Twelve wires each having resistance 2Ω are joined to form a cube. A battery of emf 6 V is joined across point a and c. The voltage difference between e and f is ______ V.

battery is connected across a and c
so points b,d and e,g at middle due to symmetry
which means b,d are at same potential , similarly points e,g are at same potential
so we can draw the simplified equivalent circuit as above
 apply Kirchhoff's Current Law (KCL) at each of the unknown nodes. Let us define the potentials at the various nodes.


The battery applies a $$ 6\text{ V} $$ potential difference across the circuit. Let us set the negative terminal node $$ c $$ to $$ 0\text{ V} $$ as our reference.
$$ V_a = 6\text{ V} $$
$$ V_c = 0\text{ V} $$

Let the potentials at the remaining nodes be:
 $$ V_{bd} $$ for the combined node $$ (b, d) $$
 $$ V_{eg} $$ for the combined node $$ (e, g) $$
 $$ V_h $$ for node $$ h $$
$$ V_f $$ for node $$ f $$

Now, we set up KCL equations for each of these nodes (sum of currents leaving the node equals zero).

1. Node $$ (b, d) $$:
It connects to $$ a $$ ($$ 1\,\Omega $$), $$ c $$ ($$ 1\,\Omega $$), and $$ (e, g) $$ ($$ 1\,\Omega $$).
$$ \frac{V_{bd} - 6}{1} + \frac{V_{bd} - 0}{1} + \frac{V_{bd} - V_{eg}}{1} = 0 $$
$$ 3V_{bd} - V_{eg} = 6 \quad \text{--- (Equation 1)} $$

2. Node $$ h $$:
It connects to $$ a $$ ($$ 2\,\Omega $$) and $$ (e, g) $$ ($$ 1\,\Omega $$).
$$ \frac{V_h - 6}{2} + \frac{V_h - V_{eg}}{1} = 0 $$
Multiply by 2 to clear the denominator:
$$ V_h - 6 + 2V_h - 2V_{eg} = 0 $$
$$ 3V_h - 2V_{eg} = 6 $$
$$ V_h = 2 + \frac{2}{3}V_{eg} \quad \text{--- (Equation 2)} $$

3. Node $$ f $$:
It connects to $$ (e, g) $$ ($$ 1\,\Omega $$) and $$ c $$ ($$ 2\,\Omega $$).
$$ \frac{V_f - V_{eg}}{1} + \frac{V_f - 0}{2} = 0 $$
Multiply by 2:
$$ 2V_f - 2V_{eg} + V_f = 0 $$
$$ 3V_f = 2V_{eg} $$
$$ V_f = \frac{2}{3}V_{eg} \quad \text{--- (Equation 3)} $$

4. Node $$ (e, g) $$:
It connects to $$ (b, d) $$ ($$ 1\,\Omega $$), $$ h $$ ($$ 1\,\Omega $$), and $$ f $$ ($$ 1\,\Omega $$).
$$ \frac{V_{eg} - V_{bd}}{1} + \frac{V_{eg} - V_h}{1} + \frac{V_{eg} - V_f}{1} = 0 $$
$$ 3V_{eg} - V_{bd} - V_h - V_f = 0 \quad \text{--- (Equation 4)} $$

Substitute Equations 2 and 3 into Equation 4:
$$ 3V_{eg} - V_{bd} - \left(2 + \frac{2}{3}V_{eg}\right) - \left(\frac{2}{3}V_{eg}\right) = 0 $$
$$ 3V_{eg} - \frac{4}{3}V_{eg} - V_{bd} - 2 = 0 $$
$$ \frac{5}{3}V_{eg} - V_{bd} = 2 $$
$$ V_{bd} = \frac{5}{3}V_{eg} - 2 $$

Now, substitute this expression for $$ V_{bd} $$ back into Equation 1:
$$ 3\left(\frac{5}{3}V_{eg} - 2\right) - V_{eg} = 6 $$
$$ 5V_{eg} - 6 - V_{eg} = 6 $$
$$ 4V_{eg} = 12 $$
$$ V_{eg} = 3\text{ V} $$

Since node $$ e $$ is part of the combined node $$ (e, g) $$, its potential is:
$$ V_e = 3\text{ V} $$

Now, substitute $$ V_{eg} $$ into Equation 3 to find the potential at node $$ f $$:
$$ V_f = \frac{2}{3}(3) $$
$$ V_f = 2\text{ V} $$

The voltage difference between nodes $$ e $$ and $$ f $$ is:
$$ V_e - V_f = 3\text{ V} - 2\text{ V} = 1\text{ V} $$