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An AC circuit is defined with L = 20 mH, C = 50 μF, R = 40 Ω, and voltage V = 10 sin 340t. Determine the power dissipated in this circuit.

1. 50 multiplied by 40 divided by 4304 equals 0.51 W
2. 10 divided by square root of 2 multiplied by 50
3. 314 multiplied by 20 multiplied by 10 to the power of minus 3
4. 1 divided by 314 multiplied by 100 multiplied by 10 to the power of minus 6

Correct answer: 50 multiplied by 40 divided by 4304 equals 0.51 W

Solution

The power loss in an AC circuit is given by P = I²R, where I is the RMS current. First, calculate the impedance Z = √(R² + (XL - XC)²), where XL = ωL and XC = 1/(ωC). Using ω = 340 rad/s, XL = 6.8 Ω, and XC = 9.35 Ω, Z = √(40² + (6.8 - 9.35)²) ≈ 40.1 Ω. RMS current I = V_RMS / Z = (10/√2) / 40.1 ≈ 0.176 A. Finally, P = I²R ≈ (0.176)² × 40 ≈ 0.51 W.

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