Physics Notes Class 11th 12th

Here's a compiled list of all the important mathematics (formulas and derivations) from Chapter 8: Electromagnetic Waves (Class 12 Physics, Part 4) that are especially important for NEET and JEE: 📘 Chapter 8: Electromagnetic Waves – Important Maths (NEET & JEE) 🔶 1. Speed of Electromagnetic Wave in Vacuum c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} Where: c=3×108 m/sc = 3 \times 10^8 \, \text{m/s} μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \, \text{H/m} (permeability of free space) ε0=8.854×10−12 F/m\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} (permittivity of free space) 🔶 2. Electromagnetic Wave Equations (in Free Space) Electric field: E⃗(x,t)=E0sin⁡(kx−ωt)j^\vec{E}(x,t) = E_0 \sin(kx - \omega t) \hat{j} Magnetic field: B⃗(x,t)=B0sin⁡(kx−ωt)k^\vec{B}(x,t) = B_0 \sin(kx - \omega t) \hat{k} E⃗⊥B⃗⊥k⃗\vec{E} \perp \vec{B} \perp \vec{k} (Wave vector direction) 🔶 3. Relation between Electric and Magnetic Fields E0B0=c⇒B0=E0c\frac{E_0}{B_0} = c \quad \Rightarrow \quad B_0 = \frac{E_0}{c} 🔶 4. Wavelength, Frequency, Wave Number, Angular Frequency Wavelength: λ=2πk\lambda = \frac{2\pi}{k} Frequency: f=ω2πf = \frac{\omega}{2\pi} Wave number: k=2πλk = \frac{2\pi}{\lambda} Angular frequency: ω=2πf\omega = 2\pi f Speed: v=fλ=ωkv = f \lambda = \frac{\omega}{k} 🔶 5. Poynting Vector (Energy Flow in EM Wave) S⃗=1μ0(E⃗×B⃗)\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) Magnitude: S=E0B0μ0S = \frac{E_0 B_0}{\mu_0} Time-Averaged Poynting Vector (Intensity): ⟨S⟩=E022μ0c=12ε0cE02\langle S \rangle = \frac{E_0^2}{2\mu_0 c} = \frac{1}{2} \varepsilon_0 c E_0^2 🔶 6. Energy Density in EM Waves Electric field energy density: uE=12ε0E2u_E = \frac{1}{2} \varepsilon_0 E^2 Magnetic field energy density: uB=12B2μ0u_B = \frac{1}{2} \frac{B^2}{\mu_0} Total energy density: u=uE+uB=ε0E2=B2μ0u = u_E + u_B = \varepsilon_0 E^2 = \frac{B^2}{\mu_0} 🔶 7. Intensity of Electromagnetic Waves I=⟨S⟩=12ε0cE02I = \langle S \rangle = \frac{1}{2} \varepsilon_0 c E_0^2 Intensity ∝ E02E_0^2 🔶 8. Displacement Current Id=ε0dΦEdtorId=ε0ddt∫E⃗⋅dA⃗I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \quad \text{or} \quad I_d = \varepsilon_0 \frac{d}{dt} \int \vec{E} \cdot d\vec{A} 🔶 9. Maxwell's Equations (Integral Form) Gauss’s Law (Electric): ∮E⃗⋅dA⃗=qinε0\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{in}}}{\varepsilon_0} Gauss’s Law (Magnetic): ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0 Faraday’s Law (Induction): ∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = - \frac{d\Phi_B}{dt} Ampere-Maxwell Law: ∮B⃗⋅dl⃗=μ0I+μ0ε0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 I + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} 🔶 10. Radiation Pressure For perfect absorption: P=IcP = \frac{I}{c} For perfect reflection: P=2IcP = \frac{2I}{c} ✅ Tips for NEET / JEE: Memorize c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}, ε0\varepsilon_0, and μ0\mu_0 values. Practice vector directions: E⃗\vec{E}, B⃗\vec{B}, and k⃗\vec{k} form a right-handed system. Focus on intensity and Poynting vector questions — frequently asked. Derivations of energy density and wave equations are important for JEE Mains/Adv.

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Here is a focused summary of important formulas and mathematical concepts from Chapter 8 – Electromagnetic Waves (Class 12 Physics) relevant for NEET & JEE (Part 3: Math-based Concepts). 📘 Chapter 8: Electromagnetic Waves – Important Maths (NEET + JEE) 🔹 1. Speed of Electromagnetic Waves in Vacuum c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} cc = speed of light in vacuum = 3×108 m/s3 \times 10^8 \, \text{m/s} μ0\mu_0 = permeability of free space = 4π×10−7 H/m4\pi \times 10^{-7} \, \text{H/m} ε0\varepsilon_0 = permittivity of free space = 8.854×10−12 F/m8.854 \times 10^{-12} \, \text{F/m} 🔹 2. Relationship Between E and B in EM Waves E0B0=c\frac{E_0}{B_0} = c Where: E0E_0 = amplitude of electric field B0B_0 = amplitude of magnetic field cc = speed of light ⚠️ This is frequently asked in NEET & JEE as a conceptual + numerical question. 🔹 3. Energy Density of EM Waves Total energy density: u=uE+uB=12ε0E2+12μ0B2u = u_E + u_B = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2\mu_0} B^2 But since EB=c\frac{E}{B} = c, the electric and magnetic energy densities are equal: uE=uB=12ε0E2u_E = u_B = \frac{1}{2} \varepsilon_0 E^2 🔹 4. Poynting Vector (Power Flow Per Unit Area) S⃗=1μ0E⃗×B⃗\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} Magnitude: S=E02μ0c=ε0cE02S = \frac{E_0^2}{\mu_0 c} = \varepsilon_0 c E_0^2 Unit: W/m2\text{W/m}^2 Direction: Along the propagation of the wave 🔹 5. Intensity of Electromagnetic Wave I=⟨S⟩=Average Poynting vectorI = \langle S \rangle = \text{Average Poynting vector} I=12ε0cE02=12cμ0B02I = \frac{1}{2} \varepsilon_0 c E_0^2 = \frac{1}{2} \frac{c}{\mu_0} B_0^2 ✅ Used in both NEET (for bio-optics & radiation) and JEE problems. 🔹 6. Maxwell’s Equations in Free Space (Conceptual Basis of EM Waves) Gauss’s Law (Electric): ∇⋅E⃗=ρε0\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} Gauss’s Law (Magnetic): ∇⋅B⃗=0\nabla \cdot \vec{B} = 0 Faraday’s Law (EM Induction): ∇×E⃗=−∂B⃗∂t\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} Ampere-Maxwell Law: ∇×B⃗=μ0ε0∂E⃗∂t\nabla \times \vec{B} = \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t} 🔍 These are usually asked conceptually, but are the basis of deriving the wave equation. 🔹 7. Wavelength and Frequency Relation c=λνc = \lambda \nu λ\lambda = wavelength (in m) ν\nu = frequency (in Hz) Useful for calculating EM spectrum ranges (radio, microwave, infrared, visible, UV, X-rays, gamma). 🔹 8. EM Wave Equation (1D Form) ∂2E∂x2=μ0ε0∂2E∂t2\frac{\partial^2 E}{\partial x^2} = \mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2} Same for BB: ∂2B∂x2=μ0ε0∂2B∂t2\frac{\partial^2 B}{\partial x^2} = \mu_0 \varepsilon_0 \frac{\partial^2 B}{\partial t^2} 🔹 9. EM Spectrum Ranges (for Numerical Matching) RegionWavelength (approx.)UseRadio waves> 1 mCommunicationMicrowaves1 m – 1 mmRadar, Microwave ovensInfrared1 mm – 700 nmRemote controls, heat sensingVisible Light700 nm – 400 nmHuman visionUV400 nm – 10 nmSterilizationX-rays10 nm – 0.01 nmMedical imagingGamma rays< 0.01 nmCancer treatment 📌 Top 5 Must-Know NEET/JEE Questions Patterns Find the magnetic field given electric field amplitude Use: B0=E0cB_0 = \frac{E_0}{c} Calculate intensity from E0E_0 Use: I=12ε0cE02I = \frac{1}{2} \varepsilon_0 c E_0^2 Find energy flow direction Use direction of E⃗×B⃗\vec{E} \times \vec{B} Given frequency, find wavelength Use: λ=cν\lambda = \frac{c}{\nu} Ratio of electric to magnetic energy density Always 1:1

📘 Chapter 8: Electromagnetic Waves – Important Maths (Part 2) 🧮 1. Energy Density of EM Waves Total energy density uu: u=uE+uB=12ε0E2+12B2μ0u = u_E + u_B = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} For electromagnetic waves: uE=uB=12ε0E2=12B2μ0⇒u=ε0E2=B2μ0u_E = u_B = \frac{1}{2} \varepsilon_0 E^2 = \frac{1}{2} \frac{B^2}{\mu_0} \Rightarrow u = \varepsilon_0 E^2 = \frac{B^2}{\mu_0} 🌟 2. Relation Between E and B in EM Wave EB=c=1μ0ε0≈3×108 m/s\frac{E}{B} = c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8 \, \text{m/s} 🔆 3. Intensity of EM Wave Average intensity (I): I=Power/Area=⟨u⟩c=ε0cE02I = \text{Power/Area} = \langle u \rangle c = \varepsilon_0 c E_0^2 Alternatively: I=12ε0cE02=cB022μ0I = \frac{1}{2} \varepsilon_0 c E_0^2 = \frac{c B_0^2}{2 \mu_0} 📈 4. Poynting Vector Instantaneous Poynting vector: S⃗=1μ0(E⃗×B⃗)(Direction of wave propagation)\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) \quad \text{(Direction of wave propagation)} Magnitude of average Poynting vector: ⟨S⟩=1μ0E0B0=ε0cE02=I\langle S \rangle = \frac{1}{\mu_0} E_0 B_0 = \varepsilon_0 c E_0^2 = I 🌊 5. EM Wave Equation in Free Space For electric field E(x,t)E(x, t): ∂2E∂x2=μ0ε0∂2E∂t2⇒∂2E∂x2=1c2∂2E∂t2\frac{\partial^2 E}{\partial x^2} = \mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2} \Rightarrow \frac{\partial^2 E}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 E}{\partial t^2} Similarly for B(x,t)B(x, t) 📡 6. Wave Nature of EM Radiation Electric field (in z-direction): E⃗(x,t)=E0sin⁡(kx−ωt)j^\vec{E}(x,t) = E_0 \sin(kx - \omega t) \hat{j} Magnetic field (in y-direction): B⃗(x,t)=B0sin⁡(kx−ωt)k^\vec{B}(x,t) = B_0 \sin(kx - \omega t) \hat{k} k⃗\vec{k} is the wave vector, ω\omega is angular frequency 📏 7. Wave Parameters Wavelength: λ\lambda Frequency: ff Angular frequency: ω=2πf\omega = 2\pi f Wave number: k=2πλk = \frac{2\pi}{\lambda} Wave speed: v=λf=ωk=cv = \lambda f = \frac{\omega}{k} = c ⚛️ 8. Permittivity and Permeability of Free Space ε0=8.85×10−12 C2/Nm2\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 μ0=4π×10−7 Tm/A\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} 🧲 9. Electromagnetic Spectrum Basics You don't need derivations here, but memorize wavelength and frequency ranges: Radiation TypeWavelengthFrequencyRadio> 1 m< 10810^8 HzMicrowaves10−310^{-3}–1 m108−101110^8 - 10^{11} HzInfrared10−6−10−310^{-6} - 10^{-3} m1011−101410^{11} - 10^{14} HzVisible400−700400 - 700 nm4.3×1014−7.5×10144.3 \times 10^{14} - 7.5 \times 10^{14} HzUV, X-rays, Gammaprogressively shorter ✅ Tips for NEET & JEE: NEET: Focus on basic formulas, units, and concept of wave propagation, energy, and spectrum. JEE: Derive energy density, wave equations, and practice vector relations (e.g., E⃗⊥B⃗⊥k⃗\vec{E} \perp \vec{B} \perp \vec{k}).

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📘 Chapter 8: Electromagnetic Waves – Important Math (NEET & JEE) 🔹 1. Speed of Electromagnetic Waves in Vacuum c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \ \text{H/m} (Permeability of free space) ε0=8.85×10−12 F/m\varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} (Permittivity of free space) c=3×108 m/sc = 3 \times 10^8 \ \text{m/s} 🔹 2. Wave Equation for EM Wave Electric field: E⃗(x,t)=E0sin⁡(kx−ωt)j^\vec{E}(x, t) = E_0 \sin(kx - \omega t) \hat{j} Magnetic field: B⃗(x,t)=B0sin⁡(kx−ωt)k^\vec{B}(x, t) = B_0 \sin(kx - \omega t) \hat{k} E⃗⊥B⃗⊥k⃗\vec{E} \perp \vec{B} \perp \vec{k} → All are mutually perpendicular (Transverse wave) 🔹 3. Relation Between Electric and Magnetic Fields E0B0=c\frac{E_0}{B_0} = c E0=E_0 = peak electric field, B0=B_0 = peak magnetic field Units check: V/m/T=m/s\text{V/m} / \text{T} = \text{m/s} 🔹 4. Wave Parameters Wavelength: λ=2πk\lambda = \frac{2\pi}{k} Frequency: f=ω2πf = \frac{\omega}{2\pi} Wave speed: v=fλ=ωkv = f \lambda = \frac{\omega}{k} 🔹 5. Energy Density Electric energy density: uE=12ε0E2u_E = \frac{1}{2} \varepsilon_0 E^2 Magnetic energy density: uB=12B2μ0u_B = \frac{1}{2} \frac{B^2}{\mu_0} Total energy density: u=uE+uB=ε0E2=B2μ0u = u_E + u_B = \varepsilon_0 E^2 = \frac{B^2}{\mu_0} 🔹 6. Poynting Vector (Energy Flow per Unit Area) S⃗=1μ0(E⃗×B⃗)\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) Magnitude: S=EBμ0=ε0cE2S = \frac{E B}{\mu_0} = \varepsilon_0 c E^2 SI unit: W/m2\text{W/m}^2 🔹 7. Intensity of EM Wave I=⟨S⟩=12ε0cE02=cB022μ0I = \langle S \rangle = \frac{1}{2} \varepsilon_0 c E_0^2 = \frac{c B_0^2}{2 \mu_0} Average value of Poynting vector over time. 🔹 8. Momentum of Electromagnetic Waves Momentum density: p=ucp = \frac{u}{c} Radiation pressure: For complete absorption: P=IcP = \frac{I}{c} For total reflection: P=2IcP = \frac{2I}{c} 🔹 9. Electromagnetic Spectrum Order Increasing frequency (↓ wavelength): Radio < Microwave < Infrared < Visible < UV < X-ray < Gamma Join For More https://whatsapp.com/channel/0029Va9492L3QxRv5MtNFF0Y/151

📘 Chapter 8: Electromagnetic Waves – Important Maths (Part 2) 🧮 1. Energy Density of EM Waves Total energy density uu: u=uE+uB=12ε0E2+12B2μ0u = u_E + u_B = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} For electromagnetic waves: uE=uB=12ε0E2=12B2μ0⇒u=ε0E2=B2μ0u_E = u_B = \frac{1}{2} \varepsilon_0 E^2 = \frac{1}{2} \frac{B^2}{\mu_0} \Rightarrow u = \varepsilon_0 E^2 = \frac{B^2}{\mu_0} 🌟 2. Relation Between E and B in EM Wave EB=c=1μ0ε0≈3×108 m/s\frac{E}{B} = c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8 \, \text{m/s} 🔆 3. Intensity of EM Wave Average intensity (I): I=Power/Area=⟨u⟩c=ε0cE02I = \text{Power/Area} = \langle u \rangle c = \varepsilon_0 c E_0^2 Alternatively: I=12ε0cE02=cB022μ0I = \frac{1}{2} \varepsilon_0 c E_0^2 = \frac{c B_0^2}{2 \mu_0} 📈 4. Poynting Vector Instantaneous Poynting vector: S⃗=1μ0(E⃗×B⃗)(Direction of wave propagation)\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) \quad \text{(Direction of wave propagation)} Magnitude of average Poynting vector: ⟨S⟩=1μ0E0B0=ε0cE02=I\langle S \rangle = \frac{1}{\mu_0} E_0 B_0 = \varepsilon_0 c E_0^2 = I 🌊 5. EM Wave Equation in Free Space For electric field E(x,t)E(x, t): ∂2E∂x2=μ0ε0∂2E∂t2⇒∂2E∂x2=1c2∂2E∂t2\frac{\partial^2 E}{\partial x^2} = \mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2} \Rightarrow \frac{\partial^2 E}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 E}{\partial t^2} Similarly for B(x,t)B(x, t) 📡 6. Wave Nature of EM Radiation Electric field (in z-direction): E⃗(x,t)=E0sin⁡(kx−ωt)j^\vec{E}(x,t) = E_0 \sin(kx - \omega t) \hat{j} Magnetic field (in y-direction): B⃗(x,t)=B0sin⁡(kx−ωt)k^\vec{B}(x,t) = B_0 \sin(kx - \omega t) \hat{k} k⃗\vec{k} is the wave vector, ω\omega is angular frequency 📏 7. Wave Parameters Wavelength: λ\lambda Frequency: ff Angular frequency: ω=2πf\omega = 2\pi f Wave number: k=2πλk = \frac{2\pi}{\lambda} Wave speed: v=λf=ωk=cv = \lambda f = \frac{\omega}{k} = c ⚛️ 8. Permittivity and Permeability of Free Space ε0=8.85×10−12 C2/Nm2\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 μ0=4π×10−7 Tm/A\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} 🧲 9. Electromagnetic Spectrum Basics You don't need derivations here, but memorize wavelength and frequency ranges: Radiation TypeWavelengthFrequencyRadio> 1 m< 10810^8 HzMicrowaves10−310^{-3}–1 m108−101110^8 - 10^{11} HzInfrared10−6−10−310^{-6} - 10^{-3} m1011−101410^{11} - 10^{14} HzVisible400−700400 - 700 nm4.3×1014−7.5×10144.3 \times 10^{14} - 7.5 \times 10^{14} HzUV, X-rays, Gammaprogressively shorter ✅ Tips for NEET & JEE: NEET: Focus on basic formulas, units, and concept of wave propagation, energy, and spectrum. JEE: Derive energy density, wave equations, and practice vector relations (e.g., E⃗⊥B⃗⊥k⃗\vec{E} \perp \vec{B} \perp \vec{k}). Share this channel link - https://t.me/Physics_Notes_Class_11th_12th Join For More https://whatsapp.com/channel/0029Va9492L3QxRv5MtNFF0Y/151

Chapter 8 Electromagnetic Waves part 1 Join For More Part 2 Part 3 Part 4

Here are some important and frequently asked numerical questions from the chapter AC Current (Alternating Current) for NEET and JEE preparation. These cover key concepts such as reactance, impedance, resonance, power, RMS values, and LC/RLC circuits. 🔢 Important Numerical Questions – AC Current 1. RMS & Peak Values Q1. A sinusoidal AC voltage has a peak value of 325 V. Calculate the RMS value. Formula: Vrms=Vpeak2V_{\text{rms}} = \dfrac{V_{\text{peak}}}{\sqrt{2}}Vrms​=2​Vpeak​​ 2. Power in AC Circuit Q2. An AC circuit has Vrms=220V_{\text{rms}} = 220Vrms​=220 V, Irms=5I_{\text{rms}} = 5Irms​=5 A, and a power factor of 0.8. Find the power consumed. Formula: P=VIcos⁡ϕP = VI \cos\phiP=VIcosϕ 3. Reactance & Impedance Q3. A 100 Ω resistor and a 100 mH inductor are connected in series across a 50 Hz AC supply. Calculate: (a) Inductive reactance XLX_LXL​ (b) Impedance of the circuit (c) Current in the circuit if Vrms=220V_{\text{rms}} = 220Vrms​=220 V XL=2πfLX_L = 2\pi f LXL​=2πfL Z=R2+XL2Z = \sqrt{R^2 + X_L^2}Z=R2+XL2​​ I=VZI = \dfrac{V}{Z}I=ZV​ 4. Resonance in RLC Circuit Q4. A series RLC circuit has R=20 ΩR = 20\,\OmegaR=20Ω, L=0.1 HL = 0.1\,\text{H}L=0.1H, and C=100 μFC = 100\,\mu\text{F}C=100μF. Calculate the resonant frequency. Formula: f0=12πLCf_0 = \dfrac{1}{2\pi\sqrt{LC}}f0​=2πLC​1​ 5. LC Oscillations Q5. A capacitor of 10 μF10\,\mu\text{F}10μF is connected to an inductor of 0.5 H0.5\,\text{H}0.5H. Calculate the frequency of oscillation. f=12πLCf = \dfrac{1}{2\pi\sqrt{LC}}f=2πLC​1​ 6. Power in Purely Inductive or Capacitive Circuit Q6. In a purely inductive AC circuit with L=0.1L = 0.1L=0.1 H and a supply of 220 V at 50 Hz, calculate the current and average power consumed. XL=2πfLX_L = 2\pi f LXL​=2πfL, I=VXLI = \dfrac{V}{X_L}I=XL​V​, Power P=0P = 0P=0 (purely inductive) 7. Phase Difference Q7. In an RLC series circuit, R=10 ΩR = 10\,\OmegaR=10Ω, L=0.1 HL = 0.1\,\text{H}L=0.1H, C=10 μFC = 10\,\mu\text{F}C=10μF, and supply frequency is 50 Hz. Calculate the phase angle between voltage and current. XL=2πfLX_L = 2\pi f LXL​=2πfL, XC=12πfCX_C = \dfrac{1}{2\pi f C}XC​=2πfC1​, tan⁡ϕ=XL−XCR\tan\phi = \dfrac{X_L - X_C}{R}tanϕ=RXL​−XC​​ 8. Time-Averaged Power Q8. An AC source V=200sin⁡(100πt)V = 200\sin(100\pi t)V=200sin(100πt) is connected to a resistor of 50 Ω50\,\Omega50Ω. Find the average power consumed. RMS voltage Vrms=V02V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}Vrms​=2​V0​​, P=Vrms2RP = \dfrac{V_{\text{rms}}^2}{R}P=RVrms2​​ 9. Transformer Equation (JEE focus) Q9. A step-up transformer increases voltage from 220 V to 2200 V. If the primary has 100 turns, find the number of turns in the secondary. VsVp=NsNp\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}Vp​Vs​​=Np​Ns​​ 10. Energy Stored in Inductor Q10. An inductor of 2 H carries a current of 3 A. Find the energy stored in it. E=12LI2E = \dfrac{1}{2}LI^2E=21​LI2 📌 Tips for NEET/JEE Preparation NEET focuses more on formulas and direct applications. Units and basic definitions matter a lot. JEE includes conceptual + numerical problems, often requiring deeper understanding and multi-step calculations. Practice previous year questions and mock tests. Memorize standard formulas (RMS, Power, Reactance, Resonance). Join For More https://whatsapp.com/channel/0029Va9492L3QxRv5MtNFF0Y/151

Here are some important and frequently asked numerical questions from the chapter AC Current (Alternating Current) for NEET and JEE preparation. These cover key concepts such as reactance, impedance, resonance, power, RMS values, and LC/RLC circuits. 🔢 Important Numerical Questions – AC Current 1. RMS & Peak Values Q1. A sinusoidal AC voltage has a peak value of 325 V. Calculate the RMS value. Formula: Vrms=Vpeak2V_{\text{rms}} = \dfrac{V_{\text{peak}}}{\sqrt{2}}Vrms​=2​Vpeak​​ 2. Power in AC Circuit Q2. An AC circuit has Vrms=220V_{\text{rms}} = 220Vrms​=220 V, Irms=5I_{\text{rms}} = 5Irms​=5 A, and a power factor of 0.8. Find the power consumed. Formula: P=VIcos⁡ϕP = VI \cos\phiP=VIcosϕ 3. Reactance & Impedance Q3. A 100 Ω resistor and a 100 mH inductor are connected in series across a 50 Hz AC supply. Calculate: (a) Inductive reactance XLX_LXL​ (b) Impedance of the circuit (c) Current in the circuit if Vrms=220V_{\text{rms}} = 220Vrms​=220 V XL=2πfLX_L = 2\pi f LXL​=2πfL Z=R2+XL2Z = \sqrt{R^2 + X_L^2}Z=R2+XL2​​ I=VZI = \dfrac{V}{Z}I=ZV​ 4. Resonance in RLC Circuit Q4. A series RLC circuit has R=20 ΩR = 20\,\OmegaR=20Ω, L=0.1 HL = 0.1\,\text{H}L=0.1H, and C=100 μFC = 100\,\mu\text{F}C=100μF. Calculate the resonant frequency. Formula: f0=12πLCf_0 = \dfrac{1}{2\pi\sqrt{LC}}f0​=2πLC​1​ 5. LC Oscillations Q5. A capacitor of 10 μF10\,\mu\text{F}10μF is connected to an inductor of 0.5 H0.5\,\text{H}0.5H. Calculate the frequency of oscillation. f=12πLCf = \dfrac{1}{2\pi\sqrt{LC}}f=2πLC​1​ 6. Power in Purely Inductive or Capacitive Circuit Q6. In a purely inductive AC circuit with L=0.1L = 0.1L=0.1 H and a supply of 220 V at 50 Hz, calculate the current and average power consumed. XL=2πfLX_L = 2\pi f LXL​=2πfL, I=VXLI = \dfrac{V}{X_L}I=XL​V​, Power P=0P = 0P=0 (purely inductive) 7. Phase Difference Q7. In an RLC series circuit, R=10 ΩR = 10\,\OmegaR=10Ω, L=0.1 HL = 0.1\,\text{H}L=0.1H, C=10 μFC = 10\,\mu\text{F}C=10μF, and supply frequency is 50 Hz. Calculate the phase angle between voltage and current. XL=2πfLX_L = 2\pi f LXL​=2πfL, XC=12πfCX_C = \dfrac{1}{2\pi f C}XC​=2πfC1​, tan⁡ϕ=XL−XCR\tan\phi = \dfrac{X_L - X_C}{R}tanϕ=RXL​−XC​​ 8. Time-Averaged Power Q8. An AC source V=200sin⁡(100πt)V = 200\sin(100\pi t)V=200sin(100πt) is connected to a resistor of 50 Ω50\,\Omega50Ω. Find the average power consumed. RMS voltage Vrms=V02V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}Vrms​=2​V0​​, P=Vrms2RP = \dfrac{V_{\text{rms}}^2}{R}P=RVrms2​​ 9. Transformer Equation (JEE focus) Q9. A step-up transformer increases voltage from 220 V to 2200 V. If the primary has 100 turns, find the number of turns in the secondary. VsVp=NsNp\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}Vp​Vs​​=Np​Ns​​ 10. Energy Stored in Inductor Q10. An inductor of 2 H carries a current of 3 A. Find the energy stored in it. E=12LI2E = \dfrac{1}{2}LI^2E=21​LI2 📌 Tips for NEET/JEE Preparation NEET focuses more on formulas and direct applications. Units and basic definitions matter a lot. JEE includes conceptual + numerical problems, often requiring deeper understanding and multi-step calculations. Practice previous year questions and mock tests. Memorize standard formulas (RMS, Power, Reactance, Resonance). Join For More https://whatsapp.com/channel/0029Va9492L3QxRv5MtNFF0Y/151

Chapter 7 Alternating Current Part 3

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