Question

Deciding current in a circuit?

Decide `I_1`, `I_2` and `I_3` in the circuit below

Answer 1

`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
`1/R_23= 1/R_3 + 1/R_2 = 1/5 +1/3 = (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 =(8V)/(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= 3/8 xxI_11 = 0,44 A`
`I_21= 5/8 xxI_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