A)
2\[\Omega \]
B) 4 \[\Omega \]
C) 1\[\Omega \]
D) 8 \[\Omega \]
Correct Answer: A
Solution :
The given circuit is balanced Wheatstone Bridge hence, middle resistor is ineffective. Now equivalent resistance of upper arm which are connected in series \[{{R}_{eq(U)}}=2+2=4\,\Omega \] Similarly the equivalent resistance of lower arm which are also connected in series \[{{R}_{eq(L)}}=2+2=4\,\Omega \] Now \[{{R}_{eq(U)}}\] and \[{{R}_{eq(L)}}\] are connected in parallel \[{{R}_{\text{resultant}}}=\frac{4\times 4}{4+4}=\frac{16}{8}=2\,\Omega \]
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