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In the figure shown the power generated by `y` is maximum when `y=5Omega`. Then `R` is A. `2 Omega`B. `6 Omega`C. `5 Omega`D. `3 Omega` |
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Answer» Correct Answer - D `P=I^(2)y=(10/(2+y+R))^(2)Y=100/((2+y+R)^(2)) y` For `P` to be maximum `(dP)/(dy)=0 implies d/(dy)(100y)/((2+y+R)^(2))=0` `R=y-z` put `y=5implies R=3 Omega` |
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