"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 $10 \mathrm{~V}, 50 \mathrm{~Hz}$. Average power dissipated by the circuit is ___________ W."

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To calculate the average power dissipated in the circuit, we follow these steps:

Formula for Power:

$ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi $

Expressing Power in Terms of Voltage and Impedance:

$ P = V_{\text{rms}} \times \frac{V_{\text{rms}}}{Z} \times \frac{R}{Z} $

Simplifying the Expression:

$ P = \frac{V_{\text{rms}}^2 \times R}{Z^2} $

Calculate Impedance (Z):

The impedance of the circuit (Z) is calculated as:

$ Z = \sqrt{R^2 + (X_L - X_C)^2} $

Given:

- $ X_L = 100 \, \Omega $

- $ X_C = 50 \, \Omega $

- $ R = 50 \, \Omega $

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

Thus:

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

Substitute Values to Find Power (P):

Given $ V_{\text{rms}} = 10 \, \text{V} $, we calculate $ P $ as follows:

$ P = \frac{10^2 \times 50}{(50\sqrt{2})^2} $

Simplifying further:

$ P = \frac{100 \times 50}{2500 \times 2} = 1 \, \text{W} $

Therefore, the average power dissipated by the circuit is $ 1 \, \text{W} $.

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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 $10 \mathrm{~V}, 50 \mathrm{~Hz}$. Average power dissipated by the circuit is ___________ W.

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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 $10 \mathrm{~V}, 50 \mathrm{~Hz}$. Average power dissipated by the circuit is ___________ W.

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