(a) Explain the working of a series $LCR$ alternating current circuit. Derive the expressions for the impedance ($Z$), phase angle ($\phi$), and resonant frequency ($fr$) using the phasor diagram method. (b) A series $LCR$ circuit consists of a resistor $R = 10\ \Omega$, an inductor $L = 20 ext{ mH}$, and a capacitor $C = 100\ \mu ext{F}$ connected across an AC voltage source given by $V = 200 \sin(100\pi t) ext{ V}$. Calculate: The resonant frequency of the circuit. The impedance of the circuit at resonance. The peak current ($I0$) flowing through the circuit at resonance.

Question

(a) Explain the working of a series $LCR$ alternating current circuit. Derive the expressions for the impedance ($Z$), phase angle ($\phi$), and resonant frequency ($fr$) using the phasor diagram method. (b) A series $LCR$ circuit consists of a resistor $R = 10\ \Omega$, an inductor $L = 20	ext{ mH}$, and a capacitor $C = 100\ \mu	ext{F}$ connected across an AC voltage source given by $V = 200 \sin(100\pi t)	ext{ V}$. Calculate: The resonant frequency of the circuit. The impedance of the circuit at resonance. The peak current ($I0$) flowing through the circuit at resonance.

Accepted Answer

Part (a): Derivation for Series $LCR$ Circuit Introduction and Circuit Description
In a series $LCR$ circuit, a resistor ($R$), an inductor ($L$), and a capacitor ($C$) are connected in series across an alternating voltage source given by: $$V = V0 \sin(\omega t)$$
Since all components are connected in series, the instantaneous current $I$ passing through each component is the same. --- Voltage Relations and Phasor Diagram
Let the current in the circuit be represented as the reference phasor along the X-axis. Voltage across Resistor ($VR$): $VR = I \cdot R$. It is in the same phase as the current $I$. Voltage across Inductor ($VL$): $VL = I \cdot XL$ (where $XL = \omega L$). It leads the current $I$ by a phase angle of $\pi/2$ ($90^\circ$). Voltage across Capacitor ($VC$): $VC = I \cdot XC$ (where $XC = \frac{1}{\omega C}$). It lags behind the current $I$ by a phase angle of $\pi/2$ ($90^\circ$). 
Since $VL$ and $VC$ are in opposite directions (phase difference of $180^\circ$), the net reactive voltage is $(VL - VC)$ (assuming $VL > VC$). --- Derivation of Impedance ($Z$)
Using the Pythagorean theorem on the phasor right-angled triangle formed by $VR$, $(VL - VC)$, and the resultant voltage $V$: $$V^2 = VR^2 + (VL - VC)^2$$ 
Substitute $VR = I R$, $VL = I XL$, and $VC = I XC$: $$V^2 = (I R)^2 + (I XL - I XC)^2$$ $$V^2 = I^2 \left R^2 + (XL - XC)^2 ight$$ $$V = I \sqrt{R^2 + (XL - XC)^2}$$ 
The total effective resistance offered by the $LCR$ circuit is called Impedance ($Z$): $$Z = \frac{V}{I} = \sqrt{R^2 + (XL - XC)^2}$$ $$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}ight)^2}$$ --- Phase Angle ($\phi$)
The angle $\phi$ by which the applied voltage leads or lags the current is given by: $$	an \phi = \frac{VL - VC}{VR} = \frac{I XL - I XC}{I R} = \frac{XL - XC}{R}$$ $$\phi = 	an^{-1}\left( \frac{\omega L - \frac{1}{\omega C}}{R} ight)$$ --- Resonant Frequency ($fr$)
Electrical resonance occurs when the capacitive reactance equals the inductive reactance ($XL = XC$), making the impedance minimum ($Z = R$) and current maximum. $$\omegar L = \frac{1}{\omegar C}$$ $$\omegar^2 = \frac{1}{L C} \implies \omegar = \frac{1}{\sqrt{L C}}$$ 
Since $\omegar = 2\pi fr$: $$fr = \frac{1}{2\pi \sqrt{L C}}$$ --- Part (b): Numerical Solution Given Data: Resistance, $R = 10\ \Omega$ Inductance, $L = 20	ext{ mH} = 20 	imes 10^{-3}	ext{ H} = 0.02	ext{ H}$ Capacitance, $C = 100\ \mu	ext{F} = 100 	imes 10^{-6}	ext{ F} = 10^{-4}	ext{ F}$ Peak voltage, $V0 = 200	ext{ V}$ Resonant Frequency ($fr$): $$fr = \frac{1}{2\pi \sqrt{L C}}$$ $$fr = \frac{1}{2\pi \sqrt{0.02 	imes 10^{-4}}} = \frac{1}{2\pi \sqrt{2 	imes 10^{-6}}} = \frac{1}{2\pi 	imes 10^{-3} 	imes \sqrt{2}}$$ $$fr = \frac{1000}{2 	imes 3.1416 	imes 1.414} = \frac{1000}{8.886} \approx 112.54	ext{ Hz}$$ Impedance at Resonance ($Z{res}$):
At resonance, $XL = XC$, so: $$Z{res} = R = 10\ \Omega$$ Peak Current at Resonance ($I0$): $$I0 = \frac{V0}{Z{res}} = \frac{200	ext{ V}}{10\ \Omega} = 20	ext{ A}$$

Part (a): Derivation for Series $LCR$ Circuit

1. Introduction and Circuit Description\nIn a series $LCR$ circuit, a resistor ($R$), an inductor ($L$), and a capacitor ($C$) are connected in series across an alternating voltage source given by:

2. Voltage Relations and Phasor Diagram\nLet the current in the circuit be represented as the reference phasor along the X-axis.

3. Derivation of Impedance ($Z$)\nUsing the Pythagorean theorem on the phasor right-angled triangle formed by $V_R$, $(V_L - V_C)$, and the resultant voltage $V$:

4. Phase Angle ($\phi$)\nThe angle $\phi$ by which the applied voltage leads or lags the current is given by:

5. Resonant Frequency ($f_r$)\nElectrical resonance occurs when the capacitive reactance equals the inductive reactance ($X_L = X_C$), making the impedance minimum ($Z = R$) and current maximum.

Part (b): Numerical Solution

1. Resonant Frequency ($f_r$):

2. Impedance at Resonance ($Z_{res}$):\nAt resonance, $X_L = X_C$, so:

3. Peak Current at Resonance ($I_0$):