1. Given data

• Inductive reactance: $X_L = 100\,\Omega$
• Capacitive reactance: $X_C = 50\,\Omega$
• Resistance: $R = 50\,\Omega$
• AC source voltage: $V = 10\,\text{V}$

For a series $RLC$ circuit, average power dissipated is

$P = I_{\text{rms}}^2 R$

So first we find the current.

2. Net reactance

In series,

$X = X_L - X_C = 100 - 50 = 50\,\Omega$

3. Impedance of the circuit

$Z = \sqrt{R^2 + X^2} = \sqrt{50^2 + 50^2}$

$Z = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2}\,\Omega$

4. Current in the circuit

Assuming the given AC voltage $10\,\text{V}$ is the rms value,

$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}}\,\text{A}$

5. Average power dissipated

Only the resistor dissipates average power:

$P = I_{\text{rms}}^2 R$

$P = \left(\frac{1}{5\sqrt{2}}\right)^2 \times 50$

$P = \frac{1}{50} \times 50 = 1\,\text{W}$

6. Final answer

$\boxed{1\,\text{W}}$

7. Comparison with stored answer

Stored correct answer: $1$

Our derived answer is also $1$, so it agrees.