Question

Using source conversion, reduce the circuit shown in the figure into a single voltage source in series with a single resistance.

Answer 1

Source Transformation Explained: A Step-by-Step Circuit Simplification Example

Source transformation is a powerful circuit analysis technique used to simplify complex circuits. It allows you to convert a voltage source in series with a resistor into an equivalent current source in parallel with the same resistor, and vice-versa. This method is fundamental for finding the Thevenin or Norton equivalent of a circuit.

The core principle is based on Ohm's Law:
Voltage Source to Current Source: I = V / R
Current Source to Voltage Source: V = I * R

In both conversions, the resistance value remains the same, but its connection changes from series (with a voltage source) to parallel (with a current source), or vice versa.

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Problem Statement

Q. Using source conversion, reduce the circuit shown in the figure into a single voltage source in series with a single resistance.

Initial Circuit:
The circuit consists of five branches in parallel connected between terminals a and b:
1. A 3A ideal current source.
2. A 3Ω resistor.
3. A 2A ideal current source.
4. A 6Ω resistor.
5. Another 6Ω resistor.

(Note: The solution in the image interprets the first two branches—the 3A source and the 3Ω resistor—as a single practical current source. We will follow this interpretation first.)

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Solution Walkthrough (As Shown in the Image)

Here is a breakdown of the step-by-step solution provided in the image.

Step 1: Convert the 3A Current Source to a Voltage Source
The leftmost 3A current source is in parallel with a 3Ω resistor. These can be converted into a voltage source.
Calculation: V = I × R = 3A × 3Ω = 9V
Result: A 9V voltage source in series with a 3Ω resistor.
Circuit Update: The original circuit is modified. However, the diagram in the solution incorrectly combines this new 3Ω series resistor with the other 3Ω parallel resistor, treating them as if they are in series to get 6Ω. This is an error.* For the sake of following the provided solution, we will proceed with their (incorrect) 6Ω resistor.

Step 2: Combine the Two 6Ω Resistors
The two 6Ω resistors on the right are in parallel. They can be combined into a single equivalent resistor.
Calculation: R_eq = (6 × 6) / (6 + 6) = 36 / 12 = 3Ω
Result: A single 3Ω resistor.

Step 3: Convert the 9V Voltage Source Back to a Current Source
Now, the 9V source (in series with the incorrect 6Ω resistor) is converted back to a current source to simplify it with the other parallel elements.
Calculation: I = V / R = 9V / 6Ω = 1.5A
Result: A 1.5A current source in parallel with a 6Ω resistor.

Step 4: Combine Parallel Current Sources
The circuit now has two current sources (1.5A and 2A) in parallel, both pointing up. Since they are in the same direction, their values add up.
Calculation: I_total = 1.5A + 2A = 3.5A
Result: A single 3.5A current source.

Step 5: Combine Parallel Resistors
The remaining resistors (6Ω and 3Ω) are in parallel and can be combined.
Calculation: R_eq = (6 × 3) / (6 + 3) = 18 / 9 = 2Ω
Result: A single 2Ω equivalent resistor.

Step 6: Final Conversion to a Voltage Source
The simplified circuit now consists of a 3.5A current source in parallel with a 2Ω resistor. This is the Norton equivalent. To get the final answer, we convert it to a voltage source (Thevenin equivalent).
Calculation: V_eq = I_total × R_eq = 3.5A × 2Ω = 7V
Result: A 7V voltage source in series with a 2Ω resistor.

Final Answer (from image): 7V in series with 2Ω

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Analysis and Corrected Solution

The solution in the image contains a critical error in Step 1. When the 3A source and its parallel 3Ω resistor are converted, the resulting 9V source and 3Ω series resistor form a new branch. This entire branch is parallel to the other 3Ω resistor, not in series with it.

Here is the correct and more efficient method to solve this problem.

Correct Step 1: Combine All Parallel Elements First
A simpler approach is to combine all parallel current sources and all parallel resistors at the beginning.

• Combine Parallel Current Sources:
  The 3A and 2A sources are in parallel and point in the same direction.
  I_total = 3A + 2A = 5A

• Combine Parallel Resistors:
  All three resistors (3Ω, 3Ω, and 6Ω, 6Ω) are in parallel.
  R_eq1 = 3Ω || 3Ω = (3 × 3) / (3 + 3) = 1.5Ω
  R_eq2 = 6Ω || 6Ω = (6 × 6) / (6 + 6) = 3Ω
  R_total = R_eq1 || R_eq2 = 1.5Ω || 3Ω = (1.5 × 3) / (1.5 + 3) = 4.5 / 4.5 = 1Ω

Correct Step 2: Final Source Transformation
The entire circuit has been simplified to a 5A current source in parallel with a 1Ω resistor. Now, perform the final source transformation to get the required voltage source.

• Calculation: V_Thevenin = I_Norton × R_Norton = 5A × 1Ω = 5V
• Thevenin Resistance: The series resistance is the same as the parallel resistance, which is 1Ω.

Correct Final Answer

The correct equivalent circuit is a 5V voltage source in series with a 1Ω resistor.

Key Takeaways

• Identify Parallel Elements: Before applying source transformation, always check if you can simplify the circuit by combining sources or resistors that are already in series or parallel.
• Check Circuit Topology: After each transformation, carefully redraw the circuit to ensure you have the correct connections. A common mistake is misinterpreting new series/parallel relationships, as seen in the original image's solution.
• Thevenin vs. Norton: A voltage source in series with a resistor is a Thevenin equivalent circuit. A current source in parallel with a resistor is a Norton equivalent circuit. Source transformation is the method to convert between them.