Exercises — LC circuit — oscillations (electrical analog of SHM)

WHAT hum compare kar rahe hain: mechanical $m\ddot{x}+kx=0$ vs electrical $L\ddot{q}+\frac{1}{C}q=0$. Term by term line up karo: second derivative ke aage jo coefficient hai woh "inertia" hai, variable ke aage jo coefficient hai woh "stiffness" hai.

• Inertia (mass): $m\leftrightarrow ==L==$ (inductor current mein change resist karta hai, bilkul waise jaise mass velocity mein change resist karta hai).
• Stiffness (spring): $k\leftrightarrow ==1/C==$ (chhota $C$ ek stiff spring hai — woh apna voltage change karne ke against zyada ladata hai).

Formula: $\omega =\frac{1}{\sqrt{LC}}$. Yahi kyun? Yeh seedha $\ddot{q}=-\frac{1}{LC}q$ ko $\ddot{x}=-{\omega }^{2}x$ se match karne par milta hai. $LC=(4\times 10^{-3})(1\times 10^{-6})=4\times 10^{-9}{\text{s}}^2$ $\sqrt{LC}=\sqrt{4\times 10^{-9}}=2\times 10^{-4.5}=6.32\times 10^{-5}\text{s}$ $\omega =\frac{1}{6.32\times 10^{-5}}\approx 1.58\times 10^4\text{rad/s}$

Jhooth. Current $i=\dot{q}$, $q(t)=q_0\cos \omega t$ ka slope hai; cosine ki peak par slope zero hota hai, isliye wahan $i=0$ hota hai. Current tab peak karti hai jab $q=0$ hota hai (jahan cosine zero cross karta hai aur sabse steep hota hai). Yeh dono $90^{\circ }$ apart hain.

$LC=(2\times 10^{-3})(8\times 10^{-6})=1.6\times 10^{-8}{\text{s}}^2,\sqrt{LC}=1.2649\times 10^{-4}\text{s}$ $T=2\pi \sqrt{LC}=2\pi (1.2649\times 10^{-4})\approx 7.95\times 10^{-4}\text{s}(\approx 0.795\text{ms})$ $f=\frac{1}{T}\approx 1258\text{Hz}$ $T=2\pi \sqrt{LC}$ kyun aur $2\pi /\sqrt{LC}$ kyun nahi? Kyunki $T=2\pi /\omega$ aur $\omega =1/\sqrt{LC}$ hai; $1/\sqrt{LC}$ se divide karna matlab $\sqrt{LC}$ se multiply karna, isliye root numerator mein aa jaata hai.

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{1.2649\times 10^{-4}}=7905.7\text{rad/s}$ $i_0=q_0\omega =(4\times 10^{-4})(7905.7)\approx 3.16\text{A}$ $i_0=q_0\omega$ kyun? $q=q_0\cos \omega t$ differentiate karo: $-q_0\omega \sin \omega t$ ki amplitude $q_0\omega$ hai. Equivalently, saari electric energy $q_0^2/2C$, crossing par magnetic energy $\frac{1}{2}L{i}_0^2$ ban jaati hai.

$q_0=4\times 10^{-4}$ C, $\omega =7906$ rad/s, $i_0=3.16$ A ke saath: $q(t)=(4\times 10^{-4})\cos (7906t)\text{C}$ $i(t)=-(3.16)\sin (7906t)\text{A}$ Minus sign kehta hai current pehle us direction mein flow karti hai jo capacitor ko discharge karta hai.

WHAT chahiye: $U_E=U_B$. Dono likho: $U_E=\frac{q_0^2}{2C}{\cos }^2\omega t,U_B=\frac{q_0^2}{2C}{\sin }^2\omega t$ WHY yeh form leti hain: $L{\omega }^2=1/C$ use karne par $\frac{1}{2}L{i}^2$ usi prefactor $q_0^2/2C$ times ${\sin }^2$ mein aa jaata hai. Equal set karo: ${\cos }^2\omega t={\sin }^2\omega t\Rightarrow {\tan }^2\omega t=1\Rightarrow \tan \omega t=1$ (pehla positive solution) $\Rightarrow \omega t=\frac{\pi }{4}$. $t=\frac{\pi }{4\omega }=\frac{\pi }{4\times 10^4}\approx 7.85\times 10^{-5}\text{s}=\frac{T}{8}$ Figure dekho: dono curves pehli baar $T/8$ par cross karti hain, jahan har ek total energy ka exactly aadha hold karti hai.

Energy kyun, time nahi: hum pooch rahe hain "$i=i_0/2$ par kya $q$ hai," yeh $q$ aur $i$ ke beech ek relation hai — energy unhe directly link karti hai. $\underset{U_E}{\underbrace{\frac{q^2}{2C}}}+\underset{U_B}{\underbrace{\frac{1}{2}L{i}^2}}=\frac{q_0^2}{2C}$ $i=i_0/2$ par, aur $\frac{1}{2}L{i}_0^2=\frac{q_0^2}{2C}$ use karte hue toh $\frac{1}{2}L(i_0/2{)}^2=\frac{1}{4}\cdot \frac{q_0^2}{2C}$: $\frac{q^2}{2C}=\frac{q_0^2}{2C}-\frac{1}{4}\frac{q_0^2}{2C}=\frac{3}{4}\frac{q_0^2}{2C}\Rightarrow q=\frac{\sqrt{3}}{2}{q}_0$ Numerically $q=\frac{\sqrt{3}}{2}(2\times 10^{-4})\approx 1.73\times 10^{-4}$ C.

$\omega =\frac{1}{\sqrt{LC}}$ sirf product $LC$ par depend karta hai. $\omega$ fix rakhne ke liye, $LC$ fix rakho. Agar $L\to 2L$, toh $C\to C/2$ (use halve karo), kyunki $(2L)(C/2)=LC$. Capacitance half kar do.

Step 1 (WHAT): Closed loop mein ek baar ghoom jao; Kirchhoff's Voltage Law ke according voltage rises aur drops ka sum zero hona chahiye: $V_C+V_L=0\Rightarrow \frac{q}{C}+L\frac{di}{dt}=0$ WHY: koi battery nahi aur koi resistor nahi, toh sirf do voltages hain capacitor ki ($q/C$) aur inductor ki back-EMF ($Ldi/dt$); unhe cancel hona hi chahiye. Step 2 (WHAT): $i=\frac{dq}{dt}$ replace karo, toh $\frac{di}{dt}=\frac{d^{2}q}{d{t}^2}$: $L\frac{d^{2}q}{d{t}^2}+\frac{q}{C}=0\Rightarrow \frac{d^{2}q}{d{t}^2}=-\frac{1}{LC}q$ WHY: current hi charge change ki rate hai, toh uski rate of change $q$ ki second derivative hai. Step 3 (WHAT): Simple Harmonic Motion ke SHM template $\ddot{x}=-{\omega }^{2}x$ se compare karo. Coefficients match karne par, ${\omega }^2=\frac{1}{LC}$, toh $\omega =\frac{1}{\sqrt{LC}}$.

Period se $C$ nikalo. $T=2\pi \sqrt{LC}\Rightarrow \sqrt{LC}=\frac{T}{2\pi }\Rightarrow LC={\left(\frac{T}{2\pi }\right)}^2$. $\frac{T}{2\pi }=\frac{4\times 10^{-4}}{2\pi }=6.366\times 10^{-5}\text{s},LC=4.053\times 10^{-9}{\text{s}}^2$ $C=\frac{LC}{L}=\frac{4.053\times 10^{-9}}{1\times 10^{-3}}=4.053\times 10^{-6}\text{F}\approx 4.05\mu \text{F}$ $i_0$ nikalo. Pehle $q_0=C{V}_0=(4.053\times 10^{-6})(20)=8.11\times 10^{-5}$ C. Phir $\omega =\frac{2\pi }{T}=\frac{2\pi }{4\times 10^{-4}}=1.5708\times 10^4\text{rad/s}$ $i_0=q_0\omega =(8.11\times 10^{-5})(1.5708\times 10^4)\approx 1.27\text{A}$

$U_E=\frac{q_0^2}{2C}{\cos }^2\omega t,U_B=\frac{1}{2}L(q_0\omega {)}^2{\sin }^2\omega t=\frac{L{\omega }^2{q}_0^2}{2}{\sin }^2\omega t$ $L{\omega }^2=L\cdot \frac{1}{LC}=\frac{1}{C}$ use karo, toh $\frac{L{\omega }^2{q}_0^2}{2}=\frac{q_0^2}{2C}$. Phir $U_E+U_B=\frac{q_0^2}{2C}({\cos }^2\omega t+{\sin }^2\omega t)=\frac{q_0^2}{2C}$ kyunki ${\cos }^2+{\sin }^2=1$. Constant total energy hai $\boxed{U=\frac{q_0^2}{2C}}$.

Resonance kyun: $f=\frac{1}{2\pi \sqrt{LC}}$ natural frequency hai — yahi woh resonant frequency hai jis par circuit ring karta hai. $C$ ke liye solve karo: square karo aur rearrange karo. $f=\frac{1}{2\pi \sqrt{LC}}\Rightarrow \sqrt{LC}=\frac{1}{2\pi f}\Rightarrow LC=\frac{1}{4{\pi }^2{f}^2}\Rightarrow C=\frac{1}{4{\pi }^2{f}^{2}L}$ $C=\frac{1}{4{\pi }^2(1.0\times 10^6{)}^2(25\times 10^{-6})}=\frac{1}{4{\pi }^2\cdot 10^{12}\cdot 25\times 10^{-6}}$ $=\frac{1}{9.8696\times 10^8}\approx 1.013\times 10^{-9}\text{F}\approx 1.01\text{nF}$

(a) $C\to \infty$ (bahut bada capacitor = super-soft spring). $\omega =1/\sqrt{LC}\to 0$: period $T=2\pi \sqrt{LC}\to \infty$. "Spring" $1/C\to 0$ gayab ho jaata hai, toh charge ko wapas kheenchne ke liye kuch nahi — oscillation infinitely slow ho jaati hai. (b) $L\to 0$ (koi inductor nahi = koi inertia nahi). $\omega \to \infty$, $T\to 0$. Koi electrical inertia nahi toh current ko overshoot karne wali koi cheez nahi; capacitor bas instantly discharge ho jaata — oscillation degenerate ho jaati hai. (c) Resistance $R$ add karo. Energy conserve nahi rehti; circuit ek LCR Circuit ban jaata hai jo Damped Oscillations dikhata hai. Amplitude decay hoti hai aur frequency thoda shift hoti hai ${\omega }_d=\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}$ par. Ideal LC result iska $R\to 0$ limit hai.

(i) Total energy. $U=\frac{q_0^2}{2C}=\frac{(2\times 10^{-4}{)}^2}{2(5\times 10^{-6})}=\frac{4\times 10^{-8}}{1\times 10^{-5}}=4\times 10^{-3}\text{J}=4\text{mJ}$ (ii) Capacitor mein quarter energy. Chahiye $U_E=\frac{1}{4}U$: $\frac{q_0^2}{2C}{\cos }^2\omega t=\frac{1}{4}\frac{q_0^2}{2C}\Rightarrow {\cos }^2\omega t=\frac{1}{4}\Rightarrow \cos \omega t=\frac{1}{2}$ (pehle crossing ke liye positive root lo) $\Rightarrow \omega t=\frac{\pi }{3}$. $t=\frac{\pi }{3\omega }=\frac{\pi }{3\times 10^4}\approx 1.047\times 10^{-4}\text{s}$ $T$ ke fraction ke roop mein: $\frac{t}{T}=\frac{\omega t}{2\pi }=\frac{\pi /3}{2\pi }=\frac{1}{6}$. Toh yeh $\boxed{T/6}$ par hota hai.