Heat Generated Across Resistor in Circuit

To find the heat generated across the $6\ \Omega$ resistor, we first analyze the circuit using nodal analysis.

Step 1: Circuit Analysis and Node Potentials

Let the reference node (common wire on the right side) have a potential of $V_R = 0\text{ V}$.

• The bottom wire contains a $2\text{ V}$ battery with its positive terminal connected to the left node $L$. Thus, the potential at node $L$ is: $V_L = 2\text{ V}$

• Let the potential at the junction between the $3\ \Omega$, $4\ \Omega$, and $6\ \Omega$ resistors be $V_M$.

• The top branch contains a $6\ \Omega$ resistor in series with a $3\text{ V}$ battery whose positive terminal faces left (towards the $6\ \Omega$ resistor) and negative terminal faces right (connected to $V_R = 0\text{ V}$). Thus, the potential on the right side of the $6\ \Omega$ resistor is $3\text{ V}$.

Step 2: Kirchhoff's Current Law (KCL) at Node $M$

Applying KCL at node $M$: $\frac{V_M - 2}{3} + \frac{V_M - 0}{4} + \frac{V_M - 3}{6} = 0$

Multiplying the entire equation by $12$ to clear the denominators: $4(V_M - 2) + 3V_M + 2(V_M - 3) = 0$ $4V_M - 8 + 3V_M + 2V_M - 6 = 0$ $9V_M = 14 \implies V_M = \frac{14}{9}\text{ V}$

Step 3: Current through the $6\ \Omega$ Resistor

The magnitude of the current $I$ passing through the $6\ \Omega$ resistor is: $I = \left| \frac{V_M - 3}{6} \right| = \left| \frac{\frac{14}{9} - 3}{6} \right| = \left| \frac{-\frac{13}{9}}{6} \right| = \frac{13}{54}\text{ A}$

Step 4: Heat Generated in $100\text{ seconds}$

The heat $H$ generated across the $6\ \Omega$ resistor in $t = 100\text{ s}$ is given by Joule's Law: $H = I^2 R t$ $H = \left(\frac{13}{54}\right)^2 \times 6 \times 100$ $H = \frac{169}{2916} \times 600 = \frac{169 \times 100}{486} = \frac{8450}{243}\text{ J}$

Step 5: Calculating $\alpha$

Given that $H = \frac{\alpha}{100}\text{ J}$: $\frac{\alpha}{100} = \frac{8450}{243}$ $\alpha = \frac{845000}{243} \approx 3477.366$

Rounding to the nearest integer: $\alpha = 3477$