Question

12. In given network, the resultant resistance between
A and B is
R
R
-В
(2) 2R
2
er is
and
10)
RR = 2R
Posle
3R E-40
(3)
R
(4)
2
4
13
For the given network, reading of ammeter is​

Answer 1

Given :

Six resistors of equal resistances R are connected as shown in the figure.

To Find :

Equivalent resistance between A and B.

Solution :

❖ First see the attachment! I have simplified the given complex circuit.

» Here you can clearly see a balanced wheatstone bridge between two red dots.

• Therefore we can neglect resistance which is in middle part. (In red box)

In upper branch of wheatstone bridge two resistances are connected in series. Hence their equivalent resistance will be

• R + R = 2R

Similarly resistance of lower branch of the wheatstone will also be 2R.

» Finally three resistances of R, 2R and 2R come in parallel$.$

We know that the reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances.

♦ Net equivalent resistance :

➙ 1/Req = 1/R + 1/2R + 1/2R

➙ 1/Req = 1/R + 2/2R

➙ 1/Req = 1/R + 1/R

➙ 1/Req = 2/R

➙ Req = R/2 Ω

∴ (1) R/2 Ω is the correct answer!

Answer 2

Given :

Six resistors of equal resistances R are connected as shown in the figure.

To Find :

Equivalent resistance between A and B.

Solution :

❖ First see the attachment! I have simplified the given complex circuit.

» Here you can clearly see a balanced wheatstone bridge between two red dots.

Therefore we can neglect resistance which is in middle part. (In red box)

In upper branch of wheatstone bridge two resistances are connected in series. Hence their equivalent resistance will be

R + R = 2R

Similarly resistance of lower branch of the wheatstone will also be 2R.

» Finally three resistances of R, 2R and 2R come in parallel$.$

We know that the reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances.

♦ Net equivalent resistance :

➙ 1/Req = 1/R + 1/2R + 1/2R

➙ 1/Req = 1/R + 2/2R

➙ 1/Req = 1/R + 1/R

➙ 1/Req = 2/R

➙ Req = R/2 Ω

∴ (1) R/2 Ω is the correct answer!