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In a circuit containing an inductor L, a capacitor C, and a resistor R (arranged in parallel) connected to an AC source of peak voltage V and angular frequency omega, what is the peak current through the AC source?

1. V / sqrt(1/R² + (omega*L - 1/(omega*C))²)
2. V * sqrt(1/R² + (1/(omega*L) - omega*C)²)
3. V / sqrt(R² + (omega*L - 1/(omega*C))²)
4. V*R*omega*C / sqrt(omega²*C² + R*(omega²*C² - 1)²)

Correct answer: V * sqrt(1/R² + (1/(omega*L) - omega*C)²)

Solution

In a parallel RLC circuit, each element shares the same voltage V. The total current = V*(total admittance). Admittance = sqrt(1/R² + (omega*C - 1/(omega*L))²), so peak current = V*sqrt(1/R² + (1/(omega*L) - omega*C)²).

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