Question

An alternating emf source with a variable frequency ${\mathit{f}}_{\mathbf{d}}$ is connected in series with a$\mathit{R}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{Ω}}$ resistor and a$\mathit{C}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{\text{μF}}$ capacitor. The emf amplitude is${\mathbf{\in }}_{\mathbf{m}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{V}}$ . (a) Draw a phasor diagram for phasor${\mathit{V}}_{\mathbf{R}}$ (the potential across the resistor) and phasor ${\mathit{V}}_{\mathbf{C}}$(the potential across the capacitor). (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?

Short answer

1. Phasor diagram is shown below in the calculation section.
2. Driving frequency, when both the phasors have same length, is$f_d=159\text{Hz}$
3. Phase angle,when both the phasors have same length, is$\varphi =-45°$
4. Angular speed of rotation of phasor,when both the phasors have same length, is${\omega }_d=1.0\times {10}^3\text{rad/s}$
5. Current amplitude,when both the phasors have same length, is$I=170\text{mA}$

Step-by-step solution

Given Data

1. Resistance, $R=50.0\text{Ω}$
2. Capacitance,$C=20\text{μF}=20\times {10}^{-6}\text{F}$
3. Amplitude of emf, ${\in }_m=12.0\text{V}$

Concept

An electric circuit composed of a resistor and capacitor connected in series, is known as a RC circuit. By equating the phasor lengths for ${\mathit{V}}_{\mathit{R}}$and ${\mathit{V}}_{\mathit{c}}$, we can find thedriving frequency. By using the formula for phase angle in RCcircuit, we can find the phase angle corresponding to the driving frequency. We can find the angular speed of rotation of phasor by substituting the driving frequency in the formula for angular speed. We can find the current amplitude I in a series RCcircuit by using its formula.

The angular frequency,

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

The capacitive reactance of the capacitor,

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

The current equation using Ohm’s law,

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

The impedance of theLCRcircuit for the driving frequency,

$\mathit{Z}\mathbf{=}\sqrt{{\mathit{R}}^{\mathbf{2}}\mathbf{+}{\left({\mathit{X}}_{\mathit{c}}\right)}^{\mathbf{2}}}$ …(4)

Phase angle for series circuit,

$\varphi \mathbf{=}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{-}\frac{{\mathbf{\chi }}_{\mathbf{c}}}{\mathbf{R}}\right)\mathbf{}$ …(5)

Here, R is the resistance of the resistor, Cis the capacitance of the capacitor, ${\mathit{\epsilon }}_{\mathit{m}}$is the emf in the circuit and ${\mathit{f}}_{\mathit{d}}$is the frequency.

Calculations

(a) The phasor diagram for $V_R$ and $V_C$is shown below:

$V_C$lags behind $V_R$ by the phase angle $\frac{\pi }{2}\text{rad}$

(b) We have to find the driving frequency at which $V_R$and $V_C$have the same length or amplitude. Therefore,

$V_R=V_C$ ...(6)

Thus,using equations (1), (2) and (6), we get-

$$\begin{gathered} R=I{\chi }_C \\ IR=\frac{I}{{\omega }_{d}C} \\ IR=\frac{I}{2\pi f_{d}C} \end{gathered}$$

Rearranging the terms,

$f_d=\frac{1}{2\pi RC}$

For the given values, we have-

$$\begin{gathered} f_d=\frac{1}{2\pi \times \left(50\text{Ω}\right)\times \left(20\times {10}^{-6}\text{F}\right)} \\ {f}_d=159\text{Hz} \end{gathered}$$

The driving frequency of two phasors is $f_d=159\text{Hz}$.

(c) Phase angle for series RC circuit is given by

$\varphi ={\tan }^{-1}\left(-\frac{V_C}{V_R}\right)$

But we have, $V_R=V_C$, hence we get,

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

The phase angle is given by $\varphi =-45°$

(d) Angular speed of rotation of phasor is

$$\begin{gathered} {\omega }_d=2\pi f_d \\ {\omega }_d=2\pi \times 159\text{Hz} \\ {\omega }_d=999.03\text{rad/s} \\ \approx 1.0\times {10}^3\text{rad/s} \end{gathered}$$

(e) Current amplitude in a RC circuit is given by

$$\begin{gathered} \text{I=}\frac{{\in }_m}{Z} \\ =\frac{{\in }_m}{\sqrt{R^2+{\chi }_C^2}} \end{gathered}$$

Since$R={\chi }_C$,

$\therefore I=\frac{{\in }_m}{\sqrt{2{\chi }_C^2}}=\frac{{\in }_m}{\sqrt{2}·{\chi }_c}=\frac{{\in }_m{\omega }_{d}C}{\sqrt{2}}$

$$\begin{gathered} I=\frac{\left(\left(12.0V\right)\times \left(1.0\times {10}^3\text{rad/s}\right)\times \left(20\times {10}^{-6}\text{F}\right)\right)}{\sqrt{2}} \\ =170\text{mA} \end{gathered}$$

Maximum value of current in RC circuit is $170\text{mA}$.