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Consider the circuit shown in the Figure, where the input Vi(t) is in Volt. The average power (in mW) dissipated in the load resistance of 1 kΩ ...
Given Values Input voltage: v(t) = 12 sin(ωt) V Load resistance: R = 1 kΩ = 1000 Ω Capacitance: C = 1 μF Inductance: L = 2.2 mH Step 1: Calculate RMS Voltage The peak voltage is V m = 12 V. Hence, V rms = V m / √2 V rms = 12 / √2 V Step 2: Calculate Average Power The average power dissipated in the resistor is P = V rms ² / R P = (12 / √2)² / 1000 P = 144 / 2000 = 0.072 W Step 3: Convert to Milliwatts P = 0.072 × 1000 mW P = 72 mW Final Answer 72.00 mW Average power dissipated in the 1 kΩ load resistance at resonance.