1. Use the resonance property of a series LCR circuit

At resonance, the inductive and capacitive reactances cancel:

$X_L = X_C$

So the impedance becomes purely resistive:

$Z = R$

Hence, the circuit behaves like a simple resistor at resonance.

2. Relate power, voltage, and resistance

Given:

• RMS voltage: $V = 120\,\text{V}$
• Power consumed at resonance: $P = 16\,\text{W}$

For a purely resistive circuit,

$P = \frac{V^2}{R}$

Substitute the values:

$16 = \frac{120^2}{R}$

$R = \frac{120^2}{16}$

$R = \frac{14400}{16} = 900\,\Omega$

3. Final answer

$R = 900\,\Omega$

4. Comparison with stored answer

Stored correct answer = $900$

Our derived answer also is $900$, so they agree.