Question

Figure shows a network of eight resistors, each equal to $2\Omega$, connected to a $3V$ battery of negligible internal resistance. The current $I$ in the circuit is

• A. $0.25A$
• B. $0.50A$
• C. $0.75A$
• D. $1.0A$

Answer 1

The correct option is D $1.0A$

Due to Wheatstone bridge prinsiple no current will pass through $BE$ and $CF$.

$\therefore$Total resistance between $A$ and $D$

$=\frac{1}{\frac{1}{6}+\frac{1}{6}}\Omega$

$=\frac{1}{\frac{1}{3}}\Omega =3\Omega$

Voltage, $V=3V$

$\therefore I=\frac{V}{R}=\frac{3V}{3\Omega }=1A$