If the Wheateston's network shown in the figure, the current l in the circuit is

Options

A. `1A`
B. `2A`
C. `0.25` A
D. `0.5` A

Answer

D

Explanation

Since `(4)/(2)=(8)/(4)`, bridge is balanced,
so `5Omega` resistance can be neglected.
Therefore, effective resistance,
`R=((2+4)*4+8)/((2+4)+(4+8))=(6xx12)/(6+12)=(72)/(18)`
`=4 Omega`
So, the current in the circuit
`I=(V)/(R)=(2)/(4)=0.5A`