Question

Let a 100 V sinusoidal source be connected to a series combination of a $\mathbf{3}\mathbf{}\mathbf{\Omega }$ resistor, an$\mathbf{8}\mathbf{}\mathbf{\Omega }$ inductor, and a$\mathbf{4}\mathbf{}\mathbf{\Omega }$ capacitor. (a) Draw the circuit diagram. (b) Compute the series impedance. (c) Determine the current Idelivered by the source. Is the current lagging or leading the source voltage? What is the power factor of this circuit?

Short answer

(a) The required circuit diagram is,

(b) The value of the series impedance is $5\angle 53.{13}^0\mathrm{\Omega }$.

(c) The current delivered by the source is $20\angle -53.{13}^0\mathrm{A}$.

(d) The power factor of the circuit is 0.6 lagging.

Step-by-step solution

Determine the formulas of series RLC circuit.

Consider the impedance of the RLC circuit.

${\mathbf{Z}}_{\mathbf{s}}\mathbf{=}\mathbf{R}\mathbf{+}{\mathbf{X}}_{\mathbf{L}}\mathbf{+}{\mathbf{X}}_{\mathbf{C}}$ …… (1)

Here, ${\mathrm{X}}_{\mathrm{L}}$and ${\mathrm{X}}_{\mathrm{C}}$ are the inductive and capacitive reactance respectively.

Consider the formula for the current delivered by the source.

$\mathbf{I}\mathbf{=}\frac{\mathbf{V}}{{\mathbf{Z}}_{\mathbf{s}}}$ …… (3)

Consider the formula for the power factor of the RLC circuit.

$\mathbf{PF}\mathbf{=}\mathbf{cos}\left(\theta \right)$ …… (3)

Here, role="math" localid="1655106381074" $\theta$ is the phase angle.

Draw the circuit diagram for series RLC circuit.

Given that the resistance of $\mathrm{R}=3\mathrm{\Omega }$is in series with inductor of ${\mathrm{X}}_{\mathrm{L}}=8\mathrm{\Omega }$ and the capacitive reactance of ${\mathrm{X}}_{\mathrm{c}}=-\mathrm{j}4\mathrm{\Omega }$. The circuit has a voltage source of $\mathrm{V}=100\angle 0^0\mathrm{V}$

The required circuit is shown below.

Draw the circuit diagram for series RLC circuit.

From equation (1) the impedance of the circuit is calculated as,

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{s}}=3\mathrm{\Omega }+\mathrm{j}8\mathrm{\Omega }-\mathrm{j}4\mathrm{\Omega } \\ =3+\mathrm{j}4 \end{gathered}$$

Write the above equation in the polar form as,

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{s}}=\sqrt{3^3+{\left(\mathrm{j}4\right)}^2}{\tan }^{-1}\left(\frac{3}{4}\right) \\ =5\angle 53.{13}^0\mathrm{\Omega } \end{gathered}$$

Thus, the value of the series impedance is $5\angle 53.{13}^0\mathrm{\Omega }$.

Solve for the current delivered by the source.

From equation (2) the current delivered by the source is calculated as,

$$\begin{gathered} \mathrm{I}=\frac{100\angle 0^0\mathrm{V}}{5\angle 53.{13}^0} \\ =20\angle -53.{13}^0\mathrm{A} \end{gathered}$$

Thus, the current delivered by the source is $20\angle -53.{13}^0\mathrm{A}$.

Solve for the power factor of the circuit.

From equation (2) the power factor of the circuit is calculated as,

$$\begin{gathered} \mathrm{PF}=\cos \left(53.13\right) \\ =0.6 \end{gathered}$$

Since, the phase angle is negative so the power factor is lagging.

Thus, the power factor of the circuit is 0.6 lagging.