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To determine \( V_{ab} \) (the voltage between points \( a \) and \( b \)) from the given circuit, let's break down the steps systematically:

### Step 1: Analyze the left-hand loop (around point \( a \))
- There are two resistors in series: \( 6 \, \Omega \) and \( 10 \, \Omega \).
- A current of \( 3 \, \text{A} \) flows through the loop.

The voltage drop across each resistor can be calculated using Ohm's Law:
\[
V = I \cdot R
\]

For the \( 6 \, \Omega \) resistor:
\[
V_1 = 3 \, \text{A} \cdot 6 \, \Omega = 18 \, \text{V}
\]

For the \( 10 \, \Omega \) resistor:
\[
V_2 = 3 \, \text{A} \cdot 10 \, \Omega = 30 \, \text{V}
\]

Thus, the total voltage drop from point \( a \) to the ground is:
\[
V_{\text{total}} = V_1 + V_2 = 18 \, \text{V} + 30 \, \text{V} = 48 \, \text{V}
\]

The polarity indicates that point \( a \) is \( +48 \, \text{V} \) relative to the ground.

### Step 2: Analyze the right-hand loop (around point \( b \))
- There are two resistors in series: \( 6 \, \Omega \) and \( 4 \, \Omega \), along with a \( 10 \, \text{V} \) battery.
- The \( 5 \, \text{V} \) source is between the two loops but connected externally.

Assume the current in this loop flows from the battery (starting at the positive terminal) through the \( 4 \, \Omega \) and \( 6 \, \Omega \) resistors.

Let’s calculate the equivalent resistance in this loop:
\[
R_{\text{total}} = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega
\]

Now, using Ohm's Law, the current through the loop:
\[
I = \frac{10 \, \text{V}}{10 \, \Omega} = 1 \, \text{A}
\]

The voltage drops across the resistors are:
- Across the \( 6 \, \Omega \) resistor:
\[
V_3 = 1 \, \text{A} \cdot 6 \, \Omega = 6 \, \text{V}
\]
- Across the \( 4 \, \Omega \) resistor:
\[
V_4 = 1 \, \text{A} \cdot 4 \, \Omega = 4 \, \text{V}
\]

The total voltage drop in this loop is:
\[
V_{\text{drop}} = V_3 + V_4 = 6 \, \text{V} + 4 \, \text{V} = 10 \, \text{V}
\]
Thus, point \( b \) is at \( 0 \, \text{V} \) relative to the ground, due to the \( 10 \, \text{V} \) battery.

### Step 3: Include the \( 5 \, \text{V} \) voltage source
- The \( 5 \, \text{V} \) source is connected between the two loops.
- Its polarity is such that it raises the potential by \( 5 \, \text{V} \) when moving from \( b \) to \( a \).

### Step 4: Calculate \( V_{ab} \)
The voltage between points \( a \) and \( b \) is now the difference between their potentials:
\[
V_{ab} = V_a - V_b
\]

From the analysis:
- \( V_a = +48 \, \text{V} \)
- \( V_b = 0 \, \text{V} \)
- The \( 5 \, \text{V} \) source raises the potential between \( b \) and \( a \).

Thus:
\[
V_{ab} = 48 \, \text{V} + 5 \, \text{V} = 53 \, \text{V}
\]

### Final Answer:
\[
V_{ab} = 53 \, \text{V}
\]

---

Would you like further details on any step or additional explanations? Here are 5 questions to deepen your understanding:

1. How does Ohm's Law apply to series circuits?
2. What happens if you reverse the polarity of the voltage source between two loops?
3. How would the calculation change if resistances were different in the second loop?
4. What is the impact of the \( 5 \, \text{V} \) source on the potential difference?
5. Can you explain why \( b \) is at \( 0 \, \text{V} \) in this context?

**Tip:** When analyzing complex circuits, break them down into simpler loops and apply basic principles like Ohm's Law and Kirchhoff's Voltage Law (KVL).

Learn how to calculate the voltage between points a and b (Vab) in a complex circuit featuring two loops, multiple resistors, and voltage sources. This step-by-step solution involves using Ohm's Law and Kirchhoff's Voltage Law. Suitable for high school students in Grades 10-12.

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