Question

The resistance in the four arms of a Wheatstone network in cyclic order are $5\Omega ,2\Omega ,6\Omega$ and $15\Omega$. If a current of $2.8A$ enters the junction of $5\Omega$ and $15\Omega$, then the current through $2\Omega$ resistor is

• A. $1.5A$
• B. $2.8A$
• C. $0.7A$
• D. $1.4A$
• E. $2.1A$

Answer 1

The correct option is E $2.1A$

Current through $2\Omega$ is
$=2.8\left\{\frac{(15+6)}{(5+2)+(15+6)}\right\}$
$=2.1A$