| Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be: [AIPMT (S) 2004] |
 |
A) \[\frac{3V}{R}\]
B) \[\frac{V}{R}\]
C) \[\frac{V}{2R}\]
D) \[\frac{2V}{R}\]
Correct Answer: C
Solution :
| [c] The given circuit can be redrawn as shown. |
| From circuit, |
 |
| \[\frac{FC}{CE}=\frac{FD}{DE}=1\] |
| Thus, it is balanced Whetstones bridge, so resistance in arm CD is ineffective and so, current flows in this arm. Net resistance of the circuit is |
| \[\frac{1}{R'}=\frac{1}{(R+R)}+\frac{1}{(R+R)}\] |
| \[=\frac{1}{2R}+\frac{1}{2R}=\frac{2}{2R}=\frac{1}{R}\] |
| \[\therefore \] \[R'=R\] |
| 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 |
| \[i'=\frac{i'}{2}=\frac{V}{2R}\] |
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