Remove the inductor from the circuit in Figure and set $\mathit{R}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{\text{Ω}}\mathbf{,}\mathbf{}\mathit{C}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{μF,}}$${\mathit{f}}_{\mathbf{d}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{Hz}}$and ${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{36}\mathbf{.}\mathbf{0}\mathbf{}\mathit{V}$.(a)What is the Z? (b)What is the $\mathit{\varphi }$?(c)What is the I?(d) Draw a phasor diagram.

1. Resistance of the circuit,$R=200\text{Ω}$
2. Capacitance of the circuit,$C=15\text{μF}$
3. Frequency value of the oscillation,$f_d=60\text{Hz}$
4. Amplitude of emf,${\epsilon }_m=36\text{V}$

A circuit composed of a resistor, inductor and a capacitor, connected in series is known as a RLC circuit. If we remove the inductor from the circuit, it will become a RC circuit. The resistance by the resistor and the reactance by the capacitor, together form the impedance of the circuit.

Formulae:

The impedance of thecircuit for the driving frequency,

$Z=\sqrt{R^2+{X_c}^2}$ ...(i)

The capacitive reactance of the circuit,

$X_C=\frac{1}{{\omega }_{d}C}$ ...(ii)

The angular frequency of the LC oscillation,

${\omega }_d=2\pi f_d$ ...(iii)

The phase angle of RLC circuit,

$\tan \varphi =\left(\frac{-X_C}{R}\right)$ ...(iv)

The current equation using Ohm’s law,

$I=\frac{{\epsilon }_m}{Z}$ ...(v)

Here, $f_d$is the driving frequency, $Z$is the impedance, $C$is the capacitance of the capacitor and${\omega }_d$is the driving angular frequency.

The value of the capacitive reactance can be given using equation (iii) in equation (ii) as follows:

$X_C=\frac{1}{2\pi f_{d}C}$ ...(a)

Now, using this value in equation (i), we can get the impedance value of the circuit as follows:

$$\begin{gathered} Z=\sqrt{{\left(R\right)}^2+{\left(\frac{1}{2\pi f_{d}C}\right)}^2} \\ =\sqrt{{\left(200\right)}^2+\frac{1}{{\left(2\pi \left(60\text{Hz}\right)\left(15\times {10}^{-6}\text{F}\right)\right)}^2}} \\ =266.97\text{Ω} \\ =267\text{Ω} \end{gathered}$$

Hence, the value of the impedance is $267\text{Ω}$.

Now, using the given data in equation (a), we can get the capacitive reactance as follows:

$$\begin{gathered} X_C=\frac{1}{2\pi \left(60\text{Hz}\right)\left(15\times {10}^{-6}\text{F}\right)} \\ =177\text{Ω} \end{gathered}$$

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

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{-177\text{Ω}}{200\text{Ω}}\right) \\ =-41.5° \end{gathered}$$

Hence, the value of the angle is $-41.5°$.

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

$$\begin{gathered} I=\frac{36\text{V}}{267\text{Ω}} \\ =0.135\text{A} \\ =135\text{mA} \end{gathered}$$

Hence, the value of the current is $135\text{mA}$.

The voltage amplitude across the resistor and capacitor is given using equation (v) as follows:

$$\begin{gathered} V_R=\left(0.135\text{A}\right)\left(200\text{Ω}\right) \\ =27\text{V} \end{gathered}$$

$$\begin{gathered} V_C=\left(0.135\text{A}\right)\left(177\text{Ω}\right) \\ =23.9\text{V} \end{gathered}$$

The given circuit is capacitive. In a capacitive circuit, the alternating current $I$leads the alternating potential difference ${\epsilon }_m$by $90°.$

Hence, the phasor diagram is plotted accordingly.