CALCULATION OF A CURRENT IN A RESISTIVE NETWORK
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We have a resistive network consisting of five resistors. A 5-Ampere current generator, indicated with I1, is connected to this network.
The five resistors are indicated with the names R1, R2, R3, R4 and R5. However, their resistance values are not known.
Instead, the currents passing through the resistors R2 and R3 are specified: they are equal to 2.5 Amperes and 0.5 Ampere,s respectively.
At this point, we ask to calculate the value of the unknown current, Ix, that crosses the resistor R5.
SOLUTION
To solve the exercise, we refer to node, A, indicated in the circuit, and apply Kirchhoff’s law of currents which, as known from the theory, establishes that assigned a node of the circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.
It follows that in R4 circulates a current equal to the sum of the currents circulating in R2 and in R3.
We therefore have that I4 is equal to 2.5 Amperes plus 0.5 Amperes, which is equal to 3 Amperes.
Let us now consider node X and apply, in this case too, Kirchhoff’s law of currents. In particular, we can note that the incoming currents are represented by Ix (which circulates in R5) and by a current of 3 Ampere (which circulates in R4). We then have a single outgoing current equal to the 5 Amperes of the current generator.
In formulas, we have that Ix plus 3 Amperes must equal 5 Amperes.
Hence, it turns out that Ix must be equal to 5 Amperes minus 3 Amperes, which in turn equals 2 Amperes.
Therefore, the unknown current is equal to 2 Amperes.
Well, the exercise has been solved!
Thanks for your attention by Online Technologies For Beginners.
