Question

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

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

Answer 1

Answer: (D)

Here, $2\Omega$ and $2\Omega$ are in parallel
$\therefore \frac{1}{R}=\frac{1}{2}+\frac{1}{2}$
$R=\frac{2\times 2}{2+2}=1\Omega$
Now, internal resistance $(1\Omega ),2\Omega ,4\Omega$ and resistance $R$ are in series
$\therefore R_{net}=1\Omega +2\Omega +4\Omega +\lfloor \Omega =8\Omega$
Hence, current $I=\frac{V}{R}=\frac{4}{8}=0.5A$