Source Transformation - Voltage to Current Source & Current to Voltage Source Conversion

What is a Source Transformation?

A source transformation or source conversion technique is a method employed in analyzing complex electrical circuits where an energy source (voltage or current) is converted or replaced by an equivalent source (voltage or current).

In source transformation technique, a voltage source in series with a resistor is converted to an equivalent current source in parallel with the same resistor, and similarly, a current source in parallel with a resistor is converted to an equivalent voltage source in series with the same resistor.

Voltage to Current Source Conversion:

An ideal voltage source delivers constant voltage across its terminals. It has zero internal resistance due to which there is no voltage drop and thus provides constant terminal voltage at any time irrespective of the amount of current drawn in the circuit. However, an ideal voltage source is a theoretical concept because in real life it is very difficult to achieve an ideal voltage source.

A practical or real voltage source has some internal resistance which causes a voltage drop in the source itself when connected to a load. Thus whatever the voltage generated by the source will not be equal to the voltage available across the load terminals.

A practical voltage source is represented by an ideal voltage source in series with a resistance R_int (equal to its internal resistance) as shown in the above figure. From the above voltage source figure, the load current I_L is given by,

$$I_L=\frac{V}{R_{\mathrm{int}}+R_L}=\frac{\frac{V}{R_{\mathrm{int}}}}{\frac{R_{\mathrm{int}}+R_L}{R_{\mathrm{int}}}}=\frac{V}{R_{\mathrm{int}}}\times \frac{R_{\mathrm{int}}}{R_{\mathrm{int}}+R_L}$$

$$I_L=I_s\times \frac{R_{\mathrm{int}}}{R_{\mathrm{int}}+R_L}$$

$$I_s=\frac{V}{R_{\mathrm{int}}}$$

From the above equation, we can say that a practical voltage source (an ideal voltage source in series with internal resistance) is equivalent to a current source of current I_s = V / R_int (where R_int is in parallel with the current source) as shown in the above figure. Note that the value of the internal resistance of practical voltage source is equal to the internal resistance of equivalent current source.

Current to Voltage Source Conversion:

Similar to an ideal voltage source, an ideal current source delivers constant current to the load. It supplies same current regardless of the value of load resistance connected due to its infinite internal resistance.

A real or non-ideal current source has some finite internal resistance due to which load current varies based on the load resistance. The circuit diagram of a practical current source can be constructed as an ideal current source connected in parallel with its internal resistance as shown in the below figure.

The above figure also shows an equivalent voltage source, where the value of internal resistance is equal to the value of parallel resistance of current source. The value of voltage of equivalent voltage source is given as,

$$V_s=I\times R_{\mathrm{int}}$$

Examples on Source Transformation:

Example-1:

A voltage source of 50 V having internal resistance (R_int) of 5 Ω is connected to a resistance (R) of 5 Ω as shown in the below figure. Find the current in the resistor R using source transformation.

Solution:

Converting voltage source to current source, the value of current source will be,

$$I_s=\frac{V}{R_{\mathrm{int}}}=\frac{50}{5}=10A$$

The equivalent current source is also shown in the above circuit. The current in resistance R will be,

$$I=I_s\times \frac{R_{\mathrm{int}}}{R_{\mathrm{int}}+R}=10\times \frac{5}{5+5}=5A$$

Example-2:

A current source of 5 A having internal resistance (R_int) of 10 Ω is connected to a resistance (R) of 5 Ω as shown in the below figure. Find the current in the resistor R using source transformation.

Solution:

Converting current source to voltage source, the value of voltage source will be,

$$V_s=I\times R_{\mathrm{int}}=5\times 10=50V$$

The equivalent voltage source is also shown in the above circuit. The current in resistance R will be,

$$I=\frac{V_s}{R_{\mathrm{int}}+R}=\frac{50}{10+5}=3.33A$$

Conclusion:

In this article, we have learned about source transformation technique, converting a voltage source into a current source and vice-versa which simplifies the process of complex circuit analysis. Therefore, we can include any resistance (internal or external) in series with a voltage source while converting it to an equivalent current source, also we can include any resistance in parallel with the current source while converting it to an equivalent voltage source.