Calculate the amount of heat generated in an external load of resistance 8Ω if an alternating current of peak value 5 A is passed through it for 100s.

Analyze the question: An AC with peak value 5 A flows through an 8 Ω resistor for 100 s. We need the heat produced. Heating depends on the rms (effective) current, not the peak current.

Concept:

• I_rms = I_peak / √2
• Heat: H = I_rms² R t

Solution:

1. I_rms = 5/√2, so I_rms² = 25/2 = 12.5 A²
2. H = I_rms² × R × t = 12.5 × 8 × 100
3. H = 100 × 100 = 1.0 × 10⁴ J

Check against the options: 1.

(If you wrongly used the peak value, you would get 25 × 8 × 100 = 2.0 × 10⁴ J, which is option A. That is the trap in this question.)

Tip: For AC heating questions, always convert peak to rms first. I_rms = 0.707 × I_peak. Equivalently, the average power is ½ I₀² R, so the energy is ½ I₀² R t.