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Electromagnetic Induction and Alternating Currents - Alternating Current (LCR Circuits, Resonance)

Grade 12ICSEPhysics

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

πŸ”‘Concepts

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Alternating Current (AC) is the current whose magnitude changes continuously with time and direction reverses periodically, represented as I=I0sin⁑ωtI = I_0 \sin \omega t.

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RMS (Root Mean Square) value, also known as the effective or virtual value, is given by Irms=I02β‰ˆ0.707I0I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0.

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Phasors are rotating vectors used to represent sinusoidally varying voltages and currents in AC circuits to show phase relationships.

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Inductive Reactance (XLX_L) is the opposition offered by an inductor to AC, where XL=Ο‰LX_L = \omega L. It increases with frequency.

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Capacitive Reactance (XCX_C) is the opposition offered by a capacitor to AC, where XC=1Ο‰CX_C = \frac{1}{\omega C}. It decreases as frequency increases.

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In a series LCRLCR circuit, the total opposition to current is called Impedance (ZZ), which is the phasor sum of RR, XLX_L, and XCX_C.

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Resonance occurs in a series LCRLCR circuit when XL=XCX_L = X_C. At this condition, the impedance is minimum (Z=RZ = R) and the current is maximum.

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The Quality Factor (QQ-factor) measures the sharpness of resonance. A higher QQ indicates a narrower and sharper resonance peak.

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Power Factor (cos⁑ϕ\cos \phi) is the ratio of real power to apparent power. In a purely resistive circuit cos⁑ϕ=1\cos \phi = 1, while in purely inductive or capacitive circuits, cos⁑ϕ=0\cos \phi = 0.

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Wattless Current is the component of AC that consumes no power in the circuit, occurring when the phase difference between voltage and current is Ο€2\frac{\pi}{2}.

πŸ“Formulae

V=V0sin⁑ωtV = V_0 \sin \omega t

Irms=I02,Vrms=V02I_{rms} = \frac{I_0}{\sqrt{2}}, \quad V_{rms} = \frac{V_0}{\sqrt{2}}

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€…β€ŠβŸΉβ€…β€Šfr=12Ο€LC\omega_r = \frac{1}{\sqrt{LC}} \implies f_r = \frac{1}{2\pi \sqrt{LC}}

Q=Ο‰rLR=1RLCQ = \frac{\omega_r L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

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

PowerΒ Factor=cos⁑ϕ=RZ\text{Power Factor} = \cos \phi = \frac{R}{Z}

πŸ’‘Examples

Problem 1:

A series LCRLCR circuit contains a resistor of R=10 ΩR = 10 \, \Omega, an inductor of L=2 HL = 2 \, H, and a capacitor of C=18 μFC = 18 \, \mu F. Calculate the resonant frequency frf_r of the circuit.

Solution:

Given: L=2 HL = 2 \, H, C=18Γ—10βˆ’6 FC = 18 \times 10^{-6} \, F. Using the formula fr=12Ο€LCf_r = \frac{1}{2\pi \sqrt{LC}}: fr=12Γ—3.14Γ—2Γ—18Γ—10βˆ’6f_r = \frac{1}{2 \times 3.14 \times \sqrt{2 \times 18 \times 10^{-6}}} fr=16.28Γ—36Γ—10βˆ’6f_r = \frac{1}{6.28 \times \sqrt{36 \times 10^{-6}}} fr=16.28Γ—6Γ—10βˆ’3f_r = \frac{1}{6.28 \times 6 \times 10^{-3}} fr=100037.68β‰ˆ26.54 Hzf_r = \frac{1000}{37.68} \approx 26.54 \, Hz.

Explanation:

Resonant frequency is the frequency at which the inductive reactance equals capacitive reactance, causing the circuit to behave as purely resistive.

Problem 2:

In an LCRLCR circuit, R=30 ΩR = 30 \, \Omega, XL=80 ΩX_L = 80 \, \Omega, and XC=40 ΩX_C = 40 \, \Omega. Find the impedance ZZ and the power factor.

Solution:

Z=R2+(XLβˆ’XC)2Z = \sqrt{R^2 + (X_L - X_C)^2} Z=302+(80βˆ’40)2=302+402Z = \sqrt{30^2 + (80 - 40)^2} = \sqrt{30^2 + 40^2} Z=900+1600=2500=50 ΩZ = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega Power factor cos⁑ϕ=RZ=3050=0.6\cos \phi = \frac{R}{Z} = \frac{30}{50} = 0.6.

Explanation:

The impedance is the total resistance of the LCRLCR circuit. The power factor is the cosine of the phase angle, indicating how much of the supplied power is actually dissipated as heat.

Alternating Current (LCR Circuits, Resonance) Revision - Class 12 Physics ICSE