Question

find the equivalent resistance between points A and B of the network shown in the given diagram​

Answer 1

Answer:

First of , we need to simplify the complex circuit and then assign parallel or series combinations for resistances.

Only then we can find the equivalent resistance of a circuit.

First we shall simplify the rectangular portion of the circuit.

Following points to be noted :

• 3 ohm and 8 ohm in series
• 4 ohm alone
• 5 ohm and 10 ohm in series

All of them in parallel with each other.

So net resistance of that part will be :

$\dfrac{1}{r} = \dfrac{1}{(3 + 8)} + \dfrac{1}{4} + \dfrac{1}{(5 + 10)}$

$= > \dfrac{1}{r} = \dfrac{1}{11} + \dfrac{1}{4} + \dfrac{1}{15}$

$= > \dfrac{1}{r} = \dfrac{269}{660}$

$= > r = \dfrac{660}{269} \approx \: 2.45 \: ohm$

Now these are in series with 1 ohm and 2 ohm.

So equivalent resistance will be :

$R_{eq} = 2.45 + 1 + 2$

$R_{eq} = 5.45 \: ohm$

Answer 2

..

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QUESTION :

find the equivalent resistance between points A and B of the network shown in the given diagram....

SOLUTION :

See the above attachment..

First we will try to simplify the given circuit diagram.

We are going to express this circuit as a series and parallel connections.

Now,

For the first :

3r Ohm + 8r Ohm = 11r Ohm

Second :

5r Ohm + 10r Ohm = 15 r Ohm.

Now, 11r , 4r and 5 r are in parallel

Their resultant is :

=> 1 / r = 220 / 269

=> r is approximately 2.45 Ohm.

Net Resistance = 2.45 + 2 + 1 = 5.45 Ohm........ [ A ]