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 I1,
I2, . . . , Im 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 R11, R12, . . . , Rmm and V1,
V2, . . . , Vm 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 I1,
I2, . . . , Im 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:
I1=0.245, I2=0.111 and I3=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:
I1 = - 4.57, I2 = 13.7 and I3 = - 1.05
- Reconstructing the branch currents from the loop currents gives the results shown in the
picture to the right.
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