In the series RLC circuit shown,C = 4 $\mu$F,L = 5 mH,and R = 1 k$\varOmega$.What is the maximum current in the circuit if V_peak = 10 V? A)0.5 mA B)1 mA C)10 mA D)50 mA

Question

Quizplus · Accepted Answer

We can use the formula for maximum current in an RLC circuit: Imax = V_peak / sqrt(R^2 + (2πfL - 1/(2πfC))^2) where f is the frequency of the source. Since the circuit is in series, the frequency of the source is the same as the frequency of the circuit. At resonance, 2πfL = 1/(2πfC), so the denominator simplifies to just R. At resonance, the maximum current is therefore: Imax = V_peak / R Imax = 10 V / 1 kΩ Imax = 10 mA So the answer is C, 10 mA.

In the Series RLC Circuit Shown,C = 4 μ\muμ F,L = 5 MH,and R = 1 K

In the Series RLC Circuit Shown,C = 4 $\mu$ F,L = 5 MH,and R = 1 K