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This article provides a complete walkthrough for solving an electrical circuit problem to find the output voltage (Vo) using the source transformation technique. Source transformation is a powerful tool in circuit analysis that simplifies complex circuits by converting voltage sources into equivalent current sources, and vice-versa.
The Problem:
Given the following electrical circuit, calculate the output voltage Vo across the 4 kΩ resistor using source transformation.
Original Circuit:
The circuit consists of a current source, two voltage sources, and four resistors. Our goal is to find the voltage Vo across the 4 kΩ resistor.
We will simplify the circuit step-by-step by repeatedly applying the source transformation rule.
V in series with a resistor R can be transformed into a current source I = V/R in parallel with the same resistor R.I in parallel with a resistor R can be transformed into a voltage source V = I * R in series with the same resistor R.Step 1: Convert the 6 mA Current Source to a Voltage Source
The leftmost part of the circuit has a 6 mA current source in parallel with a 6 kΩ resistor. We can convert this into a voltage source in series with the 6 kΩ resistor.
V = I * R = 6 mA * 6 kΩ = (6 * 10⁻³) A * (6 * 10³) Ω = 36 VStep 2: Combine Series Voltage Sources
The new 36 V source is now in series with the original 3 V source. Since their polarities align (positive-to-negative connection), we can add them to create a single equivalent voltage source.
V_total = 36 V + 3 V = 39 VThe circuit now has a 39 V source in series with the 6 kΩ resistor.
Step 3: Convert Voltage Sources to Current Sources
To further simplify the circuit, we will convert both remaining voltage source branches into parallel current sources.
a) Right Branch (10 V source):
Calculation: I = V / R = 10 V / 10 kΩ = 1 mA
The positive terminal of the 10 V source is at the top, so the equivalent current source arrow points down (conventional current flows from + to - through the external circuit).
b) Left Branch (39 V source):
Calculation: I = V / R = 39 V / 6 kΩ = 6.5 mA
The positive terminal of the 39 V source is at the top, so the equivalent current source arrow points up.
Step 4: Combine Parallel Sources and Resistors
The circuit now consists of several components in parallel. We can combine the parallel current sources and the parallel resistors (excluding the 4 kΩ resistor, as our target Vo is across it).
a) Combine Current Sources:
The two sources are in opposite directions. We take "up" as positive.
Calculation: I_total = 6.5 mA (up) - 1 mA (down) = 5.5 mA (up)
b) Combine Parallel Resistors (6 kΩ and 10 kΩ):
Calculation: `Req = (6 kΩ 10 kΩ) / (6 kΩ + 10 kΩ) = 60 / 16 kΩ = 3.75 kΩ`
The simplified circuit now has a 5.5 mA current source in parallel with a 3.75 kΩ resistor and the 4 kΩ resistor.
Step 5: Final Source Transformation
We transform the 5.5 mA current source and its parallel 3.75 kΩ resistor back into a voltage source to create a simple series circuit.
V = I * R = 5.5 mA * 3.75 kΩ = (5.5 * 10⁻³) A * (3.75 * 10³) Ω = 20.625 VThis gives us a final, single-loop circuit.
Step 6: Calculate the Final Output Voltage (Vo)
The final circuit is a simple series loop containing the 20.625 V source, the 3.75 kΩ resistor, and the 4 kΩ resistor. We can find Vo (the voltage across the 4 kΩ resistor) using the voltage divider rule or by first finding the total current.
a) Find the total current (I) in the loop:
R_total = 3.75 kΩ + 4 kΩ = 7.75 kΩ
I = V / R_total = 20.625 V / 7.75 kΩ ≈ 2.6612 mA
b) Calculate Vo using Ohm's Law:
Vo is the voltage drop across the 4 kΩ resistor.
Vo = I * R = 2.6612 mA * 4 kΩ
`Vo = (2.6612 10⁻³) A (4 10³) Ω = 10.6448 V`
The output voltage Vo across the 4 kΩ resistor is 10.6448 V.