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In the circuit What should be the value of r in ohmso that power developed in the resistor r will be maximum ? |
Answer»  Using kirchhoff's SECOND law, in closed circuit `ABEFA` we have `6 I_(1) = 3(I_(1) +I_(2)) - V = 0` or `V = 9 I_(1)+3I_(2)`….(i) Usingkirchhoff's second law , in closed circuit `BCDEB` we have `R I_(2) - 6 I_(2) = 0 or I_(1) = (I_(2))/(6) r` From (i) `V = 9 xx (I_(2))/(6) r + 3I_(2) = (3)/(2) I_(2) r + 3I_(2) = (3I_(2))/(2) (r + 2)` or `I_(2) = (2V)/(3(r+2))` Power develeped in resisteor `r`, `P = I_(2)^(2) = (4V^(2))/(9(r+2)^(2)) xx r` Power developed is maximum, when `(r+ 2)^(2)` is minimum or `(r+ 2)^(2) = 0` or `r^(2) + 4R + 4 = 0` or `r^(2) - 4r + 4 + 8 r = 0` or `(r - 2)^(2) + 8r = 0 or r = 2 Omega` |
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