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 nlinear 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 inspectionmethod.

Solving the system of equations using Gaussian elimination or some other method gives thefollowing currents, all measured in amperes:

I1=0.245, I2=0.111 and I3=0.117

Reconstructing the branch currents fromthe loop currents gives the results

shown in the picture to the right.

Example 2: Find the current

flowing in each branch of thiscircuit.

Solution:

The number of loop currents required is 3.

This time we will choosethe 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 thefollowing currents, all measured in amperes:

I1 = - 4.57, I2 = 13.7 and I3 = - 1.05

Reconstructing thebranch currents from the

loop currents gives the

results shown in the

picture to the right.

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