An inductor of reactance 100 $$\Omega$$, a capacitor of reactance 50 $$\Omega$$, and a resistor of resistance 50 $$\Omega$$ are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is _____ W.

The given components are connected in series, so the circuit is a series $$R\text{-}L\text{-}C$$ combination.

Resistance: $$R = 50 \,\Omega$$
Inductive reactance: $$X_L = 100 \,\Omega$$
Capacitive reactance: $$X_C = 50 \,\Omega$$
Applied rms voltage: $$V_{\text{rms}} = 10 \text{ V}$$

Step 1 Find the net reactance.
For a series circuit, the net reactance is the algebraic difference of the individual reactances:
$$X = X_L - X_C = 100 - 50 = 50 \,\Omega$$

Step 2 Calculate the impedance.
For a series $$R\text{-}L\text{-}C$$ circuit, the impedance is
$$Z = \sqrt{R^{2} + X^{2}}$$
$$\Rightarrow Z = \sqrt{50^{2} + 50^{2}}$$
$$\Rightarrow Z = \sqrt{2500 + 2500}$$
$$\Rightarrow Z = \sqrt{5000}$$
$$\Rightarrow Z = 70.71 \,\Omega$$

Step 3 Find the rms current.
Ohm’s law for ac gives $$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{Z}$$.
$$I_{\text{rms}} = \dfrac{10}{70.71} = 0.1414 \text{ A}$$

Step 4 Determine the power factor.
Power factor in a series circuit is $$\cos\phi = \dfrac{R}{Z}$$.
$$\cos\phi = \dfrac{50}{70.71} = 0.707$$

Step 5 Calculate the average (true) power.
Average power dissipated is
$$P = V_{\text{rms}} I_{\text{rms}} \cos\phi$$
Substitute the values:
$$P = 10 \times 0.1414 \times 0.707$$
$$P = 1.0 \text{ W}$$

Hence, the average power dissipated by the circuit is 1 W.