In the following circuit, the current through the resistor `R (=2Omega)` is I amperes. The value of I is

Options

A. `1A`
B. `2A`
C. `2.5A`
D. `5.5A`

Answer

a

Explanation

Resistors above and below `8 Omega` makes balanced wheatstone bridge so circuit reduce in

Now `1 Omega and 2 Omega` are in series , total resistance
`= 1+2= 3 Omega`
`2 Omega and 4Omega` are also in series so , total resistance
`= 2+4= 6 Omega`
Now this`3 Omega and 6 Omega` are in parallel , so their resistance`R_(p) = (6xx3)/(6+3) =2 Omega`
Circuit reduces to

Resistance above and below `10 Omega` again balance wheatstone bridge Now ` 2 Omega and 4 Omega` are in series `= 2+4= 6 Omega`
`6 Omega and 12Omega` are also in series `= 6+12= 18 Omega`
Effective resistance `= (6 xx 18)/(6+18) = 4.5 Omega`
Circuit reduce as an

Total resistance `= 4.5 + 2= 6.5Omega`
Current `I= (6.5)/(6.5) = 1A`