Example 6 to An amount of energy equal to 100 J is used in 5 s. What is the power Solution P = W/ t =100 J/5 sec = 20 W Example 7 What power is dissipated in a 100 Q resistor with 5 V across it? Solution v2 P = R (5)2 = 0.25 W %D %3D 100

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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100%
a positive
elements that produce power have a negative value of P.
Example 6
to
An amount of energy equal to 100 J is used in 5 s. What is the power
Solution
P = W/ t =100 J/5 sec = 20 W
%3D
Example 7
What power is dissipated in a 100 Q resistor with 5 V across it?
Solution
(5)²
= 0.25 W
P%3D
100
Transcribed Image Text:a positive elements that produce power have a negative value of P. Example 6 to An amount of energy equal to 100 J is used in 5 s. What is the power Solution P = W/ t =100 J/5 sec = 20 W %3D Example 7 What power is dissipated in a 100 Q resistor with 5 V across it? Solution (5)² = 0.25 W P%3D 100
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