Chapter 30, Problem 93GP

A resistor R, capacitor C, and inductor L are connected in parallel across an ac generator as shown in Fig. 30–34. The source emf is V = V₀ sin ωt. Determine the current as a function of time (including amplitude and phase): (a) in the resistor, (b) in the inductor, (c) in the capacitor, (d) What is the total current leaving the source? (Give amplitude I₀ and phase) (e) Determine the impedance Z defined as Z =V₀/I₀). (f) What is the power factor?

FIGURE 30–34

Problem 93.

93.

(a)    Since the three elements are connected in parallel at any given instant in time they will all three have the same voltage drop across them. That is the voltages across each element will be in phase with the source. The current in the resistor is in phase with the voltage source with magnitude given by Ohm’s law.

$I_R(t)=\boxed{\frac{V_0}{R}\sin \omega t}$

(b)    The current through the inductor will lag behind the voltage by π/2, with magnitude equal to the voltage source divided by the inductive reactance.

$L_C(t)=\boxed{\frac{V_0}{X_L}\sin \left(\omega t-\frac{\pi }{2}\right)}$

(c)    The current through the capacitor leads the voltage by π/2, with magnitude equal to the voltage source divided by the capacitate reactance.

$I_C(t)=\boxed{\frac{V_0}{X_C}\sin \left(\omega t+\frac{\pi }{2}\right)}$

(d)    The total current is the sum of the currents through each element. We use a phasor diagram to add the currents, as was used in Section 30-8 to add the voltages with different phases. The net current is found by subtracting the current through the inductor from the current through the capacitor. Then using the Pythagorean theorem to add the current through the resistor. We use the tangent function to find the phase angle between the current and voltage source.

$\begin{array}{l}{I}_0=\sqrt{I_{R0}^2+{(I_{C0}-I_{L0})}^2}=\sqrt{{\left(\frac{V_0}{R}\right)}^2+{\left(\frac{V_0}{X_C}-\frac{V_0}{X_L}\right)}^2}=\frac{V_0}{R}\sqrt{1+{\left(R\omega C-\frac{1}{R\omega L}\right)}^2}\\ I(t)=\boxed{\frac{V_0}{R}\sqrt{1+{\left(R\omega C-\frac{R}{\omega L}\right)}^2\sin \left(\omega t+\varphi \right)}}\\ \tan \varphi =\frac{\frac{V_0}{X_C}-\frac{V_0}{X_L}}{\frac{V_0}{R}}\to \varphi ={\tan }^{-1}\left(\frac{R}{X_C}-\frac{R}{X_L}\right)=\boxed{{\tan }^{-1}\left(R\omega C-\frac{R}{\omega L}\right)}\end{array}$

(e) We divide the magnitude of the voltage source by the magnitude of the current to find the impedance.

$Z=\frac{V_0}{I_0}=\frac{V_0}{\frac{V_0}{R}\sqrt{1+{\left(R\omega C-\frac{R}{\omega L}\right)}^2}}=\boxed{\frac{R}{\sqrt{1+{\left(R\omega C-\frac{R}{\omega L}\right)}^2}}}$

(f) The power factor is the ratio of the power dissipated in the circuit divided by the product of the rms voltage and current.

$\frac{I_{R,\text{mrs}}^{2}R}{V_{\text{rms}}{I}_{\text{rms}}}=\frac{I_R^{2}R}{V_0{I}_0}=\frac{{\left(\frac{V_0}{R}\right)}^{2}R}{V_0\frac{V_0}{R}\sqrt{1+{\left(R\omega C-\frac{R}{\omega L}\right)}^2}}=\boxed{\frac{1}{\sqrt{1+{\left(R\omega C-\frac{R}{\omega L}\right)}^2}}}$