Question

An electric current of \(2\,A\) passes through a wire of resistance \(25\,\Omega\). How much heat will be generated in \(1\,min\)?

• A. \(6 \times 10^3 \,J\)
• B. \(3.6 \times 10^3 \,J\)
• C. \(0.6 \times 10^3 \,J\)
• D. \(0.36 \times 10^3 \,J\)

Hint:

Use \( H = I^2Rt \) for electrical heating problems.

Answer 1

The Correct Option is A

To determine the heat generated by an electric current passing through a resistor, we can use Joule's Law. Joule's Law states that the heat (\(H\)) produced is proportional to the square of the current (\(I\)), the resistance (\(R\)), and the time (\(t\)) the current flows. The formula is given by:

\(H = I^2 R t\)

Where:

• \(I\) is the current in amperes (A),
• \(R\) is the resistance in ohms (Ω),
• \(t\) is the time in seconds (s).

Let's apply the given values:

• Current, \(I = 2\, \text{A}\)
• Resistance, \(R = 25\, \Omega\)
• Time, \(t = 1\, \text{min} = 60\, \text{seconds}\) (since time should be in seconds for this formula).

Now, substituting these values into the formula:

\(H = (2)^2 \times 25 \times 60\)

\(H = 4 \times 25 \times 60\)

\(H = 100 \times 60\)

\(H = 6000\, \text{Joules}\)

Therefore, the heat generated in 1 minute is \(6000\, J\), which matches the option:

• \(6 \times 10^3 \,J\)

Thus, the correct answer is \(6 \times 10^3 \,J\).