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Source Transformation Example Problems with Solutions

Source Transformation
Source transformation is a circuit analysis technique in which we transform voltage source in series with resistor into a current source in parallel with the resistor and vice versa.

A highly valuable byproduct of Thevenin’s and Norton’s theorem is the technique of source transformation. Source transformation is based upon the observation that if a Thevenin’s network and Norton’s network are both equivalent to a particular source network, then they must also equivalent to each other. This observation allows you to simply an analysis by converting a voltage source with series resistance to an equivalent current source with parallel resistance, or vice versa.

Source conversion may be applied to portions of a circuit to simplify intermediate calculations. Going even further, repeated source conversions may reduce a circuit to all-series or all-parallel form. Conversions may also be applied to controlled sources as well as to independent sources.

In the development of practical voltage sources and current sources, many similarities should have been noticed -such as internal resistance, voltage, and current characteristics. These similarities lead one to ask if it is possible to interchange one practical source for the other and retain the same results at the load. This forms the basis of source transformation.

If the same value of load resistance RL is connected to the circuits of the following figure, the load current flowing in circuit (a) is;

    ${{I}_{L}}=\frac{{{E}_{s}}}{{{R}_{s}}+{{R}_{L}}}~~~~\text{       }\cdots \text{    }~~~~~~~\left( 1 \right)$

Source Transformation

Figure 1: Source Transformation

And the load current flowing in circuit (b) is

${{I}_{L}}=\frac{{{R}_{s}}}{{{R}_{s}}+{{R}_{L}}}*{{I}_{s}}~~~~\text{        }\cdots \text{    }~~~~~\left( 2 \right)$

To produce the same effect at the load, equations (1) and (2) must be equal.

$\frac{{{E}_{s}}}{{{R}_{s}}+{{R}_{L}}}=\frac{{{R}_{s}}}{{{R}_{s}}+{{R}_{L}}}*{{I}_{s}}$

Thus establishing the result that

${{E}_{s}}={{I}_{s}}{{R}_{s}}$

Or
${{I}_{s}}=\frac{{{E}_{s}}}{{{R}_{s}}}$

The value of internal resistance for practical voltage and the current source does not change. Although these equations have been derived for AC conditions. They are also equally applicable to DC circuits. Note that the same values of voltage, current and power may be obtained at the output terminals by either a practical voltage or a practical current source.

Impossible Source Transformation

These are two certain situations when we cannot do source transformation.

1.When voltage source has zero resistance in this case,

${{I}_{s}}=\frac{{{E}_{s}}}{R}$  

       $R=0$

So;

${{I}_{s}}=\infty $

2.When current source has infinite resistance, in this case

${{V}_{s}}={{I}_{s}}R$    

           $R=\infty $ 

So,

${{V}_{s}}=\infty $

Source Transformation Example

Find Vo using source Transformation

Source Transformation Example

We convert 250V voltage source into 10A current source.

Combining both current sources, we obtain

Adding resistances in a parallel manner will give us the following circuit

Finally,

${{V}_{0}}=20V~$

Source Transformation Example with Dependent Source

Let’s find v2 in the following circuit using source transformation.

Source Transformation Example with Dependent Source

Figure.2: Source Transformation Example with Dependent Source

Solution

The 72-V source and the 4Ω series resistance convert to a parallel structure with source current of

\[~72V/4\Omega =18A\]

The VCVS and the 12 Ω series resistance likewise convert to a parallel structure with source current of

\[~3{{v}_{2}}/12\text{ }\Omega =0.25S*{{v}_{2}}\]

So, the VCVS becomes a VCCS whose transconductance is 0.25S. The resulting diagram in figure 3 still has the control voltage v2 in place, as required.

Figure.3

But now all three resistances can be combined as 4||6||12=2 Ω. This parallel equivalent resistance carries the net current from the sources, and so

\[{{V}_{2}}=2\left( 18-0.25{{v}_{2}} \right)\]

Solving for V2 yields

\[{{V}_{2}}=36/1.5=24V\]

About Ahmad Faizan

Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA.

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