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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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