LC circuit — oscillations (electrical analog of SHM): Worked examples

1.8.30 · D3 · Physics › Electromagnetism › LC circuit — oscillations (electrical analog of SHM)

Intuition Yeh page kis liye hai

Parent note LC circuit note ne tumhe machinery di thi: $\overset{q}{¨} = - \frac{1}{L C} q$ , solution $q = q_{0} cos (ω t)$ , aur energy split. Yeh page ek workout hai. Hum har type ka question list karte hain jo ek LC circuit throw kar sakta hai, phir ek-ek solve karte hain — weird edge cases bhi shamil hain (zero current start, energy-sharing instants, limits $L \to 0$ aur $C \to \infty$ , aur ek exam twist jo energy ko timing ke saath mix karta hai).

Shuru karne se pehle, paanch symbols jo hum baar baar use karte hain, sab parent note mein banaye gaye hain:

$q$ = capacitor par charge (unit: coulomb, C). Socho "tank kitna bhara hai." Ek cycle mein iska maximum value $q_{0}$ likha jaata hai — charge oscillation ka amplitude.
$i = \frac{d q}{d t}$ = current, yaani charge flow karne ki rate (unit: ampere, A). Socho "flow speed." Iska maximum hai $i_{0} = q_{0} ω$ (peak charge times angular frequency).
$ω = \frac{1}{L C}$ = angular frequency (unit: rad/s). Sloshing kitni tez cycle karta hai.
$T = \frac{2 π}{ω} = 2 π L C$ = period (unit: s) — ek complete slosh ke liye time, taaki $ω T = 2 π$ . Hum $T$ ko neeche constantly use karte hain (jaise " $T /8$ "), isliye yahan pehle se define kar diya hai; Ex 1 ise base numbers se re-derive karega.
$U_{E} = \frac{q ^{2}}{2 C}$ = capacitor mein stored electric energy, aur $U_{B} = \frac{1}{2} L i^{2}$ = inductor mein stored magnetic energy (dono joules, J mein). Yeh do "buckets" energy aage-peechhe trade karte hain; inका sum fixed rehta hai.

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

Scenario matrix

Har LC problem in cells mein se ek mein aata hai. Neeche har worked example us cell(s) ke saath tagged hai jo wo cover karta hai. (Yahan $T$ upar define kiya gaya period hai.)

Cell	Kya test karta hai	Degenerate / edge part	Example
A Frequency & period	$L, C$ ko $ω, T, f$ mein plug karo	bada/chhota $L$ aur bada/chhota $C$	Ex 1
B Peak current	energy swap $U_{E} \to U_{B}$	—	Ex 2
C Kisi given time par value	$q (t), i (t)$ evaluate karo	har quarter-cycle mein signs	Ex 3
D Energy sharing instant	kab $U_{E} = U_{B}$ , ya diya gaya ratio	$T /8$ , general fraction	Ex 4
E Alag initial condition	switch close hona jab current pehle se flow kar rahi ho	$q (0) = 0$ start (pure sine)	Ex 5
F Limiting behaviour	$L \to 0$ , $C \to \infty$ , $R = 0$ vs $R \neq = 0$	degenerate + damped + critical/over-damped	Ex 6
G Real-world word problem	radio tuning / frequency design karo	target hit karne ke liye component chunna	Ex 7
H Exam twist	timing + energy + phase combine karo	multi-step	Ex 8

Hum A–D aur H ke liye ek base circuit reuse karte hain taaki numbers familiar rahein: $L = 2 mH = 2 \times 1 0^{- 3} H, C = 5 μ F = 5 \times 1 0^{- 6} F .$

Inse ek baar aur hamesha ke liye: $ω = \frac{1}{L C} = \frac{1}{( 2 \times 1 0 ^{- 3} ) ( 5 \times 1 0 ^{- 6} )} = \frac{1}{1 0 ^{- 8}} = 1 0^{4} rad/s .$

Ex 1 — Cell A: frequency & period (aur bada $L$ aur bada $C$ kya karte hain)

Worked example Statement

Base circuit ke liye ( $L = 2 mH$ , $C = 5 μ F$ ), $ω$ , $f$ , aur $T$ nikalo. Phir predict karo kya hoga $T$ ko agar tum (a) ek chaar guna bada inductor swap karo, aur (b) ek chaar guna bada capacitor swap karo (har baar doosre component ko apni base value par rakho).

Forecast: pehle guess karo — kya ek bada inductor oscillation ko faster banata hai ya slower? Aur ek bada capacitor ke baare mein? (Hint: analogy mein $L$ "mass" $m$ hai, aur $1/ C$ "spring constant" $k$ hai.)

$ω = \frac{1}{L C} = 1 0^{4}$ rad/s. Yeh step kyun? $ω$ woh hai jis par baaki sab timing quantities bani hain — pehle yahi nikalo.
$f = \frac{ω}{2 π} = \frac{1 0 ^{4}}{2 π} \approx 1591.5$ Hz. Yeh step kyun? $ω = 2 π f$ ; $f$ ek second mein full cycles count karta hai, jo "frequency" ka matlab hai aam boli mein.
$T = \frac{1}{f} = \frac{2 π}{ω} = 2 π L C \approx 6.283 \times 1 0^{- 4}$ s $\approx 0.628$ ms. Yeh step kyun? Period frequency ka reciprocal hai — ek full slosh ke liye time.
(a) Chaar guna bada $L$ : $T \propto L$ , toh $T$ badhta hai $4 = 2$ se. Naya $T \approx 1.257$ ms. Yeh step kyun? Sirf $L$ factor badla. Bada "mass" ⟹ slower ⟹ longer period. Tumhara forecast kehna chahiye tha slower.
(b) Chaar guna bada $C$ : $T \propto C$ , toh phir $T$ badhta hai $4 = 2$ se. Naya $T \approx 1.257$ ms. Yeh step kyun? $1/ C$ "spring constant" hai. Bada $C$ ⟹ softer spring ⟹ kamzor pull-back ⟹ slow oscillation. Toh bada $L$ aur bada $C$ timing par same tarah kaam karte hain — dono slow it down — jabki chhota $L$ ya chhota $C$ use speed up karta hai. Yeh "large/small $L$ ya $C$ " edge ke charo corners cover karta hai.

Verify: $L C = H \cdot F$ ke units. Kyunki $H = V\cdots/A$ aur $F = A\cdots/V$ hai, product $s^{2}$ hai, toh $L C$ seconds mein hai. ✓ Aur $f \cdot T = 1591.5 \times 6.283 \times 1 0^{- 4} \approx 1.000$ . ✓ Dono (a) aur (b) same $4 = 2$ factor dete hain ✓.

Ex 2 — Cell B: energy se peak current

Worked example Statement

Same $L, C$ . Capacitor ko $q_{0} = 2 \times 1 0^{- 4}$ C tak charge kiya jaata hai, phir switch rest par close hota hai. Peak current $i_{0}$ nikalo.

Forecast: sab energy electric shuru hoti hai. Current sabse bada kahan hota hai — jab charge full ho, ya jab wo empty ho?

Shuru mein, energy poori tarah electric hai: $U_{E}^{m a x} = \frac{q _{0}^{2}}{2 C}$ . Yeh step kyun? $t = 0$ par current zero hai, toh $U_{B} = 0$ ; sab energy capacitor mein baithti hai.
Swap point par, sab magnetic ban jaati hai: $\frac{q _{0}^{2}}{2 C} = \frac{1}{2} L i_{0}^{2}$ . Yeh step kyun? Energy conserved hai (koi resistor nahi), toh peak $U_{E}$ = peak $U_{B}$ .
Solve karo: $i_{0} = q_{0} \frac{1}{L C} = q_{0} ω$ . Yeh step kyun? $\frac{1}{2}$ 's cancel karo, $i_{0}$ isolate karo, aur $1/ L C = ω$ pehchano.
$i_{0} = (2 \times 1 0^{- 4}) (1 0^{4}) = 2$ A. Yeh step kyun? Jaana hua $q_{0}$ aur $ω$ substitute karo taaki symbolic result ko actual number mein badle — woh answer jo question ne maanga tha.

Verify: current solution $i (t) = - q_{0} ω sin (ω t)$ se cross-check karo jiska amplitude $q_{0} ω = 2 \times 1 0^{- 4} \times 1 0^{4} = 2$ A hai. Same number, koi energy argument ki zaroorat nahi. ✓ Units: C·(rad/s) = C/s = A. ✓

Ex 3 — Cell C: kisi given time par value, aur har quarter mein sign

Yahan hume signs ke baare mein careful rehna hoga, kyunki ek full cycle chaar quarter-turns se guzarta hai aur $q, i$ alag tarah sign change karte hain. Neeche ki figure unhe track karti hai.

Worked example Statement

$q_{0} = 2 \times 1 0^{- 4}$ C, $ω = 1 0^{4}$ rad/s ke saath, $q$ aur $i$ nikalo $t = \frac{T}{8}$ aur $t = \frac{3 T}{8}$ par.

Forecast: $T /8$ par hum cosine mein ek-aathwa andar hain. Kya $q$ abhi bhi positive hai? Kya $i$ negative (draining) hai ya positive?

$t = T /8$ par $ω t$ : kyunki $ω T = 2 π$ hai, hume milta hai $ω t = \frac{2 π}{8} = \frac{π}{4}$ . Yeh step kyun? Sab kuch angle $ω t$ par depend karta hai, $t$ par directly nahi. Ek baar convert karo.
$q = q_{0} cos (π /4) = 2 \times 1 0^{- 4} \cdot \frac{1}{2} \approx 1.414 \times 1 0^{- 4}$ C (positive — capacitor abhi bhi partially charged hai, same polarity). Yeh step kyun? $π /4$ pehle quarter mein hai jahan cosine positive hai.
$i = - q_{0} ω sin (π /4) = - 2 \cdot \frac{1}{2} \approx - 1.414$ A (negative — current capacitor ko discharge kar raha hai). Yeh step kyun? $i = - q_{0} ω sin (ω t)$ mein minus sign kehta hai current pehle quarter mein capacitor empty karne ke liye flow karta hai. Figure mein orange curve ko negative dip karte dekhو.
$t = 3 T /8$ par: $ω t = \frac{3 π}{4}$ (second quarter). $q = q_{0} cos (3 π /4) \approx - 1.414 \times 1 0^{- 4}$ C (ab negative — capacitor doosri taraf charge ho raha hai). $i = - q_{0} ω sin (3 π /4) \approx - 1.414$ A (abhi bhi negative, abhi bhi same flow direction). Yeh step kyun? Yeh case-coverage point hai: $T /4$ ke baad charge sign flip karta hai lekin current abhi tak flip nahi kiya. Yeh exactly 90° phase offset hai.

Verify: dono times par energy conserved honi chahiye. $T /8$ par: $U_{E} + U_{B} = \frac{( 1.414 \times 1 0 ^{- 4} ) ^{2}}{2 \cdot 5 \times 1 0 ^{- 6}} + \frac{1}{2} (2 \times 1 0^{- 3}) (1.414)^{2} \approx 2 \times 1 0^{- 3} + 2 \times 1 0^{- 3} = 4 \times 1 0^{- 3}$ J. Total $\frac{q _{0}^{2}}{2 C} = \frac{( 2 \times 1 0 ^{- 4} ) ^{2}}{1 0 ^{- 5}} = 4 \times 1 0^{- 3}$ J se compare karo. ✓

Page ke upar se yaad karo: $U_{E} = \frac{q ^{2}}{2 C}$ capacitor ki electric energy hai aur $U_{B} = \frac{1}{2} L i^{2}$ inductor ki magnetic energy hai. Yahan hum poochhte hain kab do buckets specified fractions hold karte hain. Upar ellipse figure par, " $U_{E} = U_{B}$ " woh point hai jahan state 45° ghoom chuki hai.

Worked example Statement

(a) Pehli baar $U_{E} = U_{B}$ . (b) Pehli baar capacitor total energy ka ek-chautha hold karta hai, $U_{E} = \frac{1}{4} U_{total}$ .

Forecast: (a) $T$ ke kisi clean fraction par land karna chahiye. (b) ke liye, kam capacitor energy matlab wave aage swing hui hai — (a) se later ya earlier?

Total energy $U_{total} = \frac{q _{0}^{2}}{2 C}$ , aur $U_{E} = U_{total} cos^{2} (ω t)$ ; similarly $U_{B} = U_{total} sin^{2} (ω t)$ . Yeh step kyun? $U_{E} = \frac{q ^{2}}{2 C} = \frac{q _{0}^{2}}{2 C} cos^{2} (ω t)$ — cosine-squared saari time dependence carry karta hai; bacha hua fraction sine-squared hai, jo exactly $U_{B}$ hai.
(a) $U_{E} = U_{B} \Rightarrow cos^{2} ω t = sin^{2} ω t \Rightarrow tan^{2} ω t = 1 \Rightarrow ω t = π /4$ . Yeh step kyun? Equal shares matlab $cos^{2} = sin^{2}$ ; divide karne se $tan^{2} = 1$ milta hai. Hum $tan$ use karte hain kyunki yeh cosine/sine ratio ko ek condition mein package karta hai.
(a) $t = \frac{π /4}{ω} = \frac{π}{4 ω} = \frac{T}{8}$ . Yeh step kyun? $ω t = π /4$ aur $ω T = 2 π$ dete hain $t / T = 1/8$ .
(b) $U_{E} = \frac{1}{4} U_{total} \Rightarrow cos^{2} ω t = \frac{1}{4} \Rightarrow cos ω t = \frac{1}{2} \Rightarrow ω t = π /3$ . Yeh step kyun? Hum pehla positive time chahte hain, toh woh chhota angle lo jiska cosine $+ \frac{1}{2}$ hai.
(b) $t = \frac{π /3}{ω} = \frac{T}{6}$ . Yeh step kyun? $ω t = π /3 = \frac{2 π}{6}$ , toh $t = T /6$ . Kyunki $T /6 > T /8$ hai, capacitor ek-chautha baad mein girta hai half hit karne ke baad — forecast se match karta hai (wave aage swing hui).

Verify: $t = T /6$ par, $ω t = π /3$ : $cos^{2} (π /3) = (0.5)^{2} = 0.25$ ✓ aur $sin^{2} (π /3) + cos^{2} (π /3) = 1$ ✓. $t = T /8$ par, $cos^{2} (π /4) = 0.5 = sin^{2} (π /4)$ ✓.

Ex 5 — Cell E: alag initial condition ( $t = 0$ par current flow kar rahi hai)

Worked example Statement

Ab switch us instant close hota hai jab capacitor poori tarah discharged hai ( $q (0) = 0$ ) lekin current already apni peak value $i_{0}$ par flow kar rahi hai. (Yaad karo $q_{0}$ charge oscillation ka amplitude hai, aur $i_{0} = q_{0} ω$ corresponding peak current hai.) $q (t)$ aur $i (t)$ likho, aur nikalo kab capacitor pehli baar full charge reach karta hai.

Forecast: parent note ne $cos$ use kiya kyunki charge maximum se shuru hua tha. Agar charge zero se shuru ho, toh $q (t)$ describe karne ke liye kaun sa function use karna chahiye?

General solution: $q (t) = q_{0} cos (ω t + ϕ)$ kisi phase $ϕ$ ke liye, jahan $q_{0}$ charge ka (abhi bhi unknown) amplitude hai. Yeh step kyun? $\overset{q}{¨} = - ω^{2} q$ ka har solution ek shifted cosine hai; initial conditions $ϕ$ aur $q_{0}$ pick karte hain.
$q (0) = 0$ apply karo: $q_{0} cos ϕ = 0 \Rightarrow ϕ = - π /2$ (taaki charge $t = 0$ ke baad rise kare, fall nahi). Yeh step kyun? Hum $ϕ = - π /2$ choose karte hain $+ π /2$ ki jagah taaki current initially positive ho (charging up).
Toh $q (t) = q_{0} cos (ω t - π /2) = q_{0} sin (ω t)$ . Yeh step kyun? Identity $cos (θ - π /2) = sin θ$ . Toh ek pure sine charge describe karta hai jo empty se shuru hota hai. Yeh parent ke cosine case ka mirror image hai. Phase ellipse par, hum simply same ellipse par ek alag point se shuru karte hain.
Current: $i (t) = \frac{d q}{d t} = q_{0} ω cos (ω t) = i_{0} cos (ω t)$ jahan $i_{0} = q_{0} ω$ hai. $t = 0$ par: $i = i_{0}$ , aur kyunki problem deta hai peak current $i_{0}$ , yeh amplitude fix karta hai via $q_{0} = i_{0} / ω$ . Yeh step kyun? $sin$ differentiate karne se $cos$ milta hai, toh ab current cosine se lead karta hai. Page ke upar se link $i_{0} = q_{0} ω$ woh hai jo given peak current ko charge amplitude mein convert karta hai — dono independent nahi hain. Charge aur current abhi bhi exactly 90° apart hain.
Capacitor pehli baar full jab $sin (ω t) = 1 \Rightarrow ω t = π /2 \Rightarrow t = \frac{π}{2 ω} = \frac{T}{4}$ . Yeh step kyun? Empty se shuru karke, fill hone mein quarter cycle lagta hai — same quarter-cycle timing jaise SHM equilibrium se maximum displacement tak jaata hai.

Verify: $q = q_{0} sin ω t$ ke liye $\overset{q}{¨} = - ω^{2} q$ check karo: $\overset{q}{¨} = - q_{0} ω^{2} sin ω t = - ω^{2} q$ ✓. Aur $q (0) = q_{0} sin 0 = 0$ ✓, $i (0) = i_{0} cos 0 = i_{0}$ ✓, aur $i_{0} = q_{0} ω$ ✓.

Ex 6 — Cell F: limiting behaviour ( $L \to 0$ , $C \to \infty$ , aur poora damped picture)

Worked example Statement

Explain karo kya hota hai oscillation ko jaise (a) $L \to 0$ , (b) $C \to \infty$ , (c) ek resistance $R$ add hoti hai. (c) mein, equation of motion se damped frequency $ω_{d}$ derive karo, critical-damping threshold do, aur over-/critically-damped regimes describe karo jahan circuit oscillate karna band kar deta hai.

Forecast: kaun sa limit oscillation ko infinitely fast banata hai, aur kaun sa use bilkul band kar deta hai? Aur jab $R$ badhta hai, kis value par "oscillation" completely khatam hoti hai sirf fade hone ki jagah?

(a) $L \to 0$ : $T = 2 π L C \to 0$ , toh $ω = 1/ L C \to \infty$ . Yeh step kyun? $L$ "mass" hai. Zero mass ⟹ swap ko slow karne ke liye koi inertia nahi ⟹ infinitely fast oscillation. Physically current instantly change ho sakta hai. Degenerate: yeh real oscillator nahi hai, yeh ek limit hai.
(b) $C \to \infty$ : $T = 2 π L C \to \infty$ , $ω \to 0$ . Yeh step kyun? $1/ C$ "spring constant" hai. Infinite $C$ ⟹ zero springiness ⟹ kuch bhi charge wapas nahi kheechta ⟹ oscillation period infinity tak stretch ho jaati hai (kabhi return nahi hota). Degenerate: effectively koi restoring force nahi.
(c) $R$ add karo: loop ke around Kirchhoff's Voltage Law ab deta hai $L \overset{q}{¨} + R \overset{q}{˙} + \frac{q}{C} = 0$ — yeh damped case hai (dekho LCR Circuit). Yeh step kyun? Resistor ek voltage drop $R \overset{q}{˙}$ add karta hai (uska $V = i R$ jahan $i = \overset{q}{˙}$ hai), jo ideal circuit mein absent tha.
$ω_{d}$ derive karo: $q = e^{s t}$ try karo. Substitute karne par, $L s^{2} + R s + \frac{1}{C} = 0$ , characteristic equation. Iske roots hain $s = - \frac{R}{2 L} \pm (\frac{R}{2 L})^{2} - \frac{1}{L C} .$ Jab root ke neeche ka term negative ho, use $\pm j ω_{d}$ likho jahan $ω_{d} = \frac{1}{L C} - (\frac{R}{2 L})^{2}, T_{d} = \frac{2 π}{ω _{d}} .$ Yeh step kyun? $q = e^{s t}$ guess karna differential equation ko ordinary quadratic mein badal deta hai (isliye hum exponential use karte hain — yeh woh ek function hai jiske derivatives khud ki copies hain). Real part $- R /2 L$ amplitude decay $e^{- R t /2 L}$ deta hai; imaginary part $ω_{d}$ par oscillation deta hai. Kyunki $ω_{d} < ω$ hai, damping swing ko thoda slower banati hai ( $T_{d} > T$ ).
Critical threshold: root exactly tab real hota hai (koi oscillation nahi) jab $(\frac{R}{2 L})^{2} = \frac{1}{L C}$ , yaani $R_{crit} = 2 \frac{L}{C} .$ Yeh step kyun? Yeh boundary case hai jahan $ω_{d} = 0$ . $R < R_{crit}$ ke liye circuit decay karte hue oscillate karta hai (underdamped). $R = R_{crit}$ par yeh critically damped hai — overshoot kiye bina jitna tez ho sake zero par return karta hai. $R > R_{crit}$ ke liye yeh overdamped hai — charge bina kisi oscillation ke zero par wapas aata hai. Yeh important non-oscillatory edge cases hain.
Sanity ordering: jaise $R \to 0$ , $ω_{d} \to 1/ L C = ω$ aur $T_{d} \to 2 π L C$ , ideal LC result recover hota hai. Yeh step kyun? Ek limit check — koi bhi correct damped formula zaroor undamped par collapse hona chahiye jab naya ingredient ( $R$ ) band ho; yeh confirm karta hai hum $R = 0$ case nahi tooda.

Verify: hamare base circuit ke liye, $R_{crit} = 2 L / C = 2 (2 \times 1 0^{- 3}) / (5 \times 1 0^{- 6}) = 2400 = 40 Ω$ . $R = 8 Ω$ lo (kafi neeche, toh underdamped): $\frac{1}{L C} = 1 0^{8}$ , $(\frac{R}{2 L})^{2} = (2000)^{2} = 4 \times 1 0^{6}$ , toh $ω_{d} = 1 0^{8} - 4 \times 1 0^{6} = 9.6 \times 1 0^{7} \approx 9798$ rad/s $< 1 0^{4}$ ✓ (slower). $T_{d} = 2 π / ω_{d} \approx 6.413 \times 1 0^{- 4}$ s $> 6.283 \times 1 0^{- 4}$ s ✓. $R = 40 Ω$ par: $(\frac{R}{2 L})^{2} = (1 0^{4})^{2} = 1 0^{8} = \frac{1}{L C}$ , toh $ω_{d} = 0$ — exactly critical ✓.

Ex 7 — Cell G: real-world word problem (radio tuning)

Worked example Statement

Ek radio ke tuning circuit mein fixed inductor $L = 0.25 μ H$ hai. Kaun sa capacitance $C$ use $f = 100$ MHz (ek FM station) par resonate karata hai? (Dekho Resonance and AC Circuits — resonance exactly $ω = 1/ L C$ hai.)

Forecast: high frequency hit karne ke liye, kya $C$ bada hona chahiye ya tiny?

$f = \frac{1}{2 π L C}$ se shuru karo aur $C$ ke liye solve karo. Yeh step kyun? Humein $f$ aur $L$ pata hai, $C$ chahiye — tuning formula rearrange karo.
Square karo: $f^{2} = \frac{1}{4 π ^{2} L C} \Rightarrow C = \frac{1}{4 π ^{2} f ^{2} L}$ . Yeh step kyun? $L C$ ek reciprocal ke neeche appear karta hai; squaring root cleanly remove karta hai.
$f = 1 0^{8}$ Hz, $L = 0.25 \times 1 0^{- 6}$ H plug karo: $C = \frac{1}{4 π ^{2} ( 1 0 ^{8} ) ^{2} ( 0.25 \times 1 0 ^{- 6} )} = \frac{1}{4 π ^{2} \cdot 1 0 ^{16} \cdot 0.25 \times 1 0 ^{- 6}} .$ Denominator $= 4 π^{2} \cdot 2.5 \times 1 0^{9} \approx 9.870 \times 1 0^{10}$ , toh $C \approx 1.013 \times 1 0^{- 11}$ F $\approx 10.1$ pF. Yeh step kyun? Direct substitution; powers of ten carefully track karo.
Interpretation: ek tiny ~10 pF capacitor — high frequency ko chhota $C$ chahiye, forecast se match karta hai. Yeh step kyun? Number ko physics ke roop mein wapas padhna design intuition confirm karta hai aur gross power-of-ten slips pakadta hai.

Verify: exact $C = \frac{1}{4 π ^{2} f ^{2} L}$ ko $f = \frac{1}{2 π L C}$ mein wapas feed karo: exactly $1 0^{8}$ Hz $= 100$ MHz return karta hai ✓. Units: $1/ H \cdot F$ yaani $1/ s$ , matlab hertz. ✓

Ex 8 — Cell H: exam twist (timing + energy + phase saath mein)

Worked example Statement

Base circuit, $q_{0} = 2 \times 1 0^{- 4}$ C. Full charge par rest se shuru karte hue, pehla time $t^{*}$ nikalo jis par inductor ki magnetic energy capacitor ki electric energy ke teen guna ho, aur wahan current do.

Forecast: $U_{B}$ ka $U_{E}$ par dominate karna matlab hum halfway sharing point ( $T /8$ ) ke baad hain lekin swap ( $T /4$ ) se pehle. Toh $t^{*}$ $T /8$ aur $T /4$ ke beech hai — agree karte ho?

Condition: $U_{B} = 3 U_{E} \Rightarrow sin^{2} ω t = 3 cos^{2} ω t \Rightarrow tan^{2} ω t = 3$ . Yeh step kyun? $U_{B} \propto sin^{2}$ , $U_{E} \propto cos^{2}$ ; inका ratio $tan^{2}$ hai. $tan$ use karna do conditions ko ek mein collapse karta hai.
$tan ω t = 3 \Rightarrow ω t = \frac{π}{3}$ (pehla positive solution). Yeh step kyun? $tan (π /3) = 3$ ; hum pehle time ke liye sabse chhota positive angle lete hain.
$t^{*} = \frac{π /3}{ω} = \frac{π}{3 \cdot 1 0 ^{4}} \approx 1.047 \times 1 0^{- 4}$ s. Bracket check karo: $T /8 = 0.785 \times 1 0^{- 4}$ s, $T /4 = 1.571 \times 1 0^{- 4}$ s — haan $t^{*}$ inke beech baithta hai ✓ (forecast confirmed). Yeh step kyun? Angle ko time mein convert karo aur do landmark instants ke against sanity-check karo.
Current: $i = - q_{0} ω sin (π /3) = - (2) \cdot \frac{3}{2} = - 3 \approx - 1.732$ A. Yeh step kyun? $ω t = π /3$ par current solution use karo; sign negative hai (pehle quarter mein discharging).

Verify: $ω t = π /3$ par: $U_{B} / U_{E} = sin^{2} (π /3) / cos^{2} (π /3) = 0.75/0.25 = 3$ ✓. Energy check: $U_{B} = \frac{1}{2} L i^{2} = \frac{1}{2} (2 \times 1 0^{- 3}) (1.732)^{2} = 3 \times 1 0^{- 3}$ J, aur total $4 \times 1 0^{- 3}$ J hai, toh $U_{B}$ hona chahiye $\frac{3}{4} \cdot 4 \times 1 0^{- 3} = 3 \times 1 0^{- 3}$ J ✓.

Recall Matrix par quick self-test

"100 MHz radio ke liye $C$ nikalo" kaun se cell mein aata hai? ::: Cell G (real-world design). Ex 3 mein, $3 T /8$ par $q < 0$ kyun hai lekin current abhi bhi negative hai? ::: Charge ne polarity flip ki (2nd quarter) lekin current abhi bhi 90° peeche hai, abhi bhi us direction mein draining kar raha hai. $L \to 0$ infinite frequency kyun deta hai? ::: $L$ inertia hai; zero inertia matlab current instantly swap ho sakta hai, $T = 2 π L C \to 0$ . Pehli baar $U_{E} = \frac{1}{4} U_{total}$ ? ::: $t = T /6$ ( $cos ω t = \frac{1}{2}$ se). Circuit kis $R$ par oscillate karna bilkul band karta hai? ::: $R_{crit} = 2 L / C$ par (critical damping); uske upar, overdamped. Chhota $R$ add hone par, period longer hota hai ya shorter? ::: Longer, kyunki $ω_{d} = 1/ L C - (R /2 L)^{2} < ω$ hai, toh $T_{d} = 2 π / ω_{d} > T$ .

Connections

Parent topic (Hinglish) — woh theory jo yeh examples drill karte hain.
Simple Harmonic Motion — quarter-cycle timings (Ex 5) seedha SHM se aate hain.
Capacitance and Energy in Capacitors — $U_{E} = q^{2} /2 C$ Ex 2, 4, 8 mein use hua.
Inductance and Self-Induction — $U_{B} = \frac{1}{2} L i^{2}$ Ex 2, 8 mein use hua.
Kirchhoff's Voltage Law — har solution ke peeche loop equation.
Damped Oscillations / LCR Circuit — $R \neq = 0$ limit, damped period, aur critical damping Ex 6 mein.
Resonance and AC Circuits — Ex 7 mein tuning formula.