Consider the circuit shown in the Figure, where the input v_i(t) = 12 sin(ωt) is in Volt. The average power (in mW) dissipated in the load resistance of 1 kΩ at the resonant frequency is _________. (rounded off to two decimal places)

The correct answer is 72.00.

1. Circuit Analysis from the Given Image:
By analyzing the provided circuit diagram, we can identify the following components connected in parallel across the sinusoidal input voltage source:
- Input voltage source:

$v_i\left(t\right)=12\sin \left(\omega t\right)\text{V}$

- A capacitor with capacitance: _{C = 1 μF}
- An inductor with inductance: _{L = 2.2 mH}
- A load resistor with resistance: _{R = 1 kΩ = 1000 Ω}

2. Determination of the Voltage Across the Load Resistance:
Since all three passive elements (capacitor, inductor, and resistor) are connected directly in parallel across an ideal voltage source _vi(t), the voltage across each component is equal to the source voltage at any frequency (including the resonant frequency). Therefore, the voltage across the load resistor _R is:

$v_R\left(t\right)=v_i\left(t\right)=12\sin \left(\omega t\right)\text{V}$

Thus, the peak voltage amplitude across the resistor is:

$V_m=12\text{V}$

The root-mean-square (RMS) value of this voltage is:

$V_{\mathrm{rms}}=\frac{V_m}{\sqrt{2}}=\frac{12}{\sqrt{2}}\text{V}$

3. Calculation of Average Power Dissipated in the Load:
The average power _P dissipated in the load resistor _R is given by:

$P=\frac{V_{\mathrm{rms}}^2}{R}$

Substituting the values into the formula:

$P=\frac{{\left(\frac{12}{\sqrt{2}}\right)}^2}{1000}=\frac{144}{2\times 1000}=\frac{72}{1000}\text{W}$

Converting this value to milliwatts (mW):

$P=72\text{mW}$