Convert the circuit shown in fig. into a single voltage source in series with resistance.
1 Answer
Mastering Source Transformation: A Step-by-Step Circuit Simplification Example
Source transformation is a fundamental and powerful technique in electrical circuit analysis. It allows you to simplify a complex circuit by replacing a voltage source in series with a resistor with an equivalent current source in parallel with the same resistor, and vice-versa. This method, rooted in Thevenin's and Norton's theorems, makes solving for currents and voltages in a network significantly easier.
In this guide, we will walk through a complete example of simplifying a circuit into a single equivalent voltage source and series resistor.
The Problem
Question: Convert the circuit shown in the figure into a single voltage source in series with a resistor.
Original Circuit Diagram:
The initial circuit contains:
A 50V voltage source in series with a 5Ω resistor on the left branch.
A 10V voltage source in series with a 3Ω resistor on the right branch.
* A 10A independent current source and a 2Ω resistor in the central bridge structure.
Step-by-Step Solution
To solve this, we will convert all voltage sources into current sources, combine the parallel components, and then convert the final simplified circuit back into a voltage source.
Step 1: Convert the 50V Voltage Source to a Current Source
First, we'll transform the leftmost branch. A voltage source (V) in series with a resistor (R) is equivalent to a current source (I) in parallel with the same resistor (R).
- Formula: I = V / R
- Calculation: I = 50V / 5Ω = 10A
The 5Ω resistor is now placed in parallel with this new 10A current source. The direction of the current arrow points towards the positive terminal of the original voltage source (upwards).
Step 2: Convert the 10V Voltage Source to a Current Source
Next, we apply the same transformation to the rightmost branch of the circuit.
- Formula: I = V / R
- Calculation: I = 10V / 3Ω = 10/3 A
The 3Ω resistor is now in parallel with the 10/3 A current source. The arrow also points upwards, toward the original positive terminal.
Step 3: Combine All Parallel Resistors
After the transformations, our circuit consists of only current sources and resistors, all connected between the same two nodes. This means all three resistors (5Ω, 2Ω, and 3Ω) are in parallel. We can combine them into a single equivalent resistor (Req).
- Formula for Parallel Resistors: 1/Req = 1/R₁ + 1/R₂ + 1/R₃
- Calculation:
1/Req = 1/5 + 1/2 + 1/3
To add these fractions, we find a common denominator, which is 30.
1/Req = (6/30) + (15/30) + (10/30)
1/Req = 31/30 - Equivalent Resistance: Req = 30/31 Ω (approx. 0.97 Ω)
Step 4: Combine All Parallel Current Sources
Just like the resistors, all three current sources are now in parallel. Since their current arrows all point in the same direction (upwards), we can simply add their values to find the total equivalent current (I_total).
- Sources: 10A (from the 50V source), 10A (original source), and 10/3 A (from the 10V source).
- Calculation:
I_total = 10A + 10A + (10/3)A
I_total = 20 + 10/3
I_total = (60/3) + (10/3) = 70/3 A - Simplified Circuit (Norton Equivalent):
The circuit is now reduced to a single current source of 70/3 A in parallel with a single resistor of 30/31 Ω.
Step 5: Convert to the Final Equivalent Voltage Source
The final step is to convert this simplified Norton equivalent circuit back into a Thevenin equivalent circuit (a voltage source in series with a resistor), as requested by the problem.
- The series resistance (R_th) is the same as the equivalent parallel resistance: R_eq = 30/31 Ω
- The equivalent voltage (V_eq) is found using Ohm's Law: V = I * R
- Calculation:
V_eq = (70/3 A) × (30/31 Ω)
V_eq = (70 × 10) / 31
V_eq = 700 / 31 V - Equivalent Voltage: V_eq ≈ 22.58 V
Final Answer
The original circuit can be simplified into a single equivalent circuit consisting of a 22.58 V voltage source in series with a 30/31 Ω (or approximately 0.97 Ω) resistor.
Final Circuit Diagram: