# Question 7bd35

Dec 17, 2016

${A}_{1} = 19 A$

${A}_{2} = 17 A$

${A}_{3} = 14 A$

${A}_{4} = 3 A$

#### Explanation:

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

The emerging current ${A}_{1} = 19 A$ is divided in two paths in the first branch as ${A}_{2} \mathmr{and} 2 A$ ,

So ${A}_{2} + 2 A = 19 A \implies {A}_{2} = \left(19 - 2\right) A = 17 A$

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

${A}_{3} \mathmr{and} 3 A$. So A_3 +3A=17A=>A_3= (17-3)A=14A.

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