Question

Find Current I using Source Transformation.

Answer 1

---

How to Find Current in a Circuit Using Source Transformation: A Step-by-Step Guide

Source transformation is a powerful technique in circuit analysis that simplifies complex circuits by converting voltage sources into equivalent current sources, and vice versa. This method can make it much easier to find unknown values like voltage or current.

In this guide, we will walk through a detailed example to find the current I flowing through a 6Ω resistor using the source transformation method.

The Problem

Question: Find the current I flowing through the central 6Ω resistor in the circuit below using source transformation.

Original Circuit:

The circuit consists of:
A 24V voltage source in series with a 6Ω resistor on the left.
A 12V voltage source in series with a 6Ω resistor on the right.
* A 6Ω resistor in the central branch, where we need to find the current I.

Solution: Step-by-Step Breakdown

The goal is to simplify the circuit by converting the voltage sources into current sources. This will place all components in parallel, making them easier to combine.

Step 1: Apply Source Transformation to Both Voltage Sources

A voltage source (V) in series with a resistor (R) can be transformed into a current source (I) in parallel with the same resistor (R). The value of the current source is calculated using Ohm's Law: I = V / R.

• Left Branch Transformation:
  The 24V source with its 6Ω series resistor is transformed into a current source.
  Current (I₁) = 24V / 6Ω = 4A
  The direction of the current source arrow points upwards, towards the positive terminal of the original voltage source. This 4A source is now in parallel with a 6Ω resistor.

• Right Branch Transformation:
  Similarly, the 12V source with its 6Ω series resistor is transformed.
  Current (I₂) = 12V / 6Ω = 2A
  The direction of this current source also points upwards. This 2A source is now in parallel with a 6Ω resistor.

Step 2: Redraw and Simplify the Transformed Circuit

After performing the transformations, the new circuit has all components in a parallel configuration.

We can now simplify this parallel circuit:

1. Combine the Parallel Current Sources: Since both the 4A and 2A current sources are in the same direction (upwards), we can add them to create a single equivalent current source.
   * Total Current (I_total) = 4A + 2A = 6A

2. Combine the Parallel Resistors: The two 6Ω resistors that were part of the source transformations are in parallel. We can combine them into a single equivalent resistor. (Note: We leave the central 6Ω resistor alone for now, as it's the branch where we need to find the current I).
   Equivalent Resistance (R_eq) = (6Ω 6Ω) / (6Ω + 6Ω) = 36 / 12 = 3Ω

After simplification, the circuit is reduced to a 6A current source in parallel with two resistors: the central 6Ω resistor and the equivalent 3Ω resistor.

Step 3: Apply the Current Divider Rule to Find I

Now, we can find the current I flowing through the central 6Ω resistor by using the current divider rule. This rule determines how the total current from a source is split between parallel branches.

The formula for the current in a specific branch is:

*I_branch = I_total (R_other / (R_branch + R_other))**

In our case:
I_total = 6A
R_branch (the resistor in the branch we're interested in) = 6Ω
* R_other (the resistance of the other parallel branch) = 3Ω

Plugging in the values:

I = 6A (3Ω / (6Ω + 3Ω))
I = 6 (3 / 9)
I = 6 * (1/3)
I = 2A

Final Answer

By using source transformation to simplify the circuit and then applying the current divider rule, we find that the current I flowing through the central 6Ω resistor is 2A.