An alternating emf source with a variable frequency ${\mathit{f}}_{\mathbf{d}}$ is connected in series with a$\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{Ω}}$resistor and an $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{mH}}$inductor. The emf amplitude is$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{V}}$. (a) Draw a phasor diagram for phasor ${\mathit{V}}_{\mathbf{R}}$(the potential across the resistor) and phasor ${\mathit{V}}_{\mathbf{L}}$(the potential across the inductor). (b) At what driving frequency ${\mathit{f}}_{\mathbf{d}}$do the two phasors have the same length? At that driving frequency, what are (c) the phase angle in degrees, (d) the angular speed at which the phasors rotate, and (e) the current amplitude?

1. Resistance,$R=80.0\text{Ω}$
2. Inductance,$L=40.0\text{mH}=40.0\times {10}^{-3}\text{H}$
3. Amplitude of emf,${\epsilon }_m=6.0\text{V}$

The RL circuit consists of a combination of resistors and inductors. We can find the required quantities by using the corresponding formulae for the RL circuit and substituting the given values.

The angular frequency of the LC oscillation,

${\mathit{\omega }}_{\mathit{d}}\mathbf{=}\mathbf{2}\mathit{\pi }{\mathit{f}}_{\mathit{d}}$ …(1)

The inductive reactance of the inductor,

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

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{\epsilon }}_{\mathit{m}}}{\mathit{Z}}$ …(3)

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

$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{{\mathbf{X}}_{\mathbf{L}}}^{\mathbf{2}}}$ …(4)

The phase angle of RLC circuit,

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

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

Phasor diagram for$V_R$and$V_L$ is shown below.

Here, $V_R$lags behind $V_L$ by phase angle $\frac{\pi }{2}\text{rad}$.

Hence, the phasor diagram is drawn.

Now we have to find the driving frequency at which $V_R$ and $V_L$ have the same length or amplitude, i.e., $V_R=V_L$.

Thus, using equation (3), we can get the above equation reduced to resistance as follows:

$$\begin{gathered} IR=I{X}_L \\ IR=I{\omega }_{d}L \\ R={\omega }_{d}L \end{gathered}$$

Substituting the above equation in equation (1), we get the frequency of the oscillation as follows:

$$\begin{gathered} f_d=\frac{R}{2\pi L} \\ =\frac{80.0\text{Ω}}{2\pi \times \left(40.0\times {10}^{-3}\right)\text{H}} \\ =\left(\frac{1}{\pi \times {10}^{-3}}\right)\text{Hz} \\ =318\text{Hz} \end{gathered}$$

Hence, the value of the frequency is $318\text{Hz}$.

The phase angle for series RL circuit is given using equation (3) in equation (4) as follows:

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{I{X}_L}{IR}\right) \\ ={\tan }^{-1}\left(1\right) \\ =\frac{\pi }{4}\text{rad} \\ =45° \end{gathered}$$

Hence, the value of the phase is $45°$.

Angular speed of phasor rotation is given using equation (2) as follows: (for resonance condition $X_L=R$)

$$\begin{gathered} {\omega }_d=\frac{R}{L} \\ =\frac{80.0\text{Ω}}{\left(40.0\times {10}^{-3}\right)\text{H}} \\ =2.0\times {10}^3\text{rad/s} \end{gathered}$$

Hence, the value of the angular speed is $2.0\times {10}^3\text{rad/s}$.

Impedance for series RL circuit is given by equation (4) as follows: (for resonance condition $X_L=R$)

$$\begin{gathered} Z=\sqrt{2}·R\text{Ω} \\ =\sqrt{2}·80.0\text{Ω} \\ =113.13\text{Ω} \end{gathered}$$

Thus, the current amplitude is given using equation (3) as follows:

$$\begin{gathered} I=\frac{6.0\text{V}}{113.13\text{Ω}} \\ =0.0530\text{A} \\ =53\text{mA} \end{gathered}$$

Hence, the value of the current is $53\text{mA}$.