Question

A battery of 10V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of 3 ohm resistance. Use Kirchhoff’s rules to determine: (a) the equivalent resistance of the network. (b) the total current in the network.


urgent pls solve​

Answer 1

Answer:

refer to the attachment given above

Explanation:

Answer 2

Use KVL and for p.d. across each component and solve for current though each component. The network is not reducible to a simple series and parallel combinations of resistors. There is however, clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network. The path AA’, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be 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 Kirchhoff’s a first rule and the symmetry in the problem. Next take a closed loop, say, ABCC’EA, and apply Kirchhoff’s second rule: where R is the resistance of each edge and ε the emf of battery. Thus, ε = 5/2 IR The equivalent Req of the network is Total current (=3I) in the network is 3I = 10V/(5/6)Ω = 12A, i.e., I = 4A The current flowing in each edgRead more on Sarthaks.com - https://www.sarthaks.com/454571/battery-of-and-negligible-internal-resistance-connected-across-the-diagonally-opposite