The equivalent resistance of the given circuit between the terminals $$A$$ and $$B$$ is:

Here is the formatted, step-by-step solution to find the equivalent resistance of the given circuit network.

Problem Statement

The equivalent resistance of the given circuit between terminals A and B is:

• (1) $$0\ \Omega$$
• (2) $$3\ \Omega$$
• (3) $$\frac{9}{2}\ \Omega$$
• (4) $$1\ \Omega$$
• Look closely at the $$5\ \Omega$$ resistor on the far left. Both of its ends are connected directly together by a continuous conducting wire (short-circuit path).
• Because both ends are connected to the same potential wire (the bottom wire line), the potential difference across it is zero ($$V = 0$$), so no current flows through it ($$i = 0$$).
• We can completely remove this $$5\ \Omega$$ resistor from our calculation.
• First Parallel Combination:
• Series Combination:
• Second Parallel Combination:

Step-by-Step Solution

To find the total equivalent resistance between terminals A and B, we analyze the circuit starting from the leftmost side and simplify it step-by-step moving towards the right.

Step 1: Analyze the Leftmost $$5\ \Omega$$ Resistor

Step 2: Parallel and Series Reductions

After removing the $$5\ \Omega$$ resistor, the first series $$2\ \Omega$$ resistor on the top branch is now connected in parallel across the middle vertical $$2\ \Omega$$ resistor.

$$R_{p1} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

This $$1\ \Omega$$ equivalent resistance is now connected directly in series with the second top-branch $$2\ \Omega$$ resistor.

$$R_{\text{series}} = 1\ \Omega + 2\ \Omega = 3\ \Omega$$

On the far right, near terminals A and B, we have two vertical $$3\ \Omega$$ resistors connected in parallel with each other.

$$R_{p2} = \frac{3 \times 3}{3 + 3} = \frac{9}{6} = \frac{3}{2}\ \Omega$$

Step 3: Final Equivalent Resistance ($$R_{\text{eq}}$$)

Now, the remaining circuit is simply our simplified series branch ($$3\ \Omega$$) connected in parallel across the terminal pair's combined resistance ($$\frac{3}{2}\ \Omega$$):

$$R_{\text{eq}} = \frac{3 \times \frac{3}{2}}{3 + \frac{3}{2}}$$

$$R_{\text{eq}} = \frac{\frac{9}{2}}{\frac{9}{2}} = 1\ \Omega$$

Correct Answer

The correct option is (4) $$1\ \Omega$$.