A) \[\frac{3V}{R}\]
B) \[\frac{V}{R}\]
C) \[\frac{V}{2R}\]
D) \[\frac{2V}{R}\]
Correct Answer: C
Solution :
The given circuit can be redrawn as shown. From circuit, \[\frac{FC}{CE}=\frac{FD}{DE}=1\] Thus, it is balanced Wheatstone's bridge, so resistance in arm CD is ineffective and so, current flows in this arm. Net resistance of the circuit is \[\frac{1}{R'}=\frac{1}{(R+R)}+\frac{1}{(R+R)}\] \[=\frac{1}{2R}+\frac{1}{2R}=\frac{2}{2R}=\frac{1}{R}\] \[R'=R\] So, net current drawn from the battery \[i=\frac{V}{R'}=\frac{V}{R}\] As from symmetry, upper circuit AFCEB is half of the whole circuit and is equal to AFDEB. So, in both the halves half of the total current will flow. Hence, in AFCEB, the current flowing is. \[i=\frac{i'}{2}=\frac{V}{2R}\] Aliter: The given circuit can be redrawn as follows Equivalent resistance between A and B is R and current \[i=\frac{V}{R}\]You need to login to perform this action.
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