Question
Download Solution PDFDetermine the load resistance RL that will result in maximum power delivered to the load for the given circuit. Also, determine the maximum power Pmax delivered to the load resistor.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Maximum power transfer for DC circuit:
According to the MPT the maximum power transfer to the load when the load resistance is equal to the source resistance or Thevenin resistance.
RL = Rth
RL = load resistance
Rth = Thevenin or source resistance
The power at maximum power transfer (Pmax) = Vth2 / 4Rth
The maximum power transfer theorem is used in electrical circuits.
Calculation:
Rth = RL
= ( 30 × 150 ) / 180
= 25 Ω
Vth = Vab
= ( 150 × 180 ) / (150 + 30 )
= 150 V
From above concept,
\(P_{max}=\frac{V_{th}^2}{4R_{th}}=\frac{150^2}{4\times25}=225\ W\)
Pmax = 225 W
Last updated on May 29, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.
-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31.
-> Candidates with a degree/diploma in engineering are eligible for this post.
-> The selection process includes Paper I and Paper II online exams, followed by document verification.
-> Prepare for the exam using SSC JE EE Previous Year Papers.