Question 21.6: A Circuit in Resonance GOAL Understand resonance frequency a...
A Circuit in Resonance
GOAL Understand resonance frequency and its relation to inductance, capacitance, and the rms current.
PROBLEM Consider a series RLC circuit for which R = 1.50 × 10² Ω, L = 20.0 mH, \Delta V_{\mathrm{rms}} = 20.0 V, and f = 796 \scriptstyle{s^{-1}} . (a) Determine the value of the capacitance for which the rms current is a maximum. (b) Find the maximum rms current in the circuit.
STRATEGY The current is a maximum at the resonance frequency f_{\mathrm{0}}, which should be set equal to the driving frequency, 796 \scriptstyle{s^{-1}}. The resulting equation can be solved for C. For part (b), substitute into Equation 21.18
I_{\mathrm{rms}}={\frac{\Delta V_{\mathrm{rms}}}{Z}}={\frac{\Delta V_{\mathrm{rms}}}{\sqrt{R^{2}+(X_{L}~-~X_{C})^{2}}}} [21.18]
to get the maximum rms current.
(a) Find the capacitance giving the maximum current in the circuit (the resonance condition).
Solve the resonance frequency for the capacitance:
f_{0}=\frac{1}{2\pi\sqrt{L C}}\ \ \ \rightarrow\ \ \ \sqrt{L C}=\frac{1}{2\pi f_{0}}\ \ \rightarrow\ \ \ L C=\frac{1}{4\pi^{2}f_{0}{}^{2}}
C={\frac{1}{4\pi^{2}f_{0}{}^{2}L}}
Insert the given values, substituting the source frequency for the resonance frequency, f_{\mathrm{0}}:
C={\frac{1}{4\pi^{2}(796\;\mathrm{Hz})^{2}(20.0~\times~10^{-3}\;\mathrm{H})}}=\;{{{\ }2.00\times10^{-6}\;{\mathrm{F}}}}
(b) Find the maximum rms current in the circuit.
The capacitive and inductive reactances are equal, so Z = R = 1.50 × 10² Ω. Substitute into Equation 21.18 to find the rms current:
I_{\mathrm{rms}}={\frac{\Delta V_{\mathrm{rms}}}{Z}}={\frac{20.0\,\mathrm{V}}{1.50~\times~10^{2}\,\Omega}}=\mathrm{~}0.133\,\mathrm{A}
REMARKS Because the impedance Z is in the denominator of Equation 21.18, the maximum current will always occur when X_{\mathrm{L}}=X_{\mathrm{C}} because that yields the minimum value of Z.