Answer:
Option A
Explanation:
An equivalent of the given network is as shown in figure.
If Rp be the net resistance, then
$\frac{1}{R_{p}}=\frac{1}{10}+\left(\frac{1}{10+10}\right)+\left(\frac{1}{10+10}\right)+\frac{1}{10}$
$=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\frac{6}{20}=\frac{3}{10}$
$\therefore$ $R_{p}=\frac{10}{3}$Ω
Hence, current f;owing through ammeter is
$I=\frac{V}{R_{p}}=\frac{12}{(\frac{10}{3})}=3.6 A$
