alleryStart learning →An RMS AC voltage of $230\,\text{V}$ drives a $100\Omega$ RESISTIVE load. The real power dissipated is
A$2.3\,\text{W}$
B$23\,\text{W}$
C$529\,\text{W}$
D$2300\,\text{W}$
Answer & Solution
Correct answer: C. $529\,\text{W}$
1. For a RESISTIVE load: $P = V_{RMS}^2 / R = V_{RMS} \cdot I_{RMS}$.
2. $P = 230^2/100 = 52900/100 = 529\,\text{W}$.
3. Equivalent: $I_{RMS} = 230/100 = 2.3\,\text{A}$; $P = VI = 230 \times 2.3 = 529\,\text{W}$. ✓
4. RMS values give the actual heating power — no $\sqrt{2}$ conversion needed.
5. Other options are wrong by orders of magnitude.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — AC", Vol II, Ch 11 (AC power numerical)._
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