Question

A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corner of a cubical network consisting of 12 resistors each of resistance 1$\Omega$. The total current I in the circuit external to the network is

• A. 0.83A
• B. 12 A
• C. 1A
• D. 4A

Answer 1

The correct option is B 12 A

Answer is B.
The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.The paths AA, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchoffs first rule and the symmetry in the problem.
Next, take a closed loop, say ABCCEA, and apply Kirchoffs second law:
-IR - (1/2)IR IR + emf = 0.
Where R is the resistance of each edge and the emf of battery. Thus,
emf = (5/2) IR
The equivalent resistance R of the network is = emf/3I = (5/6)R.
For R = 1 ohm, R = (5/6) ohm and for emf = 10 V, the total current (= 3I) in the network is 3I = 10V/(5/6) = 12 A.
Hence, the total current I in the circuit external to the network is : 12 A.