Question #7bd35

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Question #7bd35

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Answer 1 · 2016-12-17T14:13:52

$A_{1} = 19 A$ $A_1=19A$

$A_{2} = 17 A$ $A_2=17A$

$A_{3} = 14 A$ $A_3=14A$

$A_{4} = 3 A$ $A_4=3A$

Explanation:

drawn

As shown in figure of the circuit the main current returning from cell is $19 A$ $19A$ and emerging from cell is $A_{1}$ $A_1$ . Hence $A_{1} = 19 A$ $A_1=19A$ by conservation of charge.

The emerging current $A_{1} = 19 A$ $A_1=19A$ is divided in two paths in the first branch as $A_{2} and 2 A$ $A_2 and 2A$ ,

So $A_{2} + 2 A = 19 A \Rightarrow A_{2} = (19 - 2) A = 17 A$ $A_2 +2A=19A=>A_2= (19-2)A=17A$

Similarly $A_{2} = 17 A$ $A_2 =17A$ is divided in two paths in the 2nd branch as

$A_{3} and 3 A$ $A_3 and 3A$ . So #A_3 +3A=17A=>A_3= (17-3)A=14A.

From the figure we see $3 A$ $3A$ returns back passing through $R_{3}$ $R_3$ as $A_{4}$ $A_4$ So #A_4==3A

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Question #7bd35

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