Question

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

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

Answer 1

Answer: (A)

3.6 A

Answer 2

An equivalent of the given network is as shown in the 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
$I = \frac{V}{R_{p}} = \frac{12}{\left(\frac{10}{3}\right)} = 3.6 A$