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A parallel combination of three resistors take a currect of `7.5 A` from `30 V` supply. If the two resistor are `10 Omega` and `12 Omega`. find the third one. |
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Answer» Correct Answer - `R_(3) = 15 Omega` `R_(P) = 30//7.5 = 4 Omega` `(1)/(R_(P)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) or (1)/(4) + (1)/(10) + (1)/(12) + (1)/(R_(3))` or `(1)/(R_(1)) = (1)/(4) - ((1)/(10) + (1)/(12)) = (4)/(60) = (1)/(15)` or `R_(3) = 15 Omega` |
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