Question

Find IL for the circuit shown in fig. using Source transformation.

Answer 1

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

This article provides a detailed walkthrough of solving a DC circuit problem using the source transformation method. The goal is to find the current, denoted as I_L, flowing through the 30Ω resistor in the given circuit.

The Problem

Question: Find I_L for the circuit shown in the figure using Source Transformation.

Initial Circuit Diagram:

The circuit consists of:
A 50V voltage source in series with a 10Ω resistor on the left.
A 20V voltage source in series with a 20Ω resistor on the right.
* A 30Ω load resistor in the central branch, through which the current I_L flows.

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Step-by-Step Solution

The source transformation technique is used to simplify the circuit by converting voltage sources into equivalent current sources, or vice versa. This often makes it easier to combine elements that are in parallel.

Step 1: Apply Source Transformation to the Left Branch

We transform the 50V voltage source and its series 10Ω resistor into an equivalent current source with a parallel resistor.

• Principle: A voltage source (V) in series with a resistor (R) can be converted into a current source (I) in parallel with the same resistor (R).
• Calculation: The value of the current source is found using Ohm's Law:
  I = V / R = 50V / 10Ω = 5A
• Direction: The direction of the current source arrow points towards the positive terminal of the original voltage source (upwards).

The left branch is now a 5A current source in parallel with the 10Ω resistor.

Step 2: Apply Source Transformation to the Right Branch

Similarly, we transform the 20V voltage source and its series 20Ω resistor.

• Calculation:
  I = V / R = 20V / 20Ω = 1A
• Direction: The arrow of the new current source also points upwards, towards the positive terminal of the 20V source.

The right branch is now a 1A current source in parallel with the 20Ω resistor.

Step 3: Simplify the Transformed Circuit

After performing the transformations, all components in the circuit are now in parallel.

Transformed Circuit Diagram:

We can now simplify this parallel circuit:

1. Combine Parallel Current Sources: Since both the 5A and 1A current sources point in the same direction (upwards), we can add them to get a single equivalent current source.
   Total Current (I_total) = 5A + 1A = 6A

2. Combine Parallel Resistors: The 10Ω and 20Ω resistors are in parallel. We can combine them into a single equivalent resistor (R_eq).
   R_eq = (10Ω * 20Ω) / (10Ω + 20Ω) = 200 / 30 ≈ 6.67Ω
   (Note: The solution in the image uses the approximation 6.6Ω for this value, which we will use for consistency in the final step.)

Simplified Circuit Diagram:

Step 4: Calculate I_L using the Current Divider Rule

Now we have a simple parallel circuit with a 6A total current splitting between the 6.6Ω resistor and the 30Ω load resistor. We can find the current I_L through the 30Ω resistor using the Current Divider Rule.

• Formula: To find the current in a specific branch (I_L), multiply the total current by the resistance of the other parallel branch and divide by the sum of the two resistances.
  I_L = I_total * (R_other / (R_L + R_other))

• Calculation:
  I_L = 6A * (6.6Ω / (30Ω + 6.6Ω))
I_L = 6 * (6.6 / 36.6)
I_L = 39.6 / 36.6
I_L ≈ 1.0819 A

Final Answer

By applying source transformation and the current divider rule, the current I_L flowing through the 30Ω resistor is:

I_L = 1.081 A

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Key Concepts Explained

• Source Transformation: This is a powerful circuit analysis tool used to simplify circuits. It allows for the interchange between equivalent voltage sources and current sources.
  Voltage to Current: A voltage source V in series with a resistor R is equivalent to a current source I = V/R in parallel with the same resistor R.
  Current to Voltage: A current source I in parallel with a resistor R is equivalent to a voltage source V = I*R in series with the same resistor R.

• Current Divider Rule: This rule is used to determine how current splits between two or more parallel branches. For two parallel resistors, R1 and R2, the current through R2 (I2) from a total current (I_total) is I2 = I_total * (R1 / (R1 + R2)).