Question

Calculate the equivalent resistance of the network across the points A and B as shown​

Answer 1

Question :

Calculate the equivalent resistance of the network across the points A & B as shown ?

ANSWER

Required to find : -

• Equivalent resistance beyond the points A&B ?

Formulae used : -

To find the equivalent resistance among the resistors when connected in series is;

R_(eq) = R_1 + R_2 + R_3 ......

Similarly,

To find the equivalent resistance among the resistors when connected in parallel is;

[1]/[R_(eq)] = [1]/[R_1] + [1]/[R_2] + [1]/[R_3] ......

( or )

R_(eq) = [R_1*R_2*R_3]/[R_1+R_2+R_3]

Solution : -

We need to find the equivalent resistance of the network across the points A&B.

So,

By referring the pic we can conclude that;

The resistors i.e 4 Ω & 8 Ω are in series with each other on either sides of the points A&B .

However, the equivalent resistance of those pair ( 2 combinations across points A&B) is in parallel to each other .

Using this knowledge as the basic reference let's solve this question .

Since, the resistors of 4Ω & 8Ω are in series with each other.

So,

Let's find the equivalent resistance !

Using the formula;

• R_(eq) = R_1 + R_2 + R_3 ......

R_(eq) = 4Ω + 8Ω

R_(eq) = 12Ω

Similarly,

On the either side of points A&B the equivalent resistance is 12Ω .

Now,

These both 12Ω on the either sides are in parallel to each other .

So,

Using the formula;

• R_(eq) = [R_1*R_2*R_3]/[R_1+R_2+R_3]

R_(eq) = [12Ω*12Ω]/[12Ω + 12Ω]

R_(eq) = [144Ω²]/[24Ω]

R_(eq) = 6Ω

Therefore,

The equivalent resistance of the network across the points A&B is 6Ω .

Answer 2

To find : -

Equivalent resistance beyond the points A&B .

Formulae used : -

→The equivalent resistance among the resistors when connected in series is;

R_(eq) = R_1 + R_2 + R_3 ......

And ,

→The equivalent resistance among the resistors when connected in parallel is;

[1]/[R_(eq)] = [1]/[R_1] + [1]/[R_2] + [1]/[R_3] ......

( or )

R_(eq) = [R_1×R_2×R_3]/[R_1+R_2+R_3]

Solution : -

→The resistors i.e 4 Ω & 8 Ω are in series with each other on either sides of the points A&B .

→However, the equivalent resistance of those pair ( 2 combinations across points A&B) is in parallel to each other .

As, the resistors of 4Ω & 8Ω are in series with each other.

⇒R_(eq) = R_1 + R_2 + R_3 ......

⇒R_(eq) = 4Ω + 8Ω

⇒R_(eq) = 12Ω

Similarly,

→On the either side of points A&B the equivalent resistance is 12Ω .

Now,

These both 12Ω on the either sides are in parallel to each other .

So,

⇒R_(eq) = [R_1×R_2×R_3]/[R_1+R_2+R_3]

⇒R_(eq) = [12Ω×12Ω]/[12Ω + 12Ω]

⇒R_(eq) = [144Ω²]/[24Ω]

⇒R_(eq) = 6Ω

Hence,

The equivalent resistance of the network across the points A&B is 6Ω .

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