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To solve this circuit using the **superposition principle**, we'll analyze the contribution of each independent source (the 24V voltage source and the 3A current source) individually, then combine their effects to find \( V_{ab} \).

### Steps:
1. **Turn off the 24V voltage source** (replace it with a short circuit) and analyze the circuit with the 3A current source only.
2. **Turn off the 3A current source** (replace it with an open circuit) and analyze the circuit with the 24V voltage source only.
3. **Sum the individual contributions** to find the total voltage \( V_{ab} \).

---

### Step 1: Analyzing the circuit with only the 3A current source
- Turn off the 24V voltage source (replace it with a short circuit).
- Now, we only have the 3A current source driving current through the 16Ω and 10Ω resistors.

Since the current source is in parallel with the 16Ω and 10Ω resistors, we can use **current division** to find the current through each resistor:
\[
I_{16\Omega} = I_{total} \times \frac{10\Omega}{10\Omega + 16\Omega} = 3A \times \frac{10}{26} = \frac{30}{26} = 1.154A
\]
\[
I_{10\Omega} = 3A \times \frac{16\Omega}{10\Omega + 16\Omega} = 3A \times \frac{16}{26} = \frac{48}{26} = 1.846A
\]

Next, calculate the voltage drop across the 16Ω resistor:
\[
V_{ab} \text{ (due to current source)} = I_{16\Omega} \times 16\Omega = 1.154A \times 16\Omega = 18.46V
\]

---

### Step 2: Analyzing the circuit with only the 24V voltage source
- Turn off the 3A current source (replace it with an open circuit).
- Now, we only have the 24V voltage source driving current through the series combination of the 10Ω and 5Ω resistors.

The equivalent resistance is:
\[
R_{eq} = 10\Omega + 5\Omega = 15\Omega
\]
The current through the circuit is:
\[
I = \frac{24V}{15\Omega} = 1.6A
\]
This current flows through the 10Ω resistor. Now, calculate the voltage drop across the 10Ω resistor:
\[
V_{ab} \text{ (due to voltage source)} = I \times 10\Omega = 1.6A \times 10\Omega = 16V
\]

---

### Step 3: Combine the effects of both sources
Finally, we sum the contributions from both sources to find the total voltage \( V_{ab} \):
\[
V_{ab} = V_{ab} \text{ (due to current source)} + V_{ab} \text{ (due to voltage source)} = 18.46V + 16V = 34.46V
\]

Thus, the voltage \( V_{ab} \) is approximately **34.46V**.

---

Do you need further clarification, or have any questions?

Here are some related questions to explore further:
1. What is the principle behind the superposition theorem?
2. How does current division work in parallel circuits?
3. What happens when you short circuit a voltage source in a circuit?
4. How can you open circuit a current source in circuit analysis?
5. How does the combination of different sources affect the overall circuit behavior?

**Tip:** Superposition only works for linear circuits—those where responses are directly proportional to the inputs.

Using superposition principle, find Vab for the given circuit with a 24V voltage source and a 3A current source.

Solve for Vab in a circuit with a 24V voltage source and a 3A current source using the superposition principle. Learn how to apply current division and Ohm's Law, and how to combine contributions from multiple sources. Ideal for undergraduate electrical engineering students.

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