91Ó°ÊÓ

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

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 $\left({\mathrm{Ó\neg }}_d\right)$,

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

The inductive reactance of the inductor,

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

The angular frequency of the LC oscillation,

${\mathit{Ó\neg }}_{\mathit{d}}\mathbf{=}\mathbf{2}\mathit{Ï€}{\mathit{f}}_{\mathit{d}}$ ...(3)

The phase angle of RLC circuit,

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

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{Î\mu }}_{\mathit{m}}}{\mathit{Z}}$ ...(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.

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

$w_d=2\mathrm{Ï€}{f}_d$

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

$$\begin{gathered} {\mathrm{Ó\neg }}_d=2\mathrm{Ï€}×60\text{Hz} \\ =377.0\text{rad/s} \end{gathered}$$

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

$X_L={\mathrm{Ó\neg }}_{d}L$

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

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

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

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

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

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

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

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

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

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

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

The given circuit is purely inductive. Therefore, ${\mathrm{Î\mu }}_m$leads $I$, and the phase angle is $\mathrm{Ï•}=23.41°$

The phasor diagram is shown below.

Hence, the phasor diagram is plotted accordingly.