Four resistances `1Omega, 2Omega, 2 Omega,` and `4 Omega` are connected so as to form a Wheatstone's network. The cell of e.m.f. 2V is connected between its opposite points. The total current of bridge is

Options

A. 1A
B. 0.5A
C. 2A
D. 1.5A

Answer

A

Explanation

The Wheatstone's network is in balanced condition. Thus,
`R_(R)=(3xx6)/(3+6)=(18)/(9)= 2 Omega`
`:.R=(V)/(R)=(2)/(2)=1A`