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 \[\frac{1}{{{R}_{eq}}}=\frac{1}{2R}+\frac{1}{2R}=\frac{1}{R}\] \[\therefore \]\[{{R}_{eq}}=R\]
You need to login to perform this action.
You will be redirected in
3 sec
