InterviewSolution
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InterviewSolution
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For the resistor network shown in the potential drop between a and b s 12V. a. Current through resistance of `6Omega is ………..` b. Current through resistance `2 Omega is ………..` c. Current through resistance of 8 Omega is ...........` |
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Answer» `V_(ab) = I_1(6+(4xx6)/(4+6)) = I_1(8.4) (i)` `V_(ab) I_2(4 +(8xx8)/(8+8)) = I_2 (8) (ii)` a. From Eq. (i) we get `I_= (12)/(8.4) (10)/(7)A` `I_(1)` is the current in the `6Omega` resistor. b. `I_(2,4) = (4)/(6+4)xx(10)/(7) =(4)/(7)A` c. From eq (ii) we get `I_(2) = 3//2 A,` so current in the `4Omega` resistacne is 3//2A. It will be equally dicided inot `8Omega` resistance which are in parallel. Therefore, `I_(8) = (3)/(2xx2) = (3)/(4)A` |
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