Question 2.SA.7: Determine the value of voltage source current in the paralle......

KCL is applied to the given circuit after the node has been identified and circuit has been annotated with voltage designation, voltage polarity, branch currents, and current directions. See the circuit diagram below:l

Subscribing to the definition of a node as a point where three or more conduc-tors merge, the shaded segment in the diagram above is designated as the node for this circuit. Then, before applying KCL to determine the source current – using Ohm’s law – define the individual currents through each of the resistors, in terms of the specific resistance values and the voltages around them:

Since all of the resistors are in parallel with the voltage source,

Therefore,

I_1=\frac{20}{5}=4  A, I_2=\frac{20}{3}=6.67  A, I_3=\frac{20}{10}=2  A

Then, application of KCL at the designated node yields the following equation:

or

Ancillary Question: If one of the 5 Ω resistors is open-circuited and the other one short-circuited, the parallel resistor network would appear as follows:

The highlighted segment in the circuit above represents the short circuit that replaces R1. Since R1 is replaced by a short circuit – regardless of the dispo-sition of other circuit elements – it becomes a path of least resistance for the entire circuit. In other words, the voltage source is short-circuited. Interpreted in terms of Ohm’s law, this would mean:

Since infinite current is not practical, this means that a very large amount of current would pass through the shunt or short circuit resulting in a catastrophic failure (burning or melting) of the short-circuiting conductor, the interconnect-ing wires or a fault in the voltage source.