Question

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$\bold\red{Question:}$
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$\sf{Calculate \: the \: effective \: resistances \: between}$ $\sf{the \: points \: A \: and \: B \: in \: the \:networks \: shown }$ $\sf{in \: the \: figure}$​

Answer 1

Explanation:

effective resistance y along AB is 1.4 ohm

Answer 2

Answer:

the answer is

$\frac{5}{7} ohms$

Explanation:

As you can see in the reduced circuit diagram the wire AC and CB are in series

therefore applying the formula of equivalent resistance ins series we get,

AC = 3 ohms

CB = 2 ohms

Series formula

$Series =R1 + R2 \\ = R(AC) + R(CB) \\ = 3 + 2 = 5 \: ohms$

Therefore, we get equivalent resistance of one segment of wire as 5 ohms.

Now the wire AD and DB are in series

therefore applying the formula of series connection,

$Series =R1 + R2 \\ = R(AD ) + R( DB) \\ = 3 + 2 = 5 \: ohms$

Therefore, we get equivalent resistance of one segment of wire as 5 ohms.

Now all the three wires in diagram are in parallel connection between A and B points.

Parallel connection formula is

$\frac{1}{R \: equivalent} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} \\ \frac{1}{R \: AB}= \frac{1}{5} + 1 + \frac{1}{5} \\ = \frac{1 + 5 + 1}{5} \\ = \frac{7}{5} \\ R \: AB = \frac{5}{7} \: ohms$

therefore the equivalent resistance between AB is

$\frac{5}{7} \: ohms$