Problem 14-5    Source:   Problem created by the site's author.

In the circuit shown in Figure 14-05.1, calculate I₁ and I₂ using the Superposition method.

Solution of the Problem 14-5   -    Superposition Method

As we know, the Superposition method allows us to calculate the circuit currents using one source at a time. Then, we algebraically add the values found and we will have the final result. This is possible because the circuit is linear.

So let's eliminate one of the sources. Remember: we eliminated a voltage source short-circuiting it, as shown in Figure 14-05.2. To differentiate the calculation of currents with different sources, we will assume the index a for the calculation with the first source and the index b for the calculation with the second source.

The best way to solve this problem is by applying the Law of Meshes. So let's build the equations for each mesh.

Solving this system of three equations for three unknowns by any method, we find:

Now that we have the values of the currents in the circuit, by inspection, we easily determine that:

These values are only found when the 50 volts source is considered. Now we need to calculate the currents when only the source 30 volts is operating. To do this, we will use the circuit shown in Figure 14-05.3.

In the same way as was done previously, we will write the mesh equations in the circuit.

Solving this system of three equations for three unknowns by any method, we find:

Now that we have the values of the currents in the circuit, by inspection, we easily determine that:

With all the partial values of the currents, we can now, algebraically adding them, find the values of the currents in the circuit, that is: