Loop Analysis of Electric Circuits
In this method, we set up and solve a system of equations in which the unknowns are loop currents.
The currents in the various branches of the circuit are then easily determined from the loop currents.
(Click here for a tutorial on loop currents vs. branch currents.)
The steps in the loop current method are:

Count the number of loop currents required. Call this number m.

Choose m independent loop currents, call them I_{1},
I_{2}, . . . , I_{m} and draw them on the circuit diagram.

Write down Kirchhoff's Voltage Law for each loop.
The result, after simplification, is a system of n linear equations in the n
unknown loop currents in this form:
where R_{11}, R_{12}, . . . , R_{mm} and V_{1},
V_{2}, . . . , V_{m} are constants.
Alternatively, the system of equations can be gotten (already in simplified form) by using
the inspection method.

Solve the system of equations for the m loop currents I_{1},
I_{2}, . . . , I_{m} using Gaussian elimination
or some other method.

Reconstruct the branch currents from the loop currents.
Example 1: Find the current flowing in each branch of this circuit.
Solution:
 The number of loop currents required is 3.
 We will choose the loop currents shown to the right. In fact these loop currents are
mesh currents.
 Write down Kirchoff's Voltage Law for each loop. The result is the following system of equations:
Collecting terms this becomes:
This form for the system of equations could have been gotten immediately by using the
inspection method.
 Solving the system of equations using Gaussian elimination or some other method gives the
following currents, all measured in amperes:
I_{1}=0.245, I_{2}=0.111 and I_{3}=0.117
 Reconstructing the branch currents from the loop currents gives the results shown in
the picture to the right.
Example 2: Find the current flowing in each branch of this circuit.
Solution:
 The number of loop currents required is 3.
 This time we will choose the loop currents shown to the right.
 Write down Kirchoff's Voltage Law for each loop. The result is the following system of equations:
Collecting terms this becomes:
This form for the system of equations could have been gotten immediately by using the inspection method.
 Solving the system of equations using Gaussian elimination or some other method gives the
following currents, all measured in amperes:
I_{1} =  4.57, I_{2} = 13.7 and I_{3} =  1.05
 Reconstructing the branch currents from the loop currents gives the results shown in the
picture to the right.
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