A series LCR circuit contains an inductor of inductance $L = 5.0 ext{ H}$, a capacitor of capacitance $C = 80\ \mu ext{F}$, and a resistor of resistance $R = 40\ \Omega$ connected to a variable frequency $230 ext{ V}$ AC supply.
Calculate: The resonant angular frequency ($\omegar$) of the circuit. The impedance of the circuit and the amplitude (peak value) of current at resonance. The RMS potential drop across the inductor, capacitor, and resistor at resonance. Show that the potential drop across the $LC$ combination at resonance is zero.

Q: A series LCR circuit contains an inductor of inductance $L = 5.0\text{ H}$, a capacitor of capacitance $C = 80\ \mu\text{F}$, and a resistor of resistance $R = 40\ \Omega$ connected to a variable frequency $230\text{ V}$ AC supply. \nCalculate: The resonant angular frequency ($\omegar$) of the circuit. The impedance of the circuit and the amplitude (peak value) of current at resonance. The RMS potential drop across the inductor, capacitor, and resistor at resonance. Show that the potential drop across the $LC$ combination at resonance is zero.

Given Data: Inductance, $L = 5.0\text{ H}$ Capacitance, $C = 80\ \mu\text{F} = 80 \times 10^{-6}\text{ F}$ Resistance, $R = 40\ \Omega$ Supply RMS Voltage, $V{\text{rms}} = 230\text{ V}$ --- Step-by-Step Solution: Resonant Angular Frequency ($\omegar$): $$\omegar = \frac{1}{\sqrt{LC}}$$ $$\omegar = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{20 \times 10^{-3}}$$ $$\omegar = \frac{1000}{20} = 50\text{ rad/s}$$ --- Impedance ($Z$) and Peak Current ($I0$) at Resonance:\nAt resonance, inductive reactance equals capacitive reactance ($XL = XC$). Impedance ($Z$): $$Z = R = 40\ \Omega$$ RMS Current ($I{\text{rms}}$): $$I{\text{rms}} = \frac{V{\text{rms}}}{Z} = \frac{230}{40} = 5.75\text{ A}$$ Peak Current ($I0$): $$I0 = \sqrt{2} \times I{\text{rms}} = 1.414 \times 5.75 \approx 8.13\text{ A}$$ --- RMS Potential Drop Across Components at Resonance: Reactance Values at Resonance: $$XL = \omegar L = 50 \times 5.0 = 250\ \Omega$$ $$XC = \frac{1}{\omegar C} = \frac{1}{50 \times 80 \times 10^{-6}} = 250\ \Omega$$ Potential drop across Resistor ($VR$): $$VR = I{\text{rms}} \times R = 5.75 \times 40 = 230\text{ V}$$ Potential drop across Inductor ($VL$): $$VL = I{\text{rms}} \times XL = 5.75 \times 250 = 1437.5\text{ V}$$ Potential drop across Capacitor ($VC$): $$VC = I{\text{rms}} \times XC = 5.75 \times 250 = 1437.5\text{ V}$$ --- Potential Drop Across $LC$ Combination:\nSince $VL$ leads the current by $90^\circ$ and $VC$ lags the current by $90^\circ$, they are out of phase by $180^\circ$. $$V{LC} = |VL - VC|$$ $$V{LC} = 1437.5\text{ V} - 1437.5\text{ V} = 0\text{ V}$$ Conclusion: Hence, at resonance, the potential drop across the $LC$ combination is zero.

Question

A series LCR circuit contains an inductor of inductance $L = 5.0	ext{ H}$, a capacitor of capacitance $C = 80\ \mu	ext{F}$, and a resistor of resistance $R = 40\ \Omega$ connected to a variable frequency $230	ext{ V}$ AC supply. 
Calculate: The resonant angular frequency ($\omegar$) of the circuit. The impedance of the circuit and the amplitude (peak value) of current at resonance. The RMS potential drop across the inductor, capacitor, and resistor at resonance. Show that the potential drop across the $LC$ combination at resonance is zero.

Accepted Answer

Given Data: Inductance, $L = 5.0	ext{ H}$ Capacitance, $C = 80\ \mu	ext{F} = 80 	imes 10^{-6}	ext{ F}$ Resistance, $R = 40\ \Omega$ Supply RMS Voltage, $V{	ext{rms}} = 230	ext{ V}$ --- Step-by-Step Solution: Resonant Angular Frequency ($\omegar$): $$\omegar = \frac{1}{\sqrt{LC}}$$ $$\omegar = \frac{1}{\sqrt{5.0 	imes 80 	imes 10^{-6}}} = \frac{1}{\sqrt{400 	imes 10^{-6}}} = \frac{1}{20 	imes 10^{-3}}$$ $$\omegar = \frac{1000}{20} = 50	ext{ rad/s}$$ --- Impedance ($Z$) and Peak Current ($I0$) at Resonance:
At resonance, inductive reactance equals capacitive reactance ($XL = XC$). Impedance ($Z$): $$Z = R = 40\ \Omega$$ RMS Current ($I{	ext{rms}}$): $$I{	ext{rms}} = \frac{V{	ext{rms}}}{Z} = \frac{230}{40} = 5.75	ext{ A}$$ Peak Current ($I0$): $$I0 = \sqrt{2} 	imes I{	ext{rms}} = 1.414 	imes 5.75 \approx 8.13	ext{ A}$$ --- RMS Potential Drop Across Components at Resonance: Reactance Values at Resonance: $$XL = \omegar L = 50 	imes 5.0 = 250\ \Omega$$ $$XC = \frac{1}{\omegar C} = \frac{1}{50 	imes 80 	imes 10^{-6}} = 250\ \Omega$$ Potential drop across Resistor ($VR$): $$VR = I{	ext{rms}} 	imes R = 5.75 	imes 40 = 230	ext{ V}$$ Potential drop across Inductor ($VL$): $$VL = I{	ext{rms}} 	imes XL = 5.75 	imes 250 = 1437.5	ext{ V}$$ Potential drop across Capacitor ($VC$): $$VC = I{	ext{rms}} 	imes XC = 5.75 	imes 250 = 1437.5	ext{ V}$$ --- Potential Drop Across $LC$ Combination:
Since $VL$ leads the current by $90^\circ$ and $VC$ lags the current by $90^\circ$, they are out of phase by $180^\circ$. $$V{LC} = |VL - VC|$$ $$V{LC} = 1437.5	ext{ V} - 1437.5	ext{ V} = 0	ext{ V}$$ Conclusion: Hence, at resonance, the potential drop across the $LC$ combination is zero.

Given Data:

Step-by-Step Solution:

1. Resonant Angular Frequency ($\omega_r$):

2. Impedance ($Z$) and Peak Current ($I_0$) at Resonance:\nAt resonance, inductive reactance equals capacitive reactance ($X_L = X_C$).

3. RMS Potential Drop Across Components at Resonance:

4. Potential Drop Across $LC$ Combination:\nSince $V_L$ leads the current by $90^\circ$ and $V_C$ lags the current by $90^\circ$, they are out of phase by $180^\circ$.