Question

In Figure, set $\mathit{R}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{\text{Ω}}$, $\mathit{C}\mathbf{=}\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{μF}}$, $\mathit{L}\mathbf{=}\mathbf{230}\mathbf{}\mathbf{\text{mH}}$, ${\mathit{f}}_{\mathbf{d}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{Hz}}$, and ${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{36}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{V}}$. (a) What is Z? (b) What is $\mathit{\varphi }$?(c) What is I?(d) Draw a phasor diagram.

Fig. A single-loop circuit containing a resistor, a capacitor, and an inductor. A generator, represented by a sine wave in a circle, produces an alternating emf that establishes an alternating current; the directions of the emf and current are indicated here at only one instant.

Short answer

1. The value of the impedance is $206.0\text{Ω}$.
2. The value of phase angle is $13.7°$.
3. The value of current is $175\text{mA}$.
4. The phasor diagram is drawn using the given data.

Step-by-step solution

The given data

1. Resistance,$R=200\text{Ω}$
2. Capacitance,$C=70.0\text{μF}=70.0\times {10}^{-6}\text{F}$
3. Inductance,$L=230\text{mH}=230\times {10}^{-3}\text{H}$
4. Driving frequency,$f_d=60.0\text{Hz}$
5. Amplitude of emf${\epsilon }_m=36.\text{0}\text{V}$

Understanding the concept of oscillations of RLC circuit

An electrical circuit composed of a capacitor, a resistor and an inductor, connected in series, is known as anLCRcircuit. For a series LCR circuit, we have the relations mentioned below, to find out the inductive and capacitive reactance. By using these relations, we can find out the impedance of the circuit, phase angle, and the amplitude of the current.

The impedance of theLCRcircuit for the driving frequency,

$\mathit{Z}\mathbf{=}\sqrt{{\mathit{R}}^{\mathbf{2}}\mathbf{+}{\left({\mathit{X}}_{\mathit{L}}\mathbf{-}{\mathit{X}}_{\mathit{c}}\right)}^{\mathbf{2}}}$ ...(1)

The capacitive reactance of the capacitor,

${\mathit{X}}_{\mathit{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathit{\omega }}_{\mathit{d}}\mathit{C}}$ ...(2)

The inductive reactance of the inductor,

${\mathit{X}}_{\mathit{L}}\mathbf{=}{\mathit{\omega }}_{\mathit{d}}\mathit{L}$ ...(3)

The angular frequency of the LC oscillation,

${\mathit{\omega }}_{\mathit{d}}\mathbf{=}\mathbf{2}\mathit{\pi }{\mathit{f}}_{\mathit{d}}$ ...(4)

The phase angle of RLC circuit,

$\mathit{t}\mathit{a}\mathit{n}\varphi \mathbf{=}\left(\frac{{\mathit{X}}_{\mathit{L}}\mathbf{-}{\mathit{X}}_{\mathit{C}}}{\mathit{R}}\right)\mathbf{}$ ...(5)

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{\epsilon }}_{\mathit{m}}}{\mathit{Z}}$ ...(6)

Here, $\mathit{R}$is the resistance of the resistor,$\mathit{C}$is the capacitance of the capacitor, $\mathit{L}$is the inductance of the inductor and ${\mathit{f}}_{\mathit{d}}$is the frequency of the LC oscillation.

a) Calculation of the impedance value

First, the driving angular frequency is calculated using equation (4) as follows:

$$\begin{gathered} {\omega }_d=\left(2\times 3.14\right)\text{rad}\times \left(60.0\text{Hz}\right) \\ =376.8\text{rad/s} \end{gathered}$$

Capacitive reactance is given using equation (2) as:

$$\begin{gathered} X_c=\frac{1}{\left(376.8\text{rad/s}\right)\times \left(70.0\times {10}^{-6}\text{F}\right)} \\ =37.9\text{Ω} \end{gathered}$$

Inductive reactance is given using equation (3) as:

$$\begin{gathered} X_L=376.8\text{rad/s}\times \left(230\times {10}^{-3}\right)\text{H} \\ =86.7\text{Ω} \end{gathered}$$

Thus, the impedance of the RLC circuit is given using the above values and given data in equation (1) as follows:

$$\begin{gathered} Z=\sqrt{{\left(200\text{Ω}\right)}^2+{\left(86.7 \text{Ω}-37.9\text{Ω}\right)}^2} \\ =\sqrt{40000{\text{Ω}}^2+2381.5{\text{Ω}}^2} \\ =205.86\text{Ω} \\ \approx 206.0\text{Ω} \end{gathered}$$

Hence, the value of the impedance is $206.0\text{Ω}$.

b) Calculation of the phase angle

Now, using the calculated values from part (a) in equation (5), we can get the phase angle of the RLC circuit as follows:

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{86.7\text{Ω}-37.9\text{Ω}}{200\text{Ω}}\right) \\ ={\tan }^{-1}\left(0.244\right) \\ =13.7° \end{gathered}$$

Hence, the value of the phase is role="math" localid="1662801363373" $13.7°$.

c) Calculation of the current amplitude

Amplitude of current is calculated using the given data and impedance value in equation (6) as follows:

$$\begin{gathered} I=\frac{{\epsilon }_m}{Z} \\ =\frac{36.0\text{V}}{206\text{Ω}} \\ =0.175\text{A} \end{gathered}$$

Hence, the value of the current is $0.175\text{A}$.

d) Calculation for the diagram of the phasor

From part (a), we have,$X_L>X_C$

Phase angle, $\varphi =13.70°>0,$i.e., positive.

The circuit is more inductive than capacitive, so the corresponding phasor diagram is shown below.

Hence, the phasor diagram is plotted.