Physics Notes Class 11th 12th

📘 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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