College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:**Transcription and Explanation for Educational Website:**
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**Question:**
4. For the circuit shown in the figure below, what is the power dissipated by the 50 Ω resistor due to ohmic heating?
**Circuit Diagram Description:**
The diagram illustrates an electrical circuit consisting of a voltage source and three resistors. Here are the components:
- A 20 V voltage source.
- An 80 Ω resistor connected in series.
- Two resistors, 100 Ω and 50 Ω, connected in parallel at point A.
**Explanation of Diagram:**
In the circuit, the 20 V power source is connected to an 80 Ω resistor. This resistor is then connected to a junction labeled 'A'. At this junction, two branches of resistors are connected in parallel: one with a 100 Ω resistor and the other with a 50 Ω resistor. The goal is to find the power dissipated specifically in the 50 Ω resistor due to ohmic heating.
To calculate this, one would typically use formulas related to ohmic heating and basic circuit principles such as Ohm's Law and the power dissipation formula.
**Calculation Approach (Not Provided in the Diagram):**
1. Calculate the total resistance in the circuit.
2. Determine the total current flowing from the voltage source.
3. Calculate the current specifically through the 50 Ω resistor using parallel circuit rules.
4. Use the formula for power dissipation in a resistor, \( P = I^2 R \), where \( P \) is power, \( I \) is current, and \( R \) is resistance, to find the power dissipated by the 50 Ω resistor.
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This transcription and explanation serve as educational guidance to help understand the circuit analysis for determining power dissipation in resistors.
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