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Determine the current in each branch of the given network. |
Answer» Solution : In above circuit, we have to find currents in different branches. Applying KVL in the closed path ABDA, `- 10I_(1) -5I_(3) + 5I_(2) = 0 ` `therefore 10 I_(1) = 5I_(2) - 5I_(3)` `therefore 2I_(1)= I_(2) - I_(3) ""` ... (1) Applying KVL in the closed path BCDB, ` - 5 (I_(1) - I_(3)) + 10 (I_(2) + I_(3)) + 5I_(3) = 0 ` `therefore 5I_(1) = 20I_(3) + 10I_(2)` `therefore I_(1) = 2I_(2) + 4I_(3)` `therefore 2I_(1) = 4I_(2) + 8I_(3) "" `.... (2) From equation (1) and (2), `I_(1) - I_(3) = 4I_(2) + 8I_(3)` `therefore 3I_(2) = - 9I_(3)` `therefore I_(2) = - 3I_(3) "" `... (3) From equation (1) and (3), `2I_(1) = - 3I_(3) =- 4I_(3)` `therefore I_(1) = - 2I_(3) ""` .... (4) Applying KVL in the closed path ADCEFA , we get, `- 5I_(2) - 10 (I_(2) + I_(3)) - 10t = - 10` `therefore 15 I_(2) + 10I_(3) + 10I = 10` `therefore 15 I_(2) + 10 I_(3) + 10 (I_(1) + I_(2)) = 10 ` `therefore 10I_(1) + 25 I_(2) + 10 I_(3) = 10` `therefore 10 (-2I_(3)) + 25 (- 3I_(3)) + 10I_(3) = 10` `therefore - 85I_(3) = 10 ` `therefore I_(3) = - (10)/(85) = - (2)/(17) A ""` ... (5) From equation (4) and (5). From equation (4) and (5) , `I_(1) = - 2 (- (2)/(17))= (4)/(17) A ` ... (6) From equation (3) and (5) `I_(2)= - 3 (- (2)/(17))= (6)/(17) A "" `.... (7) now , `I = I_(1) + I_(2) = (4)/(17) + (6)/(17) = (10)/(17) A "" `....(8) From equation (5) , `| I_(3) | = (2)/(17)` A (Direction of `I_(3)` will be opposite to the direction shown in the figure ). Final answer : (i) Current in branch AB is, `I_(1) = (4)/(17) `A (ii) Current in branch BC is, `I_(1) - I_(3) = (4)/(17) - ( - (2)/(17)) = (6)/(17) A ` (iii) Current in branch CD is, `I_(2) + I_(3) = (6)/(17) + (- (2)/(17)) = (4)/(17)` A (iv) Current in branch DA is, `I_(2)= (6)/(17) `A (v) Current in branch BD is, `| I_(3)| = (2)/(17) A ` (Direction will be from D to B) (VI) Current in branch FE is, I = `(10)/(17) A ` |
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