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Question

A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1Ω in figure. Determine the equivalent resistance of the network and the current along each edge of the cube.
Refer to point no. 16[Important Terms, Definitions and Formulae]
1157560_6ccb7501d0b24eb288ae7d15797a18b7.png

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Solution

Sol. Let us number the corners as 1,2,3, and so on as shown

a(1876),b(1276),c(1856),d(12

36),e(1456),f(1436)

Now for each of these paths, we have identical resistances.

Let the current entering at point A be I. The current will be

distributed equally among all the three parts. Therefore, we can

say that the current is

I12=I14=I18=I/3

Now look at all paths from 2 to 6 . We have the following paths

g(276) and h(236) .

since they also have the same resistances, we can assume that

current in 23 and current in 27 must be same. Therefore,

I27=I23=12(I3)=I6

we can say that I43=I45=I85=I87=I/6

Therefore, to calculate the potential difference across 16

(see Fig. 5.34 ) the path A12%, we have

V1I3RI6RI3R=V6V1V6=I[R3+R6+R3]=56IR

Hence, (V1V6)I=56R

Therefore, the equivalent resistance is 56R .

I=V56R=105/6=12A

Hence, equivalent resistance is 56Ω , I12=I14=I18=4A andI43=I45=I85=I87=2A


1050149_1157560_ans_b42e0e38265b428287db015985182cd8.png

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