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QPage 329: Q10: HC Verma: Concepts of Physics - Volume 2 Alternating Current Page 329: Q10 Solution: An alternating current of peak value 14 A is used to heat a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often equate the AC peak current with the DC current needed for the same heating. The power dissipated in a resistor is proportional to the mean of $I^2$, which for a sinusoid is $(I_{ ext{peak}}/sqrt{2})^2$. Thus the equivalent DC current is the RMS value, $I_{ ext{DC}}=I_{ ext{peak}}/sqrt{2}approx10 ext{ A}$, not the peak 14 A.

An alternating current of peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used, where i is

Multiple Choice Options

  • A 14 A
  • B about 20 A
  • C 7:00 AM
  • D about 10 A

Step-by-Step Verified Solution

**Answer:** (d) about 10 A. **Step-by-Step Explanation:** For the same heating effect, DC current = rms current = 14/�2 � 10 A.

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