/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q4E Question: A 42 Ω resistor and ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question: A 42 Ω resistor and a 20 Ω 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?

Short Answer

Expert verified

(a) The resistance of the parallel combination is 13.55.

(b)The total current through the parallel combination is 17.70 A.

(c) The current through resistor R1 is 5.71 A and the current through resistor R2 is 12.00 A.x`

Step by step solution

01

Equivalent resistance

Given data:

  • R1= 42 Ω
  • R2= 20 Ω
  • V = 240 V

To find the equivalent resistance of R1 and R2:

The two resistors are in parallel and to calculate their combination we will use the equation :

Req=R1R2R1+R2

Substitute values to find Req as:

Req=R1R2R1+R2=42Ω20Ω42Ω+20Ω=13.55Ω

The resistance of the parallel combination is 13.55 Ω.

02

Total Current

The potential difference V' across resistors connected in the parallel is the same for every resistor and equals the potential difference across the combination and we can use the value of the voltage to find the current across the combination by using Ohm's law as:

It=VReq

Substitute values to find It

It=VReq=240 V13.55Ω=17.70A

The total current through the parallel combination is 17.70 A.

03

Current through each resistor

The result in Step 2 represents the total current in the resistors where the total current through resistors connected in parallel is the sum of the currents through the individual resistors R1 and R2.

I1=VR1=240 V42Ω=5.71A

But as we discussed in part (b) the voltage is the same and V1 = V2 = 240 V. Now use Ohm's law to get the current in each resistor as:

I2=VR2=240 V20Ω=12.00A

The current through resistor R1 is 5.71 A and the current through resistor R2 is 12.00 A.

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 91Ó°ÊÓ!

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

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes10.20 V . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Consider the circuit of Fig. E25.30. (a)What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors? (b) What is the power output of the 16.0-V battery? (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery? (d) Show that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Fig. E25.30.

Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500V and produce currents of 80mA(or even larger). A typical pulse lasts for 10ms. What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

An alpha particle (a He nucleus containing two protons and two neutrons and having a mass of 6.64×10-7kg) travelling horizontally at 35.6km/senter a uniform, vertical,1.80-T magnetic field.(a) What is the diameter of the path followed by this alpha particle? (b) what effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) explain why the speed of the particle does not change even though an unbalanced external force acts on it.

A particle with mass1.81×10-3kgand a charge of has1.22×10-8C, at a given instant, a velocityV⇶Ä=(3.00×104m/s).What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic fieldB=(1.63T)i+(0.980T)j^?

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.