Problem 4 = (a) Find the value of the load resistance, RÅ, such that the current through the load is i̟ = 1600μA. (b) Find the value of the load resistance, RL, such that the power dissipated in the load is P₁ = 1.8mW. W 12000 5mA ( ΙΚΩ 4ΚΩ RL Note: These two parts are separate. For part (a) the load power will not necessarily be 1.8mW and for part (b) the load current will not necessarily be 1600μA. La a) R₁₁ = 500, b) R₁₂ = 1039 or 3850.
Problem 4 = (a) Find the value of the load resistance, RÅ, such that the current through the load is i̟ = 1600μA. (b) Find the value of the load resistance, RL, such that the power dissipated in the load is P₁ = 1.8mW. W 12000 5mA ( ΙΚΩ 4ΚΩ RL Note: These two parts are separate. For part (a) the load power will not necessarily be 1.8mW and for part (b) the load current will not necessarily be 1600μA. La a) R₁₁ = 500, b) R₁₂ = 1039 or 3850.
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter4: Transmission Line Parameters
Section: Chapter Questions
Problem 4.8MCQ: An ac resistance is higher than a dc resistance. True False
Related questions
Question
Can you solve the following problem and show how the answer was found:

Transcribed Image Text:Problem 4
=
(a) Find the value of the load resistance, RÅ,
such that the current through the load is i̟ =
1600μA.
(b) Find the value of the load resistance, RL,
such that the power dissipated in the load is
P₁ = 1.8mW.
W
12000
5mA (
ΙΚΩ
4ΚΩ
RL
Note: These two parts are separate. For part (a) the load power will not necessarily be 1.8mW
and for part (b) the load current will not necessarily be 1600μA.
La
a) R₁₁ = 500, b) R₁₂ = 1039 or 3850.
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