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Alternating Current - LCR Series Circuit

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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An LCR series circuit consists of a resistor (RR), an inductor (LL), and a capacitor (CC) connected in series to an alternating voltage source V=V0sin⁑(Ο‰t)V = V_0 \sin(\omega t).

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The total opposition to the flow of current in an AC circuit is called Impedance (ZZ). It is the vector sum of resistance (RR) and net reactance (XLβˆ’XCX_L - X_C).

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The phase angle Ο•\phi represents the phase difference between the current (II) and the voltage (VV). If XL>XCX_L > X_C, voltage leads the current; if XC>XLX_C > X_L, current leads the voltage.

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Electrical Resonance occurs in an LCR circuit when inductive reactance equals capacitive reactance (XL=XCX_L = X_C). At this state, the impedance is minimum (Z=RZ = R) and the current is maximum.

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The Quality Factor (QQ) describes the sharpness of the resonance peak. A higher QQ indicates a sharper resonance and lower energy loss relative to the stored energy.

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The power factor is given by cos⁑ϕ=RZ\cos \phi = \frac{R}{Z}. At resonance, cos⁑ϕ=1\cos \phi = 1 (purely resistive), and power dissipation is maximum.

πŸ“Formulae

V=V0sin⁑(Ο‰t)V = V_0 \sin(\omega t)

XL=Ο‰L=2Ο€fLX_L = \omega L = 2\pi f L

XC=1Ο‰C=12Ο€fCX_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

Z=R2+(XLβˆ’XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

tan⁑ϕ=XLβˆ’XCR\tan \phi = \frac{X_L - X_C}{R}

Ο‰r=1LCΒ orΒ fr=12Ο€LC\omega_r = \frac{1}{\sqrt{LC}} \text{ or } f_r = \frac{1}{2\pi\sqrt{LC}}

Q=1RLC=Ο‰rLRQ = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{\omega_r L}{R}

Pavg=VrmsIrmscos⁑ϕP_{avg} = V_{rms} I_{rms} \cos \phi

πŸ’‘Examples

Problem 1:

In a series LCR circuit, R=20Ξ©R = 20 \Omega, L=1.5Β HL = 1.5 \text{ H}, and C=35ΞΌFC = 35 \mu\text{F} are connected to a 220Β V220 \text{ V}, 50Β Hz50 \text{ Hz} AC source. Calculate (i) the reactance of the circuit and (ii) the impedance.

Solution:

Given: f=50Β Hzf = 50 \text{ Hz}, L=1.5Β HL = 1.5 \text{ H}, C=35Γ—10βˆ’6Β FC = 35 \times 10^{-6} \text{ F}, R=20Ξ©R = 20 \Omega.

  1. Inductive Reactance: XL=2Ο€fL=2Γ—3.14Γ—50Γ—1.5=471Ξ©X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 1.5 = 471 \Omega.
  2. Capacitive Reactance: XC=12Ο€fC=12Γ—3.14Γ—50Γ—35Γ—10βˆ’6β‰ˆ90.95Ξ©X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 35 \times 10^{-6}} \approx 90.95 \Omega.
  3. Net Reactance: X=XLβˆ’XC=471βˆ’90.95=380.05Ξ©X = X_L - X_C = 471 - 90.95 = 380.05 \Omega.
  4. Impedance: Z=R2+(XLβˆ’XC)2=202+(380.05)2β‰ˆ380.57Ξ©Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (380.05)^2} \approx 380.57 \Omega.

Explanation:

We first calculate the individual reactances using the frequency provided. Since XL>XCX_L > X_C, the circuit is predominantly inductive. The impedance is then found using the Pythagorean relationship between resistance and net reactance.

Problem 2:

Calculate the resonant frequency and the Quality factor (QQ) of a series LCR circuit with L=2.0Β HL = 2.0 \text{ H}, C=32ΞΌFC = 32 \mu\text{F}, and R=10Ξ©R = 10 \Omega.

Solution:

Given: L=2.0Β HL = 2.0 \text{ H}, C=32Γ—10βˆ’6Β FC = 32 \times 10^{-6} \text{ F}, R=10Ξ©R = 10 \Omega.

  1. Resonant angular frequency: Ο‰r=1LC=12Γ—32Γ—10βˆ’6=164Γ—10βˆ’6=18Γ—10βˆ’3=125Β rad/s\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 32 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125 \text{ rad/s}.
  2. Quality factor: Q=Ο‰rLR=125Γ—210=25010=25Q = \frac{\omega_r L}{R} = \frac{125 \times 2}{10} = \frac{250}{10} = 25.

Explanation:

The resonant frequency is the frequency at which XL=XCX_L = X_C, making the circuit purely resistive. The QQ factor is a dimensionless quantity that characterizes the circuit's bandwidth and damping.

LCR Series Circuit - Revision Notes & Key Formulas | CBSE Class 12 Physics