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(NONE) Let the potential at the junction between $3	ext{ }\Omega$, $6	ext{ }\Omega$, and $4	ext{ }\Omega$ resistors be $V$. Let the potential at th

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(NONE) Let the potential at the junction between $3	ext{ }\Omega$, $6	ext{ }\Omega$, and $4	ext{ }\Omega$ resistors be $V$. Let the potential at the right junction be $0	ext{ V}$.Applying Kirchhoff's Current Law $(KCL)$ at the junction with potential $V$:$\frac{V - 2}{3} + \frac{V - 3}{6} + \frac{V - 0}{4} = 0$Multiply by $12$ to clear denominators:$4(V - 2) + 2(V - 3) + 3V = 0$$4V - 8 + 2V - 6 + 3V = 0$$9V = 14 \implies V = \frac{14}{9}	ext{ V}$.The current $I$ through the $6	ext{ }\Omega$ resistor is $I = \frac{V - 3}{6} = \frac{\frac{14}{9} - 3}{6} = \frac{\frac{14 - 27}{9}}{6} = \frac{-13}{54}	ext{ A}$.The magnitude of current is $|I| = \frac{13}{54}	ext{ A}$.Heat generated $H = I^2Rt = (\frac{13}{54})^2 	imes 6 	imes 100 = \frac{169}{2916} 	imes 600 = \frac{169 	imes 600}{2916} = \frac{101400}{2916} \approx 34.77	ext{ J}$.Given $H = \frac{\alpha}{100} = 34.77$, so $\alpha = 3477$.

Refer to the circuit diagram given below. The heat generated across the $6\text{ }\Omega$ resistance in $100\text{ s}$ is $\frac{\alpha}{100}\text{ J}$. The value of $\alpha$ is . . . . . . . (Nearest integer)

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