The current passing through the ideal ammeter in the dircuit given below is

Options

A. 1.25 A
B. `1" A"`
C. 0.75 A
D. 0.5 A

Answer

D

Explanation

Given,
`2Omega` and `2Omega` are in parallel
So, `(1)/( R )=(1)/(2)+(1)/(2)`
`R=(2xx2)/(4)=1Omega`
Now internal resistance `1Omega, 2Omega, 4Omega` and resistance R are in series.
`:.R_("net")=1Omega+2Omega+4Omega+1Omega=8Omega`
So, current, `l=(4)/(8)=0.5A`