Use Joule’s law of heating

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

Find the resistance using the first condition

Given:

• $H = 192\,\text{J}$
• $I = 4\,\text{A}$
• $t = 1\,\text{s}$

So, $192 = (4)^2 R (1)$ $192 = 16R$ $R = 12\,\Omega$

Now current is doubled

New current: $I' = 2 \times 4 = 8\,\text{A}$

Time: $t' = 5\,\text{s}$

Thermal energy dissipated: $H' = (I')^2 R t'$ $H' = (8)^2 \times 12 \times 5$ $H' = 64 \times 12 \times 5$ $H' = 768 \times 5$ $H' = 3840\,\text{J}$

Final Answer

$\boxed{3840}$

Comparison with stored correct answer

Stored correct answer = $3840$

My derived answer matches the stored answer.