Question

find the equivalent resistance across the two ends A and B of the circuit shown.

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Answer 1

Answer:

.

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SOLUTION:-

Clearly R1 and R2 are in parallel.

The equivalent resistance R12 of R1 and R2 will be given by,

★1/ R12 = 1/R1 + 1/R2

➡1/2+1/2 ( Using Given values of resistance)

or 1/R12 = 1/1

or 1/R12 = 1 ohm

Similarly ,

equivalent resistances , R34 ( of R3 and R4) , R56( of R5 and R6) and R78( of R7 and R8) will also be 1 ohm.

Now,

R12 and R34 are in series.

Their equivalent resistance , R1234 = R12 +R34

==> 1ohm + 1ohm = 2ohm

Also,

R78 and R56 are in series.

Their equivalent resistance, R5678 = R56 +R78

==> 1ohm +1ohm = 2ohm

So , The Given circuit can now be represented as shown.

Now,

R1234 and R5678 are in parallel

Their equivalent resistance, Re will be Given by,

1/Re= 1/R1234 + 1/ R5678

➡ 1/2 + 1/2 =1/1

➡ Re = 1 ohm

Thus, the equivalent resistance between A and B is 1 ohm.

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Answer 2

Given:

please find in attachment.

To prove:

Resistance equal between two circuit-endings A and B= ?

Solution:

$$\Therefore$$ $$\therefore$$ $$R_1 \ and R_2$$ are parallel:

$$\begin{gathered}\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}\\\\ \frac{1}{R'}= \frac{1}{2}+\frac{1}{2}\\\\ \frac{1}{R'}=\frac{1+1}{2}\\\\\frac{1}{R'}=\frac{2}{2}\\\\}{R'}=1 \ ohm\\\end{gathered}$$

if in R'' (R_3 and R_4 were also parallel) then it will be : 1 ohm

Now calculate R' and R'':

Where R' = 1 ohm and R''=1 ohm

Formula: R'+R'' = 1+1= 2 ohm

When we calculate the other R''' and R'''' So, they are also in parallel:

So, the total R =1 ohm.

The final answer is 1 ohm.