Electric Circuits Analyzing Techniques-2

Kirchhoff's Current & Voltage Laws:

Kirchhoff's Current Law-KCL: The algebraic sum of the currents entering any node is zero.

A node is not a circuit element. Therefore, it can't store, generate, or destroy a charge. As a result, the currents must sum to zero.

Example (1): To simplify this idea, consider three water pipes joined in the shape of a Y.
Imagine that three currents are flowing into each of the three pipes, we cannot have three positive water currents or the pipes would burst. Therefore, the value of either one or two of the currents must be negative.

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Example (2): Consider the node shown in figure.



The algebraic sum of the four currents entering the node must be zero:

iA + iB + (- iC) + (- iD) = 0

However, the law could be equally well applied to the algebraic sum of the currents leaving the node:

(- iA) + (- iB) + iC + iD = 0

So, in both cases the equation could be represented as:

iA + iB = iC + iD

Which clearly indicates that the sum of the currents entering the nodes must equal the sum of the currents leaving the node.

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Example (3): Consider the following circuit. Compute the current through resistor R3 if it is known that the voltage source supplies a current of 3A.



Solution:

It is not that difficult, this circuit contains only three nodes.
We have a node at the upper center & another one at the lower center, we have also a node between the 10V voltage source and the resistor R1, but in our analysis from here and so on, we don't consider a node connecting only 2 branches. Therefore they reduce to only two considered nodes.



Applying KCL to the upper center node, we get:

iR1 - 2 - i + 5 = 0

But we are given that the voltage source supplies a current of 3A. As the voltage source and the resistor R1 are in series connection, they have the same current. Substituting this value, we get:

3- 2 - i + 5 = 0

Solving for i, we get:

i = 6A

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Example (4): Count the number of nodes and branches in the circuit shown in figure.
If ix = 3A and the 18V source delivers 8A of current, what is the value of RA? (Hint: You need Ohm's law as well as KCL.)



Solution:

We have 5 branches in this circuit because we simply have 5 circuit elements.
We have 3 nodes in this circuit.

We need to find the value of RA, so to apply Ohm's law, we have to know both the current & voltage at the resistor RA.
The voltage at RA or vRA is 18V because the voltage source is parallel to RA, the voltage of 2 parallel circuit elements must be equal.
The current iRA could be computed applying KCL at the upper center node, we get:

13 + 8 - 3 - iRA = 0

Solving for iRA, we get:

iRA = 18A

We now know both vRA & iRA. Therefore, Ohm's law is applicable:

RA = vRA / iRA

RA = 18 / 18

RA = 1 Ω

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A compact expression for KCL is:

Σi = 0

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Reference: Engineering Circuit Analysis 8th Edition

To be continued ...

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