# Deciding current in a circuit?

## Decide ${I}_{1}$, ${I}_{2}$ and ${I}_{3}$ in the circuit below

May 13, 2018

${I}_{1} = 1 , 16 + 0 , 22 = 1 , 38 A$
${I}_{3} = 0 , 58 + 0 , 44 = 1 , 02 A$
${I}_{2} = - 0 , 72 + 0 , 36 = - 0 , 36 A$

#### Explanation:

You can solve the circuit by dividing the effect of the two generators and considering one at a time the effects.

Considering the only battery on the left you have two resistors in parallel on the right
$\frac{1}{R} _ 23 = \frac{1}{R} _ 3 + \frac{1}{R} _ 2 = \frac{1}{5} + \frac{1}{3} = \frac{3 + 5}{15} = 0 , 53 {\Omega}^{-} 1$
${R}_{23} = 1 , 875 \Omega$
${R}_{123} = {R}_{1} + {R}_{23} = 6 , 875 \Omega$
${I}_{11} = {V}_{1} / {R}_{123} = \frac{8 V}{6.875 \Omega} = 1 , 16 A$
By the value of the resistances, the current in 3 and in 2 caused by V1 are:
${I}_{31} = \frac{3}{8} \times {I}_{11} = 0 , 44 A$
${I}_{21} = \frac{5}{8} \times {I}_{11} = 0 , 72 A$ or better -0,72 A according with the verso on the sketch.

Now let's consider only the battery on the right
With the same reasoning, you will have
${I}_{33} = 0 , 58 A$
${I}_{23} = 0 , 36 A$positive
${I}_{13} = 0 , 22 A$
With the sum of the effects, you have
${I}_{1} = 1 , 16 + 0 , 22 = 1 , 38 A$
${I}_{3} = 0 , 58 + 0 , 44 = 1 , 02 A$
${I}_{2} = - 0 , 72 + 0 , 36 = - 0 , 36 A$
now you can see that every node is verificated