Consider the circuit shown in the figure. Taking the potential to be zero at the negative terminal of the battery :

Options

A. Potential at `F` will be higher than the potential at `C`
B. Current in `AF` will be greater than that in `BC`
C. Direction of current in `CF` will be from `C` to `F`
D. Current in `FE` will be greater than current in `CD`

Explanation

(`Sigmai=0` at point `F`)
`(V_(F)-V_(C ))/(R )+(V_(F)-V_(B))/(R )+(V_(F))/(4R)=0`…..(`1`) (`:' V_(A)=V_(B)`)
Aiso `(V_(C)-V_(F))/(R )+(V_(C)-V_(B))/(R )+(V_(C ))/(2R)=0`…….(`2`)
Using (`1`) & (`2`)
`(V_(F))/(V_(C ))=(14)/(13)`