Five resistance, each of `5Omega` are connected as shown in figure. Find the equivalent resistance between points (1) A and B , (2) A and C .

Options

A. (1) `7.5Omega` (2) `2.25Omega`
B. (1) `5Omega` (2) `2.50Omega`
C. (1) `2.5Omega` (2) ` 3.13Omega`
D. (1) `3Omega` (2) `2.5Omega`

Answer

3

Explanation

(i) Equivalent resistance between A and B the resistance connected between AD and DB are in s eries `therefore R_(1)=5Omega=5Omega=10Omega`.
The resistance connected between AC and CB are in sereis `therefore R_(2)=5Omega+5Omega=10Omega`
The resistance `R_(1),R_(2)` and the resistance b etween AB i.e., `5Omega` all are in parallel. the corresponding circuit diagram is shwon in figure (i). There fore the equivalent resistance b etween A and B is
`(1)/(R_(AB))=(1)/(R_(1))+(1)/(R_(2))+(1)/(5)`
`=(1)/(10)+(1)/(10)+(1)/(5)`
or `R_(AB)=2.5Omega`.

(ii) Equivalent resistance b etween A and C

The resistance between AD and DB are in series and corresponding circuit diagram is shown in figure (ii)
`thereforeR_(1)=5Omega+ 5Omega=10Omega`
`R_(1)=10Omega`

The resistance `R_(1)` and resistance connected between AB are in parallel and corresponding circuit diagram is shown in figure (iii)
`therefore (1)/(R_(2))=(1)/(R_(1))+(1)/(5)=(1)/(10)+(1)/(5)` or `R_(2)=(10)/(3)Omega`

the resistance `R_(2)` and resistance connected between BC are in series are corresponding circuit diagram is shown in figure (iv)
`R_(3)=R_(2)+5Omega=(10)/(3)+5=(25)/(3)Omega`

Resistance `R_(3)` and resistance connected between AC i.e. `5Omega` are parallel. therefore, the equivalent resistance between A and C is
`(1)/(R_(AC))=(1)/(R_(3))+(1)/(5Omega)=(3)/(25)+(1)/(5)` or `R_(AC)=3.125Omega`