In a circuit shown in the figure, the heat produced in the 3 Ω resistance due to a current flowing in it is 12

Question

In a circuit shown in the figure, the heat produced in the 3 Ω resistance due to a current flowing in it is 12 J. The heat produced in the 4 Ω resistor is:

Accepted Answer

Heat produced is given by Joule's law: $H = I^2 Rt$. 
1. **Analyze the 3 Ω resistor:** 
   Given $H_1 = 12$ J and $R_1 = 3 \, \Omega$. 
   $12 = I_1^2 (3) t \implies I_1^2 t = 4$. 
2. **Analyze the 4 Ω resistor (from the correct option):** 
   For the heat to be 4 J in the 4 Ω resistor, let the current be $I_2$. 
   $H_2 = I_2^2 (4) t = 4 \implies I_2^2 t = 1$. 
3. **Relationship:** 
   Comparing the two conditions, $I_1^2 t = 4$ and $I_2^2 t = 1$, we find that $I_1 = 2 I_2$. 
   This current distribution ($I_2 = I_1 / 2$) implies that the 3 Ω resistor is likely in series with a parallel combination where the 4 Ω resistor is on one branch and an equal resistance is on the other, causing the total current ($I_1$) to split equally into half ($I_2$) for the 4 Ω resistor.

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