Question

Find equivalent resistance of the circuit.

• A. $1\Omega$
• B. $2\Omega$
• C. $4\Omega$
• D. $8\Omega$

Answer 1

The correct option is C $4\Omega$


Between points $A$ and $B$, we have a wheatstone bridge which is in balanced condition
$\because \frac{10}{5}=\frac{6}{3}$

Hence, $20\Omega$ can be removed.


For the entire circuit, $16\Omega$, $8\Omega$ and $16\Omega$ are in parallel.


$R_{eq}={(\frac{1}{16}+\frac{1}{8}+\frac{1}{16})}^{-1}=4\Omega$