# Loop Analysis

We will now consider a circuit analysis technique known as Loop Analysis, in which a system of equations is solved simultaneously for all the unknown currents in the circuit. Before we go on, we need to define some terms:

Node: Any point in a circuit to which two or more circuit elements are connected. The diagram below shows circuit nodes as red dots. Notice that there is only one node along the bottom of the circuit. That is because the circuit could be redrawn with the voltage source, R2, R6, and R7 all connected together at a single point.

Loop:  A closed path in a circuit in which any node is encountered exactly one time. A typical loop is shown in blue in the circuit diagram below. Notice that each node in the loop is crossed only once.

The circuit shown below is the series parallel circuit that we analyzed in a laborious way in the last section. Our technique of continually reducing the circuit to progressively simpler equivalent circuits and then working backward once we had found the total current flowing through the entire circuit is time consuming and error-prone.

We would like to find the current flowing through R8 directly. We do this by using loop analysis. In order to use loop analysis, we must first find a set of loops in the circuit that contains each element of the circuit at least once. One possible way to do this for our example circuit is shown below. Note that in general, there are usually several ways to assign loops to the circuit such that every circuit element is included.

Once we have assigned the loops to the circuit, we then assume a direction flow for current in each loop:

It does not matter which way we assume the current flows in each loop as long as we are consistent, that is once we assume the direction, we cannot change it during the solution of the problem.

Now we use Kirchoff's Voltage Law, (KVL) which you learned in an earlier section, to relate the voltage drop around each loop to the loop current. If case you have forgotten KVL, it tells us that that sum of the voltage drops around a loop is equal to zero. Let us look at loop 1 first. When we apply KVL we get:

VR1 +VR2 - 42V = 0    Remember that voltage sources have negative voltage drops!!

but: VR1 = I1*R1 and VR2 = (I1- I3)*R2. We can substitute these into our earlier expression to get:

I1*R1 +  (I1-I3)*R2 - 42V = 0 or  I1*(R1+R2) - I3*R2  =  42V

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