1. Use Joule’s law of heating

   The thermal energy developed in a resistor is $H = I^2 R t$

   where:

   • $H$ = heat energy
   • $I$ = current
   • $R$ = resistance
   • $t$ = time

2. Given first condition

   When current $I_1 = 1.5\,\text{A}$ flows for $t = 20\,\text{s}$, the energy developed is $H_1 = 500\,\text{J}$

   So, $H_1 = I_1^2 R t$

3. Second condition

   Now current is increased to $I_2 = 3\,\text{A}$ with the same resistor and same time $t = 20\,\text{s}$.

   Since $R$ and $t$ remain constant, $H \propto I^2$

   Therefore, $\frac{H_2}{H_1} = \frac{I_2^2}{I_1^2}$

4. Substitute values

   $\frac{H_2}{500} = \frac{3^2}{1.5^2}$

   $\frac{H_2}{500} = \frac{9}{2.25} = 4$

   Hence, $H_2 = 500 \times 4 = 2000\,\text{J}$

5. Check options

   • A: $1000\,\text{J}$ ✗
   • B: $2000\,\text{J}$ ✓
   • C: $1500\,\text{J}$ ✗
   • D: $500\,\text{J}$ ✗

Final Answer: $2000\,\text{J}$