Question

Remove the capacitor from the circuit in Figure and set $\mathit{R}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{\text{Ω}}\mathbf{,}\mathit{L}\mathbf{=}\mathbf{230}\mathbf{\text{mH}}\mathbf{,}$${\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 the Z? (b)What is the $\mathit{\varphi }$? (c)What is the 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 $218\text{Ω}$.
2. The value of phase angle is $23.41°$.
3. The value of current is $0.165\text{A}$.
4. The phasor diagram is drawn using the given data.

Step-by-step solution

The given data

1. Driving frequency,$f_d=60\text{Hz}$
2. Amplitude of emf,${\epsilon }_m=36.0\text{V}$
3. Resistance,$R=200\text{Ω}$
4. Inductance,$L=230\text{mH =}230\times {10}^{-3}\text{H}$

Understanding the concept of oscillations of the LCR circuit

An electrical circuit composed of a resistor, a capacitor, and an inductor, connected in series, is known as an LCR circuit. The given circuit will become a series LR circuit when we remove the capacitor. Therefore, the impedance will only depend on resistance and inductive reactance. Using their values, we can find the impedance and phase angle. Using impedance, we can find the amplitude of the current.

The impedance of theLRcircuit for the driving frequency $\left({\omega }_d\right)$,

$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{{\mathbf{X}}_{\mathbf{L}}}^{\mathbf{2}}}$ ...(1)

The inductive reactance of the inductor,

${\mathit{X}}_{\mathit{L}}\mathbf{=}{\mathit{\omega }}_{\mathit{d}}\mathit{L}$ ...(2)

The angular frequency of the LC oscillation,

${\mathit{\omega }}_{\mathit{d}}\mathbf{=}\mathbf{2}\mathit{\pi }{\mathit{f}}_{\mathit{d}}$ ...(3)

The phase angle of RLC circuit,

$\mathit{t}\mathit{a}\mathit{n}\varphi \mathbf{=}\left(\frac{{\mathit{X}}_{\mathit{L}}}{\mathit{R}}\right)\mathbf{}\mathbf{}$ ...(4)

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{\epsilon }}_{\mathit{m}}}{\mathit{Z}}$ ...(5)

Here,Ris the resistance of the resistor, Lis the inductance of the inductor and${\mathit{f}}_{\mathit{d}}$is the frequency of the LC oscillation.

a) Calculation of the impedance value

Driving angular frequency is given using equation (3) as:

$w_d=2\pi f_d$

For$f_d=60\text{Hz}$, we get-

$$\begin{gathered} {\omega }_d=2\pi \times 60\text{Hz} \\ =377.0\text{rad/s} \end{gathered}$$

Now, the inductive reactance is given using equation (2) as follows:

$X_L={\omega }_{d}L$

$$\begin{gathered} X_L=377.0\text{rad/s}\times 230\times {10}^{-3}\text{H} \\ =86.7\text{Ω} \end{gathered}$$

Thus, the impedance for LR circuit is given using equation (1) as follows: (for $X_c=0$)

$$\begin{gathered} Z=\sqrt{{\left(200\text{Ω}\right)}^2+{\left(86.7\text{Ω}\right)}^2} \\ =217.9\text{Ω} \\ \approx 218\text{Ω} \end{gathered}$$

Hence, the value of the impedance is $218\text{Ω}$.

b) Calculation of the phase angle

From part (a) calculations, we get that$X_c=0$

Now, the phase angle can be given using equation (4) as follows:

localid="1662809694382" $$\begin{gathered} \tan \varphi =\left(\frac{86.7\text{Ω}-0}{200\text{Ω}}\right) \\ =23.41° \end{gathered}$$

Hence, the value of the phase is $23.41°$.

c) Calculation of the current amplitude

The current amplitude is given using equation (5) as follows:

$$\begin{gathered} I=\frac{{\epsilon }_m}{Z} \\ =\frac{36.0\text{V}}{218} \\ =0.165\text{A} \end{gathered}$$

Hence, the value of the current is $0.165\text{A}$.

d) Calculation for the diagram of the phasor

The given circuit is purely inductive. Therefore, ${\epsilon }_m$leads $I$, and the phase angle is $\varphi =23.41°$

The phasor diagram is shown below.

Hence, the phasor diagram is plotted accordingly.