm’s Law without needing complex math:
1. Calculate each current using \( I = \frac{V}{R} \) with \( V = 28.0 \, V \). For each resistor in this exercise:
  • \( I_1 = \frac{28.0}{1.60} = 17.5 \, A \)
  • \( I_2 = \frac{28.0}{2.40} = 11.67 \, A \)
  • \( I_3 = \frac{28.0}{4.80} = 5.83 \, A \)
2. Total current through the battery is the sum of these currents: \[ I_{total} = I_1 + I_2 + I_3 = 35 \, A \]Understanding these steps shows that even in real-world applications, current divides according to the resistance across each path in a parallel circuit. This division allows for varying current amounts, depending on each pathway's resistance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 4.60-\(\mu\)F capacitor that is initially uncharged is connected in series with a 7.50-k\(\Omega\) resistor and an emf source with \(\varepsilon =\) 245 V and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

A galvanometer having a resistance of 25.0 \(\Omega\) has a 1.00-\(\Omega\) shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a 15.0-\(\Omega\) resistor connected across the terminals of a 25.0-V battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the \(true\) current in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the \(true\) current?

A 12.4-\(\mu\)F capacitor is connected through a 0.895-M\(\Omega\) 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.

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?

The electronics supply company where you work has two different resistors, \(R_1\) and \(R_2\), in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are \(R_1\) and \(R_2\) in parallel and in series-and you can't separate these two resistor combinations. You separately connect each resistor network to a battery with emf 48.0 V and negligible internal resistance and measure the power \(P\) supplied by the battery in both cases. For the series combination, \(P =\) 48.0 W; for the parallel combination, \(P =\) 256 W. You are told that \(R_1\) \(R_2\). (a) Calculate \(R_1\) and \(R_2\). (b) For the series combination, which resistor consumes more power, or do they consume the same power? Explain. (c) For the parallel combination, which resistor consumes more power, or do they consume the same power?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free