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 Ω, 2.40 Ω, and 4.80 Ω 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 Req is given by the formula: 1Req=1R1+1R2+1R3 where R1=1.60Ω, R2=2.40Ω, and R3=4.80Ω. Substituting the given values:1Req=11.60+12.40+14.80Calculating the reciprocals:1Req=0.625+0.417+0.2081Req=1.25Finally, the equivalent resistance is:Req=11.25=0.80Ω.
02

Find the Current through Each Resistor

The current through a resistor in a parallel circuit can be calculated using Ohm's Law I=VR where V=28.0V is the voltage across each resistor.For R1=1.60Ω: I1=28.01.60=17.5AFor R2=2.40Ω: I2=28.02.40=11.67AFor R3=4.80Ω:I3=28.04.80=5.83A.
03

Find the Total Current through the Battery

The total current through the battery is the sum of the currents through each resistor:Itotal=I1+I2+I3Itotal=17.5+11.67+5.83=35A.
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 28.0V.
05

Calculate the Power Dissipated in Each Resistor

The power dissipated in a resistor can be calculated using the formula P=I2R.For R1:P1=(17.5)2×1.60=490WFor R2:P2=(11.67)2×2.40=327WFor R3:P3=(5.83)2×4.80=163W.
06

Determine which Resistor Dissipates the Most Power

The resistor with the least resistance (R1=1.60Ω) dissipates the most power since it has the highest current flow, following the relationship P=I2R. 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=VR.
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:1Req=1R1+1R2+1R3Where R1,R2,R3 are the resistances of different components. By calculating reciprocals of the resistances, we find that:Req=11.25=0.80ΩThis value, 0.80Ω, 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=I2RIn this exercise:
  • The power dissipated in the 1.60 Ω resistor is 490W, as it carries the highest current.
  • The other resistors dissipate 327W and 163W 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=VR).
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=VR with V=28.0V. For each resistor in this exercise:
  • I1=28.01.60=17.5A
  • I2=28.02.40=11.67A
  • I3=28.04.80=5.83A
2. Total current through the battery is the sum of these currents: Itotal=I1+I2+I3=35AUnderstanding 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 2.00μF capacitor that is initially uncharged is connected in series with a 6.00kΩ resistor and an emf source with E=90.0 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?

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?

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?

A circuit consists of a series combination of 6.00-kΩ and 5.00-kΩ resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00-kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ. (a) What potential difference does the voltmeter measure across the 5.00-kΩ resistor? (b) What is the true potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

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?

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