Question

In this circuit some potential difference is applied between A and B. The equivalent resistance between A and B is R, is :

• A. $R=9\Omega$
• B. $R=15\Omega$
• C. $R=12.5\Omega$
• D. $R=18/5\Omega$

Answer 1

The correct option is D $R=18/5\Omega$

Rearrangement of the circuit as shown in figure gives a balanced Wheatstone bridge, and no current flows through the $5\Omega$ resistor. It can thus be removed from the circuit.

Since no current is flowing through 5Ω so 2Ω and 4Ω are in series $R_S=2+4=6\Omega$

Further 3Ω and 6Ω are also in series so,

$R^{\prime }s=3+6=9\Omega$

Now Rs and R's are in parallel so equivalent resistance

$R=\frac{9\times 6}{9+6}=3.6\Omega$