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Cube Resistance Problem | JEE Main | JEE Advanced

Cube Resistance Problem | JEE Main | JEE Advanced :- For a cube having equal resistance along its edges, let’s determine the equivalent resistance between :-

Two vertices along the body diagonal
Two vertices along the face diagonal
Two adjacent vertices

Equivalent resistance between two vertices along the body diagonal

(Cube Resistance Problem)

Let us consider a cube as shown in figure as shown in figure below :-

Equivalent resistance between two vertices along the body diagonal (AG) is = 5R/6. To prove this let us draw the equivalent representation of the above figure :-

The vertices B,D,E are at the same potential and also the vertices C,F,H are also at the same potential due to symmetry.

Again simplifying the circuit, we get :-

Finally equivalent resistance between two vertices along the body diagonal (AG) = 5R/6.

Alternate Method

Assume that the current flowing into vertex A is i. Now because of the symmetry we can divide the current in different branches as shown in the figure below :-

Now to find potential difference difference, V_AG, i.e., potential difference between vertices A and G, let us follow the path A – B – C – G and write down the potential drop,

$\displaystyle {{V}_{AG}}=\frac{i}{3}R+\frac{i}{6}R+\frac{i}{3}R$

$\displaystyle \Rightarrow {{V}_{AG}}=\frac{5R}{6}i$

$\displaystyle {{R}_{eq}}=\frac{{{V}_{AG}}}{i}$

⇒ R_eq = 5R/6

Equivalent resistance between two vertices along the face diagonal

(Cube Resistance Problem)

Let us consider a cube as shown in figure as shown in figure below :-

Equivalent resistance between two vertices along the face diagonal (AC) is = 3R/4. To prove this let us draw the equivalent representation of the above figure :-

Note that the circuit is symmetric about B – F – H – D as indicated by red dotted line, it means all these vertices are at the same potential. This helps us to simplify the circuit again as shown below :-