A RESISTOR AND A CAPACITOR IN AN AC CIRCUIT
A 200-Ω resistor is connected in series with a 5.0-μF capacitor. The voltage across the resistor is v_R = (1.20 V) cos(2500 rad/s)t (Fig. 31.10). (a) Derive an expression for the circuit current. (b) Determine the capacitive reactance of the capacitor. (c) Derive an expression for the voltage across the capacitor.
IDENTIFY and SET UP:
Since this is a series circuit, the current is the same through the capacitor as through the resistor. Our target variables are the current i, the capacitive reactance X_C, and the capacitor voltage v_C. We use Eq. (31.6) to find an expression for i in terms of the angular frequency ω = 2500 rad/s, Eq. (31.18) to find X_C, Eq. (31.19) to find the capacitor voltage amplitude V_C, and Eq. (31.16) to write an expression for v_C.
v_{R}=i R=(I R) \cos \omega t (31.6)
v_{C}=\frac{I}{\omega C} \cos \left(\omega t-90^{\circ}\right) (31.16)
X_{C}=\frac{1}{\omega C} (31.18)
V_{C}=I X_{C} (31.19)
EXECUTE:
(a) From Eq. (31.6), v_R = iR, we find
\begin{aligned}i &=\frac{v_{R}}{R}=\frac{(1.20 \mathrm{~V}) \cos (2500 \mathrm{rad} /\mathrm{s}) t}{200 \Omega} \\&=\left(6.0 \times 10^{-3} \mathrm{~A}\right) \cos (2500\mathrm{rad} / \mathrm{s}) t\end{aligned}(b) From Eq. (31.18), the capacitive reactance at ω = 2500 rad/s is
X_{C}=\frac{1}{\omega C}=\frac{1}{(2500 \mathrm{rad} / \mathrm{s})\left(5.0 \times 10^{-6} \mathrm{~F}\right)}=80 \Omega(c) From Eq. (31.19), the capacitor voltage amplitude is
V_{C}=I X_{C}=\left(6.0 \times 10^{-3} \mathrm{~A}\right)(80 \Omega)=0.48 \mathrm{~V}(The 80-Ω reactance of the capacitor is 40% of the resistor’s 200-Ω resistance, so V_C is 40% of V_R.) The instantaneous capacitor voltage is given by Eq. (31.16):
\begin{aligned}v_{C} &=V_{C} \cos \left(\omega t-90^{\circ}\right) \\&=(0.48 \mathrm{~V})\cos [(2500 \mathrm{rad} / \mathrm{s}) t-\pi / 2 \mathrm{rad}]\end{aligned}
EVALUATE: Although the same current passes through both the capacitor and the resistor, the voltages across them are different in both amplitude and phase. Note that in the expression for v_C we converted the 90° to π/2 rad so that all the angular quantities have the same units. In ac circuit analysis, phase angles are often given in degrees, so be careful to convert to radians when necessary.