A resistor develops 500 J of thermal energy in 20 s when a current of 1.5 A is passed through it. If the current is increased from 1.5 A to 3 A, what will be the energy developed in 20 s.

The thermal energy developed in a resistor is given by $$H = I^2 R t$$, where $$I$$ is the current, $$R$$ is the resistance, and $$t$$ is the time.

From the first case, $$500 = (1.5)^2 \times R \times 20 = 2.25 \times 20R = 45R$$. Solving for resistance: $$R = \frac{500}{45} = \frac{100}{9}$$ $$\Omega$$.

When the current is increased to $$3$$ A, the energy developed in $$20$$ s is $$H_2 = (3)^2 \times \frac{100}{9} \times 20 = 9 \times \frac{100}{9} \times 20 = 100 \times 20 = 2000$$ J.

Alternatively, since $$H \propto I^2$$ (with $$R$$ and $$t$$ unchanged), $$\frac{H_2}{H_1} = \frac{(3)^2}{(1.5)^2} = \frac{9}{2.25} = 4$$, giving $$H_2 = 4 \times 500 = 2000$$ J.