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)
Numerical value type. Enter your answer.
Answer: 3477
Label the node joining the $3\ \Omega$, $4\ \Omega$ and $6\ \Omega$ as $X$, and take the right-hand wire (joined to the negative terminals of both cells) as $0$ V. The left wire is then at $+2$ V, and the $6\ \Omega$ branch ends at the $+$ terminal of the $3$ V cell, i.e. at $3$ V.
KCL at $X$:
$$\frac{V_X - 2}{3} + \frac{V_X}{4} + \frac{V_X - 3}{6} = 0 \;\Rightarrow\; 9V_X = 14 \;\Rightarrow\; V_X = \frac{14}{9}\ \text{V}$$
Current in $6\ \Omega$: $I = \dfrac{3 - 14/9}{6} = \dfrac{13}{54}$ A.
Heat in $100$ s: $I^2Rt = \dfrac{169}{2916} \times 6 \times 100 \approx 34.77$ J $= \dfrac{3477}{100}$ J, so $\alpha = 3477$.
Solution by Sreeraj P, M.Sc Physics