SSE JE Electrical Engineering 28 Oct 2020 Shift 1 Papers

Question : 109 of 200

Marks: +1, -0

A generator develops 200 V and has an internal resistance of 100 Ω. Find the maximum power delivered to a load of 100 Ω.

Solution:

Given, source voltage

(\V_\th)

= 200 volt
Internal resistance

(\R_\th)

= 100 Ω
Load resistance

(\R_\L)

= 100 Ω
For maximum power transfer to the load,

\P _{ \m }={ \V _{ \th }^{2}}/{4 \R _{ \th }}={200^{2}}/{4 ×100}=100 \W

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