Question

Find the current in branch AB of the circuit shown in fig. using Source transformation

Answer 1

Find the Current in Branch AB using Source Transformation

This article provides a detailed solution for finding the current in branch AB of the given electrical circuit. The method used is Source Transformation, a powerful technique in circuit analysis for simplifying complex circuits.

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The Problem

Question: Find the current flowing through branch AB of the circuit shown in the figure using the source transformation method.

Original Circuit Diagram:

The circuit consists of:
A 6V voltage source in series with a 15Ω resistor on the left.
A 6V voltage source in series with an 8Ω resistor on the right.
A 5Ω resistor and an 8Ω resistor in the central part of the circuit.
A 10Ω resistor connecting nodes A and B (this is branch AB).

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Solution using Source Transformation

The goal is to simplify the circuit into a single loop to easily calculate the current I flowing through the 10Ω resistor in branch AB.

Step 1: Transform the Voltage Sources into Current Sources

Source transformation allows us to convert a voltage source in series with a resistor into an equivalent current source in parallel with the same resistor. The formula is I = V / R.

• Left Side Transformation:
  The 6V source and the 15Ω series resistor are converted to a current source.
  I_left = V / R = 6V / 15Ω = 0.4A
  This results in a 0.4A current source (with the arrow pointing up) in parallel with the 15Ω resistor.

• Right Side Transformation:
  The 6V source and the 8Ω series resistor are converted to a current source.
  I_right = V / R = 6V / 8Ω = 0.75A
  This results in a 0.75A current source (with the arrow pointing up) in parallel with the 8Ω resistor.

The circuit now looks like this:

Step 2: Combine Parallel Resistors

Now, we can simplify the circuit by combining the resistors that are in parallel.

• Left Side Resistors:
  The 15Ω resistor and the 5Ω resistor are in parallel.
  R_left = 15Ω || 5Ω = (15 × 5) / (15 + 5) = 75 / 20 = 3.75Ω

• Right Side Resistors:
  The two 8Ω resistors are in parallel.
  R_right = 8Ω || 8Ω = (8 × 8) / (8 + 8) = 64 / 16 = 4Ω

The simplified circuit becomes:

Step 3: Transform Current Sources back to Voltage Sources

To create a final, single-loop circuit, we transform the current sources back into voltage sources. The formula is V = I × R.

• Left Side Transformation:
  The 0.4A current source and the 3.75Ω parallel resistor are converted back to a voltage source.
  V_left = I × R = 0.4A × 3.75Ω = 1.5V
  This results in a 1.5V voltage source in series with the 3.75Ω resistor.

• Right Side Transformation:
  The 0.75A current source and the 4Ω parallel resistor are converted back to a voltage source.
  V_right = I × R = 0.75A × 4Ω = 3V
  This results in a 3V voltage source in series with the 4Ω resistor.

This gives us the final single-loop circuit:

Step 4: Apply Kirchhoff's Voltage Law (KVL)

We can now find the current I in branch AB by applying KVL to the single loop. Let's assume the current I flows in a clockwise direction (from A to B).

Applying KVL by traversing the loop clockwise:
-1.5V + (3.75Ω × I) + (10Ω × I) + (4Ω × I) + 3V = 0

Step 5: Calculate the Final Current

Now, we solve the KVL equation for I.

1. Combine the terms with I:
   (3.75 + 10 + 4) × I + (-1.5 + 3) = 0

2. Simplify the equation:
   17.75 × I + 1.5 = 0

3. Isolate I:
   17.75 × I = -1.5

4. Solve for I:
   I = -1.5 / 17.75
I ≈ -0.0845 A

To express the answer in milliamps (mA), multiply by 1000:
I = -0.0845 A × 1000 = -84.5 mA

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Final Answer

The current flowing in branch AB is:

I = -84.5 mA

The negative sign indicates that the actual direction of the current is opposite to our assumed clockwise direction. Therefore, the current of 84.5 mA flows from node B to node A.