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)
S safeer

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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