CALCULATION OF THE UNKNOWN CURRENT.
Welcome to a new exercise from Online Technologies For Beginners.
An electrical network is assigned with 4 resistors of unspecified value, denoted by R1, R2, R3 and R4.
There are also two current generators, one with a value of 3 Amperes and one with an unknown value equal to 3 times the value of the current Ix, which passes through the resistor R1 according to the direction indicated in the figure.
It is requested to establish the value of Ix in such a way as to have a current equal to 2 Amperes flowing through R3, according to the direction indicated in the figure.
SOLUTION
To solve the exercise, let’s focus on node A of the circuit and apply Kirchhoff’s law of currents to that node.
We recall that this law establishes that the sum of the currents entering a node must be equal to the sum of the currents exiting the same node.
In the case of node A, we have an incoming current equal to 3 times Ix generated by the current generator on the left.
We then have two currents exiting the node A, one equal to Ix which circulates in the branch of resistor R1, and a second current which we denote by I2 which circulates in the branch of resistor R2.
In formulas, we have that:
Thanks to the expression just written, we can now express the value of I2 as a function of I-x.
In particular, we have that:
We report the result just obtained in the circuit, indicating with 2 times I-x the current circulating in the branch of R2.
Let us now focus on node B and also in this case we apply Kirchhoff’s law of currents.
We have two incoming currents, one of which is represented by the current 2 times Ix which circulates in the branch of R2 and a second equal to 3 Amperes generated by the current generator. We then have a current outgoing from node B, equal to 2 Amperes and circulating in the branch of resistor R3.
The result is an equation of the currents for the node B, given by the expression:
We reformulate this expression in order to obtain the value of Ix.
For this purpose, we bring the term 3 Amperes to the right.
Thus, we obtain that:
At this point it is easy to obtain that Ix is equal to -0.5 Amperes.
The value just found represents the solution of our exercise.
The value obtained allows us to observe that the current generator on the left is delivering a current of 3 times Ix which is equal to -1.5 Amperes, that is in the opposite direction with respect to the arrow indicated by the generator symbol.
Finally, known the value of Ix, we have that the currents circulating in the circuit are those indicated in blue in the figure.
In particular, the current generator on the left generates a current of 1.5 Amperes, the current circulating in R1 has an intensity equal to 0.5 Amperes, the current circulating in R2 has an intensity equal to one Ampere, the current circulating in R3 has an intensity equal to 2 Amperes and finally by algebraically adding the currents circulating in R1 and R3, we obtain the current flowing through R4 which is equal to -0.5 Amperes plus 2 Amperes, and therefore equal to 1.5 Amperes.
Well, with this last observation we have completed the exercise.
Thanks for your attention by Online Technologies For Beginners!
