SSC JE Electrical Engineering 2017 Set 2 Paper

Question : 118 of 200

Marks: +1, -0

The maximum power in the shown load is :

Solution:

For maximum power transfer

Z _L =Z*_S

Z_S = 10 + 20 j

Z*_S = 10 -20 j = Z_L

I_{\max}= 50/{10 +20j +10- 20j}

=50/20=2.5A

Maximum power

P_{\max}={ I^2}_{\max} \R _ L

= (2.5)^2 × 10 = 62.5 Ω

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