Question

The ammeter in the given figure will read

• A. $3A$
• B. $\frac{10}{3}A$
• C. $30A$
• D. $\frac{100}{3}A$

Answer 1

The correct option is A $3A$

Given circuit is equivalent to the Wheatstone bridge (ABCD in figure).

Since bridge is balanced no current will flow in branch CD, hence equivalent resistance between A and B is
$\frac{1}{R}=\frac{1}{R_{AC}+R_{CB}}+\frac{1}{R_{AD}+R_{DB}}+\frac{1}{R_{Ae}+R_{EB}}$

$\frac{1}{R}=\frac{1}{5+5}+\frac{1}{5+5}+\frac{1}{5+5}$

$\frac{1}{R}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}$

$\frac{1}{R}=\frac{3}{10}$

$R=\frac{10}{3}$

The current through the circuit (read by the ammeter) is:
$I=\frac{V}{R}$

$I=\frac{10}{\frac{10}{3}}$

$I=3A$