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Let's analyze both circuits (a) and (b) to find the equivalent resistance between points A and B in each case.

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Circuit (a)

This is a simple parallel combination of two resistors:

- One branch has two 6 Ω resistors in series, so their combined resistance is:
$$
R_{\text{series}} = 6\,\Omega + 6\,\Omega = 12\,\Omega
$$

- The other branch has a single 6 Ω resistor.

These two branches are in parallel between points A and B.

So, the equivalent resistance $ R_{eq} $ is given by:
$$
\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6}
$$

$$
\frac{1}{R_{eq}} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4}
$$

$$
R_{eq} = 4\,\Omega
$$

✔ Answer for (a): 4 Ω

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Circuit (b)

This circuit looks more complex — it has a symmetrical diamond-like structure with four 6 Ω resistors arranged in two "diamonds" connected in series.

Let’s label the nodes:

- Left side: Point A
- Right side: Point B
- Middle junction: Let's call it C (common point where the two diamonds meet)

But actually, let's look closely: There are two identical parallel paths from A to B, each consisting of two 6 Ω resistors in series?

Wait — no. Let's draw it mentally:

- From A, two 6 Ω resistors go out — one up, one down — meeting at a central node.
- Then from that central node, two 6 Ω resistors go to B — one up, one down.

But wait — this is actually two separate bridges? No, looking again:

Actually, the diagram shows:

- Two identical parallel paths from A to B, each path having two 6 Ω resistors in series?

No — let's count:

From A to B:

- Path 1: A → top left 6Ω → top right 6Ω → B
- Path 2: A → bottom left 6Ω → bottom right 6Ω → B

But there is no connection between the middle nodes — meaning the two paths are independent.

Wait — actually, the two paths share a common node in the middle — so the configuration is:

- A connects to two resistors (6Ω each), going to a central node (say C)
- From C, two 6Ω resistors go to B

So we have:

- From A to C: two 6Ω resistors in parallel
- From C to B: two 6Ω resistors in parallel

Then these two parts are in series.

Let’s compute step-by-step.

#### Step 1: Resistance from A to C

Two 6Ω resistors in parallel:
$$
R_{AC} = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3\,\Omega
$$

#### Step 2: Resistance from C to B

Same thing: two 6Ω resistors in parallel:
$$
R_{CB} = 3\,\Omega
$$

#### Step 3: Total resistance from A to B

$ R_{AB} = R_{AC} + R_{CB} = 3\,\Omega + 3\,\Omega = 6\,\Omega $

✔ Answer for (b): 6 Ω

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✔ Final Answers:

- (a): 4 Ω
- (b): 6 Ω

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🔍 Explanation Summary:

- In (a): One branch has two 6Ω resistors in series (12Ω), another has one 6Ω. These are in parallel: $ \frac{1}{R} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4} $ → $ R = 4\,\Omega $
- In (b): The network splits into two parallel sections: A→C and C→B, each made of two 6Ω resistors in parallel (3Ω each). Total: 3 + 3 = 6Ω.

Let me know if you'd like a visual explanation or simulation!