Question 21.4: An RLC Circuit Goal Analyze a series RLC AC circuit and find...
An RLC Circuit
Goal Analyze a series RLC AC circuit and find the phase angle.
Problem A series R L C AC circuit has resistance R=2.50 \times 10^{2} \Omega, inductance L=0.600 \mathrm{H}, capacitance C=3.50 \mu \mathrm{F}, frequency f=60.0 \mathrm{~Hz}, and maximum voltage \Delta V_{\max }=1.50 \times 10^{2} \mathrm{~V}. Find (a) the impedance, (b) the maximum current in the circuit, (c) the phase angle, and (d) the maximum voltages across the elements.
Strategy Calculate the inductive and capacitive reactances, then substitute them and given quantities into the appropriate equations.
(a) Find the impedance of the circuit.
First, calculate the inductive and capacitive reactances:
X_{L}=2 \pi f L=226 \Omega \quad X_{C}=1 / 2 \pi f C=758 \Omega
Substitute these results and the resistance R into Equation 21.13 to obtain the impedance of the circuit:
\begin{aligned} Z & =\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ & =\sqrt{\left(2.50 \times 10^{2} \Omega\right)^{2}+(226 \Omega-758 \Omega)^{2}}=588 \Omega \end{aligned}
(b) Find the maximum current.
Use Equation 21.12,
\Delta\,V_{\mathrm{max}}=\,I_{\mathrm{max}}\sqrt{R^{2}\,+\,(X_{L}-\,X_{C})^{2}} (21.12)
the equivalent of Ohm’s law, to find the maximum current:
I_{\max }=\frac{\Delta V_{\max }}{Z}=\frac{1.50 \times 10^{2} \mathrm{~V}}{588 \Omega}=0.255 \mathrm{~A}
(c) Find the phase angle.
Calculate the phase angle between the current and the voltage with Equation 21.15:
\tan\phi={\frac{X_{L}-X_{C}}{R}} (21.15)
\phi=\tan ^{-1} \frac{X_{L}-X_{C}}{R}=\tan ^{-1}\left(\frac{226 \Omega-758 \Omega}{2.50 \times 10^{2} \Omega}\right)=-64.8^{\circ}
(d) Find the maximum voltages across the elements.
Substitute into the “Ohm’s law” expressions for each individual type of current element:
\begin{aligned} & \Delta V_{R, \text { max }}=I_{\max } R=(0.255 \mathrm{~A})\left(2.50 \times 10^{2} \Omega\right)=63.8 \mathrm{~V} \\ & \Delta V_{L, \text { max }}=I_{\max } X_{L}=(0.255 \mathrm{~A})\left(2.26 \times 10^{2} \Omega\right)=57.6 \mathrm{~V} \\ & \Delta V_{C, \text { max }}=I_{\text {max }} X_{C}=(0.255 \mathrm{~A})\left(7.58 \times 10^{2} \Omega\right)=193 \mathrm{~V} \end{aligned}
Remarks Because the circuit is more capacitive than inductive \left(X_{C}>X_{L}\right), \phi is negative. A negative phase angle means that the current leads the applied voltage. Notice also that the sum of the maximum voltages across the elements is \Delta V_{R}+\Delta V_{L}+\Delta V_{C}=314 \mathrm{~V}, which is much greater than the maximum voltage of the generator, 150 \mathrm{~V}. As we saw in Quick Quiz 21.2, the sum of the maximum voltages is a meaningless quantity because when alternating voltages are added, both their amplitudes and their phases must be taken into account. We know that the maximum voltages across the various elements occur at different times, so it doesn’t make sense to add all the maximum values. The correct way to “add” the voltages is through Equation 21.10.
\Delta\,V_{\mathrm{max}}=\sqrt{\Delta\,V_{R}{}^{2}\,+\,(\Delta\,V_{L}-\,\Delta\,V_{C})^{2}} (21.10)