Question

In given circuit, all resistances are of $10\Omega$. Current flowing through ammeter is

• A. $3.6A$
• B. $1.8A$
• C. $2A$
• D. $1A$

Answer 1

Answer: (A)

$3.6A$

An equivalent of the given network is as shown in figure.
If $R_P$ be the net resistance, then
$\frac{1}{R_P}=\frac{1}{10}+\left(\frac{1}{10+10}\right)+\left(\frac{1}{10+10}\right)+\frac{1}{10}$
$=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}$
$=\frac{6}{20}=\frac{3}{10}$
$\therefore R_p=\frac{10}{3}\Omega$
Hence, current flowing through ammeter is
$I=\frac{V}{R_P}=\frac{12}{\left(\frac{10}{3}\right)}=3.6A$