An electric iron of 20 Ω draws 5 A for 30 s. Its heat is not 5 × 20 × 30 = 3000 J: in Joule’s law H = I²Rt the current counts twice. Solve it step by step.
Question (CBSE board question, verbatim): An electric iron of resistance 20 Ω draws a current of 5 A. The heat developed in the iron in 30 seconds is : (A) 15000 J (B) 6000 J (C) 1500 J (D) 3000 J
H = I²Rt = 5² × 20 × 30 = 25 × 20 × 30 = 500 × 30 = 15000 J, option (A).
Which heat formula: same current → H = I²Rt; same voltage → H = V²t/R.
Comment challenge: the same iron carries 2 A for 60 s. Find the heat.
Quick recall (Electricity, series circuits): same current I; V = V₁ + V₂ + V₃.
Electricity quick revision (full video):
References: NCERT Class 10 Science, Ch 11 Electricity, §11.7 Heating effect of electric current (Joule’s law H = I²Rt) and §11.6 resistors in series (same current, V = V₁ + V₂ + V₃); CBSE Class 10 board question and marking scheme ((A) 15000 J).
Narration uses a licensed AI voice (Azure AI Speech, Indian English). Script and physics research by ScoreSpark.
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Answer: H = I²Rt with the same 20 Ω iron: 2² × 20 × 60 = 4 × 1200 = 4800, so H = 4800 J (not 2 × 20 × 60 = 2400 J).
Quick recall (lenses): Lens formula and lens magnification? ⇒ 1/v − 1/u = 1/f · m = v/u
Electricity quick revision (full video): https://www.youtube.com/watch?v=ppYotzJBczM