In the circuit shown in fig. some potential difference is applied between A and B. The equivalent resistance between A and B is R.

Options

A. no current flow through `5 Omega` resistor
B. `R = 12.5 Omega `
C. `R = 15 Omega`
D. `R = 3.6 Omega`

Answer

a,d

Explanation

As it forms wheatstone bridge , since
`((2)/(3) = (4)/(6)) ` so no current will flow through the central resistance `R = 5 Omega`
(d) Since no current is flowing through `5 Omega` so `2 Omega` and `4 Omega` are in series `R_(S) = 2+4 = 6 Omega`
Further `3 Omega` and `6 Omega` are also in series so,
`R'_(s) = 3 + 6 = 9 Omega`
Now `R_(s) and R'_(s)` are is parallel so equivalent resistance
`R = (R_(S) xx R'_(S))/(R_(S) + R'_(S)) = (6xx9)/(6+9) = 3.6 Omega`