The value of equivalent resistance for the circuit shown below is,

Options

A. `2.123 Omega`
B. `5.123 Omega`
C. `4.23 Omega`
D. `6.283 Omega`

Answer

B

Explanation

Step I Find out the resistance between the nodes P and Q.
`R_(eq_(1))=R_(1)+(1)/(((1)/(R_(2))+(1)/(R_(3))))`
or `R_(eq_(1))=1+(1)/(((1)/(2)+(1)/(3)))`
or `R_(eq_(1))=1+(2xx3)/(5)`
or `R_(eq_(1))=2.2 Omega`
Step II Now, find the resistance between nodes Q and T.
`R_(eq_(2))=R_(4)+(1)/(((1)/(R_(5))+(1)/(R_(6))+(1)/(R_(7))))+R_(8)`
`= 1+(1)/(((1)/(2)+(1)/(3)+(1)/(4)))+1`
`= 1+0.923+1`
`rArr R_(eq_(2))=2.923 Omega`
Step III As from circuit diagram, the equivalent resistance must be the series combination of `R_(eq_(1))` and `R_(eq_(2))`.
`therefore " " R_(eq)=R_(eq_(1))+R_(eq_(2))`
`= 2.2+2.923`
`R_(eq)=5.123 Omega`