Shows a network of eight resistors battery of resistance `R(= 2 Omega)` connected in a `3V` battery of negligible internal resistance. The current `I` in the circuit is

Options

A. `0.25A`
B. `0.50A`
C. `0.75A`
D. `1.0 A`

Answer

d

Explanation

Due so symmetry of network, there will no current in arms `BF` and `CE` Therefore the effective resistance of arms `AB,BC and CD` in series
`R_(1)= 2+2+2 = 6 Omega`
The effective resistance of arms `AF,FE,and ED` is series `R_(1)= 2+2+2 = 6 Omega`
The effective point `A` and `D, R_(1) and R_(2)` are in parrelal, the effective resistance is
`R = (R_(1)R_(2))/(R_(1) +R_(2)) = (6xx6)/(6+6) = 3 Omega`
Current `I = (3V)/(3Omega)= 1.0A`