Deciding current in a circuit?

Question

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

Nathan Ashcroft · Accepted Answer

#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# 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

Charles Aeschliman · Answer

undefined To determine the current in a circuit, you need to apply Ohm's law, which states that current ($I$) equals voltage ($V$) divided by resistance ($R$). Mathematically, it can be represented as $I = \frac{V}{R}$. Once you have the voltage across the circuit and the resistance in the circuit, you can calculate the current flowing through it.