Table of Contents
Introduction 🙂
Source transformation is a technique that lets you replace one kind of source–resistor combination with an equivalent one that behaves the same at its terminals. It is especially useful when preparing circuits for Thevenin and Norton equivalents or when simplifying networks for node or mesh analysis.
In this chapter you will focus on how to convert between voltage sources and current sources with series or parallel resistances, why the conversions are valid, and how to apply them correctly and safely in DC circuit analysis.
Equivalent Source Forms 🔁
In basic DC circuit analysis you often meet two standard forms of a source combined with a resistor:
- A voltage source in series with a resistor.
- A current source in parallel with a resistor.
At a given pair of external terminals, these two forms can be made electrically equivalent. That means any load connected to the terminals will experience the same voltage and current in both versions, so from the outside they are indistinguishable.
The idea of source transformation is to switch back and forth between these two equivalent representations to make a circuit easier to analyze.
Voltage Source with Series Resistance 🔋
Consider a practical voltage source that is not ideal. It is modeled by an ideal voltage source $V_s$ in series with a resistor $R_s$. The terminals of this combination are where you connect the rest of the circuit or a load resistor $R_L$.
If you connect a load $R_L$ across the terminals, the circuit becomes a simple series path of $V_s$, $R_s$, and $R_L$. The current that flows is
$$
I = \frac{V_s}{R_s + R_L}
$$
and the load voltage is
$$
V_L = I R_L = \frac{V_s R_L}{R_s + R_L}.
$$
Both $I$ and $V_L$ depend on $V_s$, $R_s$, and the load resistance $R_L$.
For source transformation, you want another arrangement, a current source with parallel resistance, that will produce the same current through any $R_L$ and the same voltage across the terminals. That requirement determines how to relate the parameters of the two source forms.
Current Source with Parallel Resistance 💧
Now consider a practical current source modeled by an ideal current source $I_s$ in parallel with a resistor $R_p$. Again, the terminals are across this parallel combination, where a load $R_L$ can be connected.
When you connect a load $R_L$ across the terminals, the load is in parallel with $R_p$ and in parallel with the ideal current source $I_s$. The current divides between $R_p$ and $R_L$ according to the rules of parallel circuits, and the voltage across both resistors is the same.
Denote the terminal voltage as $V_L$. Ohm’s law gives
$$
I_p = \frac{V_L}{R_p}, \quad I_L = \frac{V_L}{R_L}
$$
for the currents through $R_p$ and $R_L$, respectively. The current source supplies the sum of these currents,
$$
I_s = I_p + I_L = \frac{V_L}{R_p} + \frac{V_L}{R_L} = V_L \left( \frac{1}{R_p} + \frac{1}{R_L} \right).
$$
For equivalence with the voltage source model, the relationship between $I_s$, $R_p$, and the original $V_s$, $R_s$ must be chosen so that the terminal behavior is identical for any $R_L$.
Core Source Transformation Relationships 📐
The key idea of equivalence is that, seen from the terminals, both representations must have the same open-circuit voltage and the same short-circuit current. If those two match and the resistance seen at the terminals is the same, then the two models will produce exactly the same voltage and current for any load.
Start with the voltage source model:
- Open circuit, no load: current is zero, so the terminal voltage is simply $V_{OC} = V_s$.
- Short circuit: the terminals are directly connected, so the current is
$$
I_{SC} = \frac{V_s}{R_s}.
$$
Now for the current source model:
- Open circuit: no current can flow out through the terminals, so all the current $I_s$ flows through the parallel resistor. The terminal voltage is then
$$
V_{OC} = I_s R_p.
$$
- Short circuit: the terminals are shorted, so the voltage is $0$, which forces the resistor current to $0$. All of $I_s$ flows through the short, so
$$
I_{SC} = I_s.
$$
To be equivalent, the open-circuit voltages and short-circuit currents must match:
$$
V_s = I_s R_p, \quad \frac{V_s}{R_s} = I_s.
$$
These two equations show that, for equivalence, the resistances must be equal and the source values must be consistent.
From $V_s = I_s R_p$ write $I_s = \dfrac{V_s}{R_p}$.
Comparing with $I_s = \dfrac{V_s}{R_s}$, you get $R_p = R_s$.
So it is customary to write just $R$ and state: the series resistor of the voltage form and the parallel resistor of the current form are the same physical resistance.
Substituting $R_p = R_s = R$ gives the central formulas:
Voltage source to current source:
$$I_s = \frac{V_s}{R}, \quad R_{\text{parallel}} = R_{\text{series}}.$$
Current source to voltage source:
$$V_s = I_s R, \quad R_{\text{series}} = R_{\text{parallel}}.$$
These relationships define the basic DC source transformations.
Step by Step: Transforming a Voltage Source 🔃
Suppose you have a Thevenin style source: an ideal voltage source of $12\ \text{V}$ in series with a $4\ \Omega$ resistor. You want to replace it with a Norton style current source plus resistor.
From the formulas, use $R = 4\ \Omega$ and
$$
I_s = \frac{V_s}{R} = \frac{12\ \text{V}}{4\ \Omega} = 3\ \text{A}.
$$
The equivalent is an ideal current source of $3\ \text{A}$ in parallel with a $4\ \Omega$ resistor. Both the original voltage source and the transformed current source will show the same voltage and current to any load attached to the terminals.
If you connect a $2\ \Omega$ load to the voltage source version, the total series resistance is $4 + 2 = 6\ \Omega$ and the load current is
$$
I_L = \frac{12}{6} = 2\ \text{A}.
$$
The load voltage is $V_L = I_L R_L = 2 \cdot 2 = 4\ \text{V}$.
If you connect the same load to the current source version, the $4\ \Omega$ and $2\ \Omega$ resistors are in parallel and share the $3\ \text{A}$ according to current division. You can compute the terminal voltage directly instead. The equivalent resistance of $4\ \Omega$ in parallel with $2\ \Omega$ is
$$
R_{\text{eq}} = \frac{4 \cdot 2}{4 + 2} = \frac{8}{6} = \frac{4}{3}\ \Omega.
$$
The total terminal voltage caused by the $3\ \text{A}$ source is
$$
V_L = I_s R_{\text{eq}} = 3 \cdot \frac{4}{3} = 4\ \text{V}.
$$
That matches the previous result, and the current through the $2\ \Omega$ load is $I_L = \dfrac{4}{2} = 2\ \text{A}$ just as before.
Step by Step: Transforming a Current Source 🔄
Now reverse the situation. Suppose you have a Norton style current source: $5\ \text{A}$ in parallel with $10\ \Omega$, and you want the equivalent Thevenin voltage source representation.
Use the relationship
$$
V_s = I_s R = 5\ \text{A} \cdot 10\ \Omega = 50\ \text{V},
$$
and keep the same resistance in series, $R = 10\ \Omega$. The equivalent is an ideal $50\ \text{V}$ voltage source in series with $10\ \Omega$.
To verify, imagine connecting a $5\ \Omega$ load across the terminals.
For the current source model, the parallel network is $10\ \Omega$ and $5\ \Omega$ in parallel, with total resistance
$$
R_{\text{eq}} = \frac{10 \cdot 5}{10 + 5} = \frac{50}{15} = \frac{10}{3}\ \Omega.
$$
The total voltage is
$$
V_L = I_s R_{\text{eq}} = 5 \cdot \frac{10}{3} = \frac{50}{3}\ \text{V}.
$$
The current through the $5\ \Omega$ load is
$$
I_L = \frac{V_L}{5} = \frac{50/3}{5} = \frac{50}{15} = \frac{10}{3}\ \text{A}.
$$
For the voltage source form, the series combination is $10\ \Omega + 5\ \Omega = 15\ \Omega$, and the load current is directly
$$
I_L = \frac{V_s}{R_s + R_L} = \frac{50}{10 + 5} = \frac{50}{15} = \frac{10}{3}\ \text{A}.
$$
The load voltage is then $V_L = I_L R_L = \dfrac{10}{3} \cdot 5 = \dfrac{50}{3}\ \text{V}$. Again, the two models are identical from the perspective of any load.
Summary Table of DC Source Transformations 📊
It is useful to collect the main relationships in one place. Let $R$ represent the resistance in either form.
| From | To | Relationship |
|---|---|---|
| Voltage source $V_s$ in series with $R$ | Current source $I_s$ in parallel with $R$ | $I_s = \dfrac{V_s}{R}$, same $R$ |
| Current source $I_s$ in parallel with $R$ | Voltage source $V_s$ in series with $R$ | $V_s = I_s R$, same $R$ |
Remember that the polarity of the voltage source and the direction of the current source must correspond. The arrow of the current source should point from the positive terminal toward the negative terminal of the equivalent voltage source.
Using Source Transformation to Simplify Circuits 🧩
Source transformation is particularly helpful when you see a source and resistor that attach to the rest of the circuit at only two nodes. That means there are only two terminals by which the combination interacts with everything else. You can then safely replace the source–resistor pair with its equivalent form without affecting the rest of the circuit.
For example, consider a circuit where a $24\ \text{V}$ source is in series with $6\ \Omega$, and this pair is directly connected to a more complicated parallel network. You might prefer to see a current source feeding the parallel network instead of a series source feeding into it.
Transform the $24\ \text{V}$ source and $6\ \Omega$ series into a current source with
$$
I_s = \frac{24}{6} = 4\ \text{A}
$$
in parallel with a $6\ \Omega$ resistor. Now the current source and its parallel resistance can be combined or grouped with the existing parallel elements, which often makes it easier to find equivalent resistances or to perform node voltage analysis.
You can also transform in the opposite direction if you have a current source feeding a set of series elements and want to see a series source instead. The choice of direction is purely for convenience and depends on the surrounding circuit.
Multiple Transformations in a Circuit 🔄🔄
In more complex circuits, you may apply source transformation several times in succession. As long as each transformation replaces a source plus its attached resistance at two terminals and you keep track of polarity and orientation, the overall terminal behavior remains unchanged.
For instance, imagine two adjacent Norton sources each with its own parallel resistance, both connected in parallel between the same two nodes. You can combine the parallel resistors into a single equivalent resistance and add the currents of the ideal sources to get a single Norton source. You can then transform that single Norton source into a Thevenin equivalent if that form is more convenient for later steps.
Similarly, you might transform several series voltage sources with series resistors into equivalent Norton sources, combine them in parallel, and then transform back to a single Thevenin source at the end. This approach is quite common when building Thevenin or Norton equivalents for a whole section of a network.
Conditions and Limitations ⚠️
Source transformation is a powerful technique, but it has important conditions.
First, it applies only when the source and the resistor form a simple series or parallel connection and share the same pair of external terminals. If other elements are connected at a point between the source and the resistor, you cannot simply combine the source and resistor into a single equivalent pair without affecting those internal connection points.
Second, the method is valid for linear resistive circuits. In basic DC analysis, this usually holds, because resistors and ideal sources are linear elements. Once you introduce dependent sources or nonlinear devices such as diodes and transistors, you must be more careful. You can still apply source transformations in some cases, but only when the dependent source and its associated elements maintain the same two-terminal behavior. That requires more advanced analysis which will be covered elsewhere.
Third, the internal resistor associated with the source is part of the source model. You should not lose track of that resistance during transformation. The same value must appear either in series with the equivalent voltage source or in parallel with the equivalent current source, never both and never changed.
Finally, transformations do not change the power delivered or absorbed by the overall source–resistor pair under any given load. The internal distribution of currents may look different, but the external behavior at the terminals, including power exchange with the rest of the circuit, remains identical.
Practical Tips for Problem Solving 🛠️
When you look at a circuit and consider source transformation, start by getting a clear view of the terminals of each source and how it connects to other elements. If you can draw a box around a source and its single series or parallel resistor that touches the rest of the circuit only at two nodes, then you can replace that box with its equivalent form.
It often helps to combine simple series or parallel resistors first, then check whether the resulting source–resistor pair can be transformed. After a transformation, redraw the circuit cleanly and look again for new combinations or simplifications, such as parallel sources that can be added together or series resistors that can be merged.
Another useful habit is to label the polarity of your voltage sources and the direction of your current sources carefully. When you transform, ensure that the direction of the current source points from the positive terminal of the equivalent voltage source toward its negative terminal. That consistency prevents sign errors when you later apply laws from basic circuit theory.
Connection with Thevenin and Norton Equivalents 🔗
The direct relationship between a voltage source with series resistance and a current source with parallel resistance is exactly the relationship between Thevenin and Norton equivalent circuits. A Thevenin equivalent is a voltage source in series with a resistance, and a Norton equivalent is a current source in parallel with a resistance.
Source transformation gives you the algebraic link between them:
Given a Thevenin equivalent $(V_{TH}, R_{TH})$, the corresponding Norton equivalent is:
$$I_N = \frac{V_{TH}}{R_{TH}}, \quad R_N = R_{TH}.$$
Given a Norton equivalent $(I_N, R_N)$, the corresponding Thevenin equivalent is:
$$V_{TH} = I_N R_N, \quad R_{TH} = R_N.$$
This close connection is why source transformation is such a useful tool when finding or using these equivalents in circuit analysis.
Closing Remarks ✅
Source transformation gives you a flexible way to recast voltage sources into equivalent current sources and back again, always preserving the same terminal behavior. By mastering the simple relationships between $V_s$, $I_s$, and $R$, and by carefully identifying where source–resistor pairs can be isolated, you gain a powerful method to simplify DC circuits and to move smoothly between Thevenin and Norton viewpoints.