Source Transformation in Electrical Circuits
Source transformation is a method used for circuit simplification by replacing the voltage source with its equivalent current source or replacing the current source with its equivalent voltage source.
For ideal electrical sources, losses will be zero. In the case of practical sources, losses will be there because of its internal resistance or impedance. In the discussion below, we have considered the practical case.
The practical voltage source consists of a voltage source in series with its internal resistance. Similarly, the practical current source consists of the current source in parallel with its internal resistance.
Now, let’s learn how the transformation is done from one source to another.
Voltage source to current source transformation
Let us consider a circuit with practical voltage source Vs having internal resistance Rs, connected to the load having resistance RL.
Now, in this circuit, let us understand, how the voltage source in series with the internal resistance is replaced with its equivalent current source.
In the circuit, the current delivered to the load is given by,
![\[I_L = \frac{V_s}{(R_s + R_L)} --->(1)\]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-1db6aa50fba0ab38b1021c43a5aa69bb_l3.png "Rendered by QuickLaTeX.com")
Since it is a series circuit, I = IL
Now, if the voltage source is to be replaced by the current source I, then the internal resistance should be connected across the current source.
By current divider rule, the load current IL will be
![\[I_L = I * \frac{R_s}{R_s + R_L} --->(2)\]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-1c22caff38235a9139bc1a121684970e_l3.png "Rendered by QuickLaTeX.com")
From equations (1) and (2),
![\[ \frac{V_s}{(R_s + R_L)} = I * \frac{R_s}{R_s + R_L} \]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-f72258c2555bb0c4185b14ddd3e39047_l3.png "Rendered by QuickLaTeX.com")
![\[V_s = I*R_s\]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-a8c5759eecdd4d230796dac235b29769_l3.png "Rendered by QuickLaTeX.com")
![\[I = \frac{V_s}{R_s} --->(3)\]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-83719d7336d7d31759c9562a1aadcee3_l3.png "Rendered by QuickLaTeX.com")
Thus is it observed that in transforming the voltage source, the obtained current source(3) is connected with internal resistance Rs in parallel with the source.
Current source to voltage source transformation
Consider a circuit with the practical current source I in parallel with its internal resistance Rs, connected to the load having resistance RL.
As discussed above and from equation (3), a current source in parallel with the internal resistance is transformed into the voltage source Vs by placing the internal resistance in series with the source, where Vs = IRs.
Solved Problem 1
In the following circuit, transform the voltage source of 50V with an internal resistance of 5 Ω into the current source.
To replace the voltage source with the current source,
![\[I = \frac{V_s}{R_s} = \frac{50}{5} = 10 Amps\]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-e6fe726cb1646ba6fab8377d399236ee_l3.png "Rendered by QuickLaTeX.com")
Thus the circuit becomes,
Solved Problem 2
For the given circuit, transform the current source into the voltage source.
In the given problem, current source I = 15 A and Rs = 10 Ω.
To replace the current source with the voltage source,
![\[V_s = I*R_s = 15 * 10 = 150 V\]](https://electrically4u.com/wp-content/ql-cache/quicklatex.com-f016fa9bedd1125fb7065c08dd0f0aff_l3.png "Rendered by QuickLaTeX.com")
So the transformed circuit becomes,
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