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Step-by-Step Solution
Step 1: Identify the given information
• Initial current, $i_1 = 1.5 \text{ A}$
• Final current, $i_2 = 3 \text{ A}$
• Time interval, $\Delta t = 20 \text{ s}$
• Heat generated (initially), $H_1 = 500 \text{ J}$
Step 2: Recall the formula for heat developed in a resistor
The thermal energy (heat) developed in a resistor due to an electric current is given by:
$$
H = i^2 R \Delta t
$$
where
• $i$ is the current,
• $R$ is the resistance,
• $\Delta t$ is the time interval.
Step 3: Relate the heats for two different currents
Let $H_2$ be the heat produced when the current is $i_2$. Using the same resistor and time interval, we have:
$$
\frac{H_1}{H_2} = \frac{i_1^2}{i_2^2}.
$$
Step 4: Substitute the known values
Substitute $H_1 = 500 \text{ J}$, $i_1 = 1.5 \text{ A}$, and $i_2 = 3 \text{ A}$ into the equation:
$$
\frac{500}{H_2} = \frac{(1.5)^2}{(3.0)^2} = \frac{2.25}{9} = \frac{1}{4}.
$$
Step 5: Solve for $H_2$
Rearranging the above equation,
$$
\frac{500}{H_2} = \frac{1}{4} \quad \Rightarrow \quad H_2 = 500 \times 4 = 2000 \text{ J}.
$$
Final Answer
The energy developed in 20 s when the current is increased to $3 \text{ A}$ is 2000 J.
