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Twelve wires each having a resistance of 3\Omega are connected to form a cubical network. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of this network. Determine its equivalent resistance and the current along each edge of the cube.

 

 

Answers (1)

The resistance of the wire is given =3\Omega 

and the battery is =10\; V

Now, arranging the 12 wires each of  3\Omega resistance to form a cubical network.

So, Applying loop rule to ABC{C}'EFA

3I+3\frac{I}{2}+3I-10=0

I=\frac{4}{3}A

R_{total}=\frac{V}{3I}=\frac{10\times 15}{3\times20}=2.5\Omega

Now, the current along each edge of the cube is given as ;

Current=I_{AB}(=I_{AA{}'}=I_{AD}=I_{{D}'{C}'}=I_{{B}'{C}'}=I_{C{C}'})=\frac{4}{3}A I_{D{D}'}(=I_{{A}'{B}'}=I_{{A}'{D}'}=I_{DC}=I_{BC}=I_{B{B}'})=\frac{2}{3}A

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Safeer PP

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