AN L-R-C SERIES CIRCUIT I
In the series circuit of Fig. 31.13a, suppose R = 300 Ω, L = 60 mH, C = 0.50 μF, V = 50 V, and ω = 10,000 rad/s. Find the reactances X_L and X_C, the impedance Z, the current amplitude I, the phase angle φ, and the voltage amplitude across each circuit element.
IDENTIFY and SET UP:
This problem uses the ideas developed in Section 31.2 and this section about the behavior of circuit elements in an ac circuit. We use Eqs. (31.12) and (31.18) to determine X_L and X_C, and Eq. (31.23) to find Z. We then use Eq. (31.22) to find the current amplitude and Eq. (31.24) to find the phase angle. The relationships in Table 31.1 then yield the voltage amplitudes.
X_{L}=\omega L (31.12)
X_{C}=\frac{1}{\omega C} (31.18)
V=I Z (31.22)
Z=\sqrt{R^{2}+[\omega L-(1 / \omega C)]^{2}} (31.23)
\tan \phi=\frac{\omega L-1 / \omega C}{ R} (31.24)
| TABLE 31.1 | Circuit elements with alternating Current | ||
| Circuit Element | Amplitude Relationship | Circuit Quantity | Phase of v |
| Resistor | V_R = IR | R | In phase with i |
| Inductor | V_L = IX_L | X_{L}=\omega L | Leads i by 90° |
| Capacitor | V_C = IX_C | X_{C}=1 / \omega C | Lags i by 90° |
EXECUTE:
The inductive and capacitive reactances are
\begin{aligned}&X_{L}=\omega L=(10,000 \mathrm{rad} / \mathrm{s})(60 \mathrm{mH})=600\Omega \\&X_{C}=\frac{1}{\omega C}=\frac{1}{(10,000 \mathrm{rad} /\mathrm{s})\left(0.50 \times 10^{-6} \mathrm{~F}\right)}=200 \Omega\end{aligned}The impedance Z of the circuit is then
\begin{aligned}Z &=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(300 \Omega)^{2}+(600\Omega-200 \Omega)^{2}} \\&=500 \Omega\end{aligned}With source voltage amplitude V = 50 V, the current amplitude I and phase angle φ are
\begin{aligned}I &=\frac{V}{Z}=\frac{50 \mathrm{~V}}{500 \Omega}=0.10 \mathrm{~A} \\\phi&=\arctan \frac{X_{L}-X_{C}}{R}=\arctan \frac{400 \Omega}{300 \Omega}=53^{\circ}\end{aligned}From Table 31.1, the voltage amplitudes V_R, V_L, and V_C across the resistor, inductor, and capacitor, respectively, are
\begin{aligned}&V_{R}=I R=(0.10 \mathrm{~A})(300 \Omega)=30 \mathrm{~V} \\&V_{L}=I X_{L}=(0.10 \mathrm{~A})(600 \Omega)=60 \mathrm{~V} \\&V_{C}=I X_{C}=(0.10 \mathrm{~A})(200\Omega)=20 \mathrm{~V}\end{aligned}
EVALUATE: As in Fig. 31.13b, X_L > X_C; hence the voltage amplitude across the inductor is greater than that across the capacitor and φ is positive. The value φ = 53° means that the voltage leads the current by 53°.
Note that the source voltage amplitude V = 50 V is not equal to the sum of the voltage amplitudes across the separate circuit elements: 50 V ≠ 30 V + 60 V + 20 V. Instead, V is the vector sum of the V_R, V_L, and V_C phasors. If you draw the phasor diagram like Fig. 31.13b for this particular situation, you’ll see that V_R, V_L – V_C, and V constitute a 3-4-5 right triangle.