2004 AIPMT Current ElectricityCombination of Resistors and Symmetry Medium Question 116 of 130 in Current Electricity

Five equal resistances each of resistance R are connected as shown in the Figure. A battery of V volts is…

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
A $\dfrac{V}{R}$
B $\dfrac{V}{2R}$
C $\dfrac{2V}{R}$
D $\dfrac{3V}{R}$

The answer is B.

Solution

The battery is effectively connected between points F (joined to A) and E (joined to B).
From F there are two routes to E: F–C–E (resistors FC and CE) and F–D–E (resistors FD and DE). The fifth resistor DC joins C and D.
Each of C and D is connected to F by one resistor R and to E by one resistor R, so by symmetry C and D are at the same potential. Hence no current flows through the resistor DC, and it can be removed.

Now the circuit is two branches, each of $R + R = 2R$, in parallel. Equivalent resistance $= \dfrac{2R \times 2R}{4R} = R$.
Total current from the battery $= \dfrac{V}{R}$, which divides equally between the two identical branches.
Current in the branch AFCEB $= \dfrac{V}{2R}$

Concept Balanced network with no current in a branch

Current Electricity in past papers

130 questions from this chapter have appeared across 27 exam years.