SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

Question : 151 of 200

Marks: +1, -0

Determine the load resistance RL that will result in maximum power delivered to the load for the given circuit. Also, determine the maximum power

\P_\max

delivered to the load resistor.

Solution:

\R _{ \th }= \R _{ \L }

=(30 ×150) / 180

= 2 5 \Ω

\V _{ \th }= \V _{ \ab }

=(150 ×180) /(150+30)

=150 \V

From above concept,

\P _{\max }={ \V _{ \th }^{2}}/{4 \R _{{\th }}}={150^{2}}/{4 ×25}=225 \W

\P _{\max }= 2 2 5 \W

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