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

The circuit of FIGURE has two resistors, with R1>R2. Which of the two resistors dissipates the larger amount of power? Explain.

Short Answer

Expert verified

The resistor that dissipates the most energy is theR2.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Step 1. Given information

The circuit that contains two resistors has relation of R1>R2

02

Step 2. Identify the resistor that dissipates the most power.

The expression for power is,

P=v2R

where, Vis the voltage and Ris the resistance.

In the diagram below, the resistors are linked in parallel:

The two resistances R1and R2are coupled in a parallel combination in this diagram. We know that when resistances are coupled in a parallel configuration, the potential difference between them is the same.

Thus can deduce from the preceding calculation of power that power is inversely related to resistance.

The power dissipated through the resistor decreases as the resistance increases, and vice versa.

R1>R2is the specified condition.

Because the resistor R1has a greater value, the power wasted through it will be reduced.

As a result, the power lost through the resistance R2will be greater.

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

What is the equivalent resistance between points a and b in FIGURE EX28.26?

The switch in FIGURE EX28.36 has been in position a for a long time. It is changed to position b att=0 . What are the charge Qon the capacitor and the current Ithrough the resistor (a)immediately after the switch is closed? (b)att=50μs ? (c)at t=200μs?

An oscillator circuit is important to many applications. A simple oscillator circuit can be built by adding a neon gas tube to an RC circuit, as shown in figureCP28.83. Gas is normally a good insulator, and the resistance of the gas tube is essentially infinite when the light is off. This allows the capacitor to charge. When the capacitor voltage reaches a value Von, the electric field inside the tube becomes strong enough to ionize the neon gas. Visually, the tube lights with an orange glow. Electrically, the ionization of the gas provides a very-low-resistance path through the tube. The capacitor very rapidly (we can think of it as instantaneously) discharges through the tube and the capacitor voltage drops. When the capacitor voltage has dropped to a value Voff, the electric field inside the tube becomes too weak to sustain the ionization and the neon light turns off. The capacitor then starts to charge again. The capacitor voltage oscillates between Voff, when it starts charging, and Von, when the light comes on to discharge it.

a. Show that the oscillation period is

T=RCinε-Voffε-Von

b. A neon gas tube has Von=80VandVoff=20V. What resistor value should you choose to go with a 10μfcapacitor and a 90Vbattery to make a 10Hzoscillator?

A battery with internal resistance r is connected to a load resistance R. If R is increased, does the terminal voltage of the battery increase, decrease, or stay the same? Explain.

The circuit in FIGURE EX28.18is called a voltage divider. What value of Rwill make Vout=Vin/10?

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