A battery of 10 volt and negligible internal resistance is connected across a diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 ohm determine the equivalent resistance of the network and the current along each edge of the cube
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(1/3+1/6+1/3) ohms is the equivalent resistance (R/3+R/6+R/3)
The equivalent resistance = 5/6 ohms
Current = 10/(5/6) = 60/5= 12 A
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