A) R
B) R/2
C) R/4
D) R/6
Correct Answer: A
Solution :
In the given circuit, \[\frac{R}{R}=\frac{R}{R}\] \[\left( \frac{P}{Q}=\frac{R}{S} \right)\] Therefore, it is a balanced Wheatstone bridge, and no current will flow through resistance between B and D. Resistance of arm ABC \[R'=R+R=2R\] Resistance of arm ADC, \[R''=R+R=2R\] \[\therefore \] Now, resistance \[R'\] and \[R'\,'\] are in parallel. Therefore, their resultant resistance \[\frac{1}{{{R}_{eq}}}=\frac{1}{2R}+\frac{1}{2R}=\frac{1}{R}\] \[\therefore \] \[{{R}_{eq}}=R\]
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