Refer to the circuit diagram given below. The heat generated across the \(6\,\Omega\) resistance in \(100\) second is \(\dfrac{\alpha}{100}\) J. The value of \(\alpha\) is _______. (Nearest integer)

Taking the right-hand wire as the reference node at \(0\text{ V}\), let the left junction (where the \(3\,\Omega\), \(4\,\Omega\) and \(6\,\Omega\) resistors meet) be Node A at potential \(V_A\).

Bottom branch: The positive terminal of the \(2\text{ V}\) battery is at the reference, so its negative terminal (and hence the left end of the \(3\,\Omega\) resistor) is at \(-2\text{ V}\).
Top branch: The positive terminal of the \(3\text{ V}\) battery is at the reference, so its negative terminal (and hence the node between the \(6\,\Omega\) resistor and the \(3\text{ V}\) battery) is at \(-3\text{ V}\).

Applying KCL at Node A (sum of currents leaving the node is zero):
\(\dfrac{V_A - 0}{4} + \dfrac{V_A - (-2)}{3} + \dfrac{V_A - (-3)}{6} = 0\)

\(\dfrac{V_A}{4} + \dfrac{V_A + 2}{3} + \dfrac{V_A + 3}{6} = 0\)

Multiplying throughout by \(12\):
\(3V_A + 4(V_A + 2) + 2(V_A + 3) = 0\)

\(3V_A + 4V_A + 8 + 2V_A + 6 = 0\)

\(9V_A + 14 = 0 \Rightarrow V_A = -\dfrac{14}{9}\text{ V}\)

Current through the \(6\,\Omega\) resistor (from Node A towards the \(-3\text{ V}\) point):
\(I_6 = \dfrac{V_A - (-3)}{6} = \dfrac{-\dfrac{14}{9} + 3}{6} = \dfrac{\dfrac{13}{9}}{6} = \dfrac{13}{54}\text{ A}\)

Heat generated in the \(6\,\Omega\) resistor in \(t = 100\text{ s}\):
\(H = I_6^2 \cdot R \cdot t = \left(\dfrac{13}{54}\right)^2 \cdot 6 \cdot 100\)

\(H = \dfrac{169}{2916} \cdot 600 = \dfrac{169 \cdot 100}{486} = \dfrac{16900}{486}\text{ J}\)

Given \(H = \dfrac{\alpha}{100}\text{ J}\):
\(\dfrac{\alpha}{100} = \dfrac{16900}{486}\)

\(\alpha = \dfrac{16900 \cdot 100}{486} = \dfrac{1690000}{486} \approx 3477.37\)

Rounding to the nearest integer:
\(\alpha = 3477\)

Hence, the answer is \(3477\).