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+1 vote
2.4k views
in Physics by (75.3k points)

Five equal resistances each of resistances R are connected as shown in the figure A battery of V volt is connected between A and B. The current flowing in AFCEB will be. 

(A) (3V/R) 

(B) (V/R) 

(C) (V/2R) 

(D) (2V/R)

1 Answer

+2 votes
by (70.6k points)
selected by
 
Best answer

The correct option (C) (V/2R)

Explanation:

The circuit can be redrawn as

Hence Resistance of AFCEB = R + R = 2R

∴ current I = (V/2R)

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