For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor: (a) Find the resonant frequency. (b) Calculate I_{ rms } at resonance if V_{ rms } is 120 V.
Strategy
The resonant frequency is found by using the expression in f_{0}=\frac{1}{2 \pi \sqrt{L C}}. The current at that frequency is the same as if the resistor alone were in the circuit.
Solution for (a)
Entering the given values for L and C into the expression given for f_{0} in f_{0}=\frac{1}{2 \pi \sqrt{L C}} yields
f_{0}=\frac{1}{2 \pi \sqrt{L C}} (23.73)
=\frac{1}{2 \pi \sqrt{\left(3.00 \times 10^{-3} H \right)\left(5.00 \times 10^{-6} F \right)}}=1.30 kHz.
Discussion for (a)
We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.
Solution for (b)
The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone.
Thus,
I_{ rms }=\frac{V_{ rms }}{Z}=\frac{120 V }{40.0 \Omega}=3.00 A. (23.74)
Discussion for (b)
At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.