The current pasing through the ideal ammeter in the circuit given below is

Options

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

Answer

D

Explanation

Here `2Omega` and `2 Omega` are in parallel.
`(1)/(R )=(1)/(2)+(1)/(2) rArr R =(2xx2)/(2+2)=1 Omega`
Now, internal resistances `1Omega, 2Omega`and resistance R are in series.
`therefore " " R_("net")=1Omega +2 Omega+4Omega+1Omega = 8 Omega`
Hence, current `l=(V)/(R )=(4)/(8)=0.5 A`