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The equivalent resistance between the points $A$ and $B$ is [[1]] Ohms.

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Detailed Solution

The equivalent resistance between the points

A

and

B

1 O h ms

.
From the given circuit diagram, it can be observed that the resistance

R_{1}

and

R_{2}

are in parallel combination and similarly,

R_{3}

and

R_{4}

R_{5}

and

R_{6}

and

R_{7}

and

R_{8}

all are in parallel combination. The equivalent resistance is equal to

\frac{1}{R_{P 12}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

\Rightarrow \frac{1}{R_{P 12}} = \frac{1}{2} + \frac{1}{2}

\Rightarrow R_{P 12} = 1 Ω

Therefore, the equivalent resistance,

R_{P 34} = R_{P 56} = R_{P 78} = 1 Ω

.
Now, the resistance

R_{P 12}

and

R_{P 34}

are in series combination, so the equivalent resistance is equal to,

R_{S 1} = R_{P 12} + R_{P 34}

\Rightarrow R_{S 1} = 1 + 1

\Rightarrow R_{S 1} = 2 Ω

The resistance

R_{P 56}

and

R_{P 78}

are in series, so equivalent resistance is equal to

R_{S 2} = R_{P 56} + R_{P 78}

\Rightarrow R_{S 2} = 1 + 1

\Rightarrow R_{S 2} = 2 Ω

Also, the resistance

R_{S 1}

and

R_{S 2}

are in parallel connection, so the equivalent resistance is equal to

\frac{1}{R_{P}} = \frac{1}{R_{S 1}} + \frac{1}{R_{S 2}}

\Rightarrow \frac{1}{R_{P}} = \frac{1}{2} + \frac{1}{2}

\Rightarrow R_{P} = 1 Ω

Hence, the equivalent resistance between the points

A

and

B

1 Ω

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