Question

In a series LCR circuit connected to an ac source of voltage $V$ and frequency $\omega$, the power loss at resonance is:

• A. $I^2\omega L$
• B. $V^2/(\omega C)$
• C. $V^2/R$
• D. $V^2/\left(\omega L - \dfrac{1}{\omega C}\right)$

Hint:

At resonance the reactances cancel, so Z equals R and the current is V/R. Only the resistor uses power, giving V squared over R.

Answer 1

The Correct Option is C

Step 1: Where is energy lost?
Only the resistor changes electrical energy to heat. An ideal inductor and an ideal capacitor store energy in each quarter cycle and return it in the next one. So $P=I_{rms}^{2}R$.

Step 2: Current at resonance:
The impedance is $Z=\sqrt{R^{2}+(\omega L-1/\omega C)^{2}}$. At resonance the bracket is zero, so $Z=R$. This is the smallest value of $Z$ and the current is the largest, $I=V/R$.

Step 3: Substitute:
\[ P=I^{2}R=\left(\frac{V}{R}\right)^{2}R=\frac{V^{2}}{R} \]

Step 4: Eliminate options by dimension and sense:
$I^{2}\omega L$ and $V^{2}/\omega C$ have the reactance of a coil or capacitor in them. Reactances never give a power loss.
$V^{2}/(\omega L-1/\omega C)$ has a zero denominator at resonance, so it would give an infinite power from a finite source. That is not physical.
Only $V^{2}/R$ has a resistance in it and stays finite.

Final Answer:
$P=V^{2}/R$, which is option (C). \[ \boxed{P=\frac{V^{2}}{R}} \]