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Determine the current in each brance of the network showin in |
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Answer» Solution :Applying KVL `( sum IR = sum EPSILON)` In the CLOSED loop ADCA, we get -4 `(I_(1) - I_(2) ) + 2 (I_(2) + I_(3)- I_(1) )= -10` `therefore 4 (I_(1) - I_(2) ) - 2 (I_(2) + I_(3) - I_(1)) + I_(1) = 10` `therefore 7 I_(1) - 6I_(2) - 2I_(3) = 10 ""` .... (1) Similarly for closed loop ABCDA, ` -4 I_(2) -2 (I_(2) + I_(3)) - I_(1) =- 10 ` `therefore 4I_(2) +I_(2) -2 (I_(2) + I_(3)) + I_(1) = 10` `therefore I_(1) + 6I_(2) + 2I_(3) = 10 ""` ... (2) For closed loop BCDEB, ` - 2 (I_(2) + I_(3)) - 2 (I_(2) + I_(3) - I_(1)) = - 5 ` ` therefore 2 (I_(2) + I_(3)) + 2 (I_(2) + I_(3) - I_(1)) =5 ` `therefore 4 I_(2) + 4I_(3) - 2I_(1) = 5 ` `therefore 2I_(1) - 4I_(2) - 4I_(3) = - 5` `therefore I_(1) - 2I_(2) - 2I_(3) = - 2.5 ""` .... (3) Adding equations (1) and (2), `8I_(1) = 20` `therefore I_(1) = (5)/(2)A ""` .... (4) Adding equations (2) and (3) `2I_(1) + 4I_(2) = 7.5` `therefore2 ((5)/(2)) + 4I_(2) = 7.5 ` `therefore 5 + 4I_(2) = 7.5` `therefore I_(2) = (2.5)/(4) = (25)/(40) = (5)/(8) A "" `... (5) Substituting values of`I_(1) and I_(2)` in EQUATION (3), 2.5 -2 `((5)/(8)) - 2I_(3) =- 2.5 ` `therefore 2.5 - (5)/(4) + 2.5 = 2 I_(3)` `therefore 10 - 5 + 10 = 8I_(3)` `therefore 15 = 8I_(3)` `therefore I_(3) = (15)/(8) A ""` .... (6) |
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