Five equal resistances each of resistance `R` are connected as shown in the figure. A battery of `V` volts is connected between `A` and `B`. The current flowing in `AFCEB` will be

Options

A. `(3 V)/(R )`
B. `(V)/(R )`
C. `(V)/(2R )`
D. `(2 V)/(R )`

Answer

C

Explanation

(b) The given circuit can be redrawn as shown. From circuit,
`(FC)/(CE) = (FD)/(DE) = 1`
Thus, it is balanced Wheatstone's bridge, so resistance in arm `CD` is indffective and so, current flows in arm.
Net resistance of the circuit is
`(1)/(R') = (1)/((R + R)) + (1)/((R + R))`
`= (1)/(2 R) + (1)/(2R) = (2)/(2R) = (1)/(R)`
`R' = R`
So, net current drawn from the battery
`i' = (V)/(R') = (V)/(R )`
Aliter : The given circuit can be redrawn as followes Equivalent resistance between `A` and `B` is `R` and current `i (V)/(R )`