Noesis
1.2.8Circuit Analysis Fundamentals

Understand RL transient behavior

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🇮🇳 Yeh note Hinglish mein padho →

WHAT is happening (the setup)

Key inductor law: the inductor's voltage depends on how fast current changes: vL=Ldidtv_L = L\frac{di}{dt}

  • If current is constant (di/dt=0di/dt=0) → vL=0v_L = 0 → inductor looks like a plain wire (short).
  • At the instant of a switch flip, the inductor forces current to be continuous — i(0+)=i(0−)i(0^+)=i(0^-) — because an instant jump would need infinite voltage.

HOW to derive the current (from first principles)

Circuit: battery VV, resistor RR, inductor LL in series, switch closed at t=0t=0.

Step 1 — Kirchhoff's Voltage Law. V=vR+vL=iR+LdidtV = v_R + v_L = iR + L\frac{di}{dt} Why this step? Sum of voltage drops around the loop equals the source — energy conservation.

Step 2 — Rearrange into a separable ODE. Ldidt=V−iR  ⇒  diV−iR=dtLL\frac{di}{dt} = V - iR \;\Rightarrow\; \frac{di}{V-iR} = \frac{dt}{L} Why? Group current terms on one side, time on the other so we can integrate.

Step 3 — Integrate both sides (from t=0t=0, i=0i=0 to time tt, current ii): −1Rln⁡(V−iR)∣0i=tL-\frac{1}{R}\ln(V-iR)\Big|_0^i = \frac{t}{L} −1R[ln⁡(V−iR)−ln⁡V]=tL-\frac{1}{R}\Big[\ln(V-iR)-\ln V\Big] = \frac{t}{L} Why? ∫diV−iR=−1Rln⁡(V−iR)\int \frac{di}{V-iR} = -\frac1R\ln(V-iR) by substitution u=V−iRu=V-iR.

Step 4 — Solve for ii. ln⁡ ⁣V−iRV=−RLt  ⇒  V−iR=Ve−Rt/L\ln\!\frac{V-iR}{V} = -\frac{R}{L}t \;\Rightarrow\; V-iR = V e^{-Rt/L}

Discharge (de-energizing): short the source at t=0t=0 with initial current I0I_0. Now V=0V=0: 0=iR+Ldidt  ⇒  i(t)=I0 e−t/τ0 = iR + L\frac{di}{dt}\;\Rightarrow\; i(t)=I_0\,e^{-t/\tau}

Figure — Understand RL transient behavior

WHY the time constant matters

tt i/Ifinali/I_{\text{final}} (rising) remaining (decay)
1τ1\tau 63.2% 36.8%
2τ2\tau 86.5% 13.5%
3τ3\tau 95.0% 5.0%
5τ5\tau 99.3% 0.7%

Worked Examples

Common Mistakes

Recall Feynman: explain to a 12-year-old

Imagine pushing a heavy shopping cart. When the light turns green (switch on), the cart doesn't zoom to full speed instantly — it takes a moment to get going, then cruises. When you stop pushing, it doesn't stop dead — it rolls a while and slows down. The inductor is that "heaviness" for electric current: it takes time to speed up and time to slow down. How long? A number called τ=L/R\tau = L/R: heavier coil (LL) = slower, more friction (RR) = quicker to settle.

Flashcards

What is the time constant of a series RL circuit?
τ=L/R\tau = L/R (seconds)
Why can't current jump instantly in an inductor?
Because vL=L di/dtv_L=L\,di/dt; an instant jump needs infinite voltage, so ii is continuous (i(0+)=i(0−)i(0^+)=i(0^-)).
Rise equation for energizing RL circuit?
i(t)=VR(1−e−t/τ)i(t)=\frac{V}{R}(1-e^{-t/\tau})
Decay equation for de-energizing RL circuit?
i(t)=I0e−t/τi(t)=I_0 e^{-t/\tau}
What fraction of final value is reached after 1τ (rising)?
63.2%
After how many τ is the transient essentially complete?
About 5τ (>99%).
Inductor voltage right after switch-on (t=0+t=0^+)?
The full source voltage VV (since i=0i=0, all drop is on LL).
Does larger R make an RL transient faster or slower?
Faster — RR is in the denominator of τ=L/R\tau=L/R.
Steady-state behavior of an ideal inductor with DC?
Acts as a short (plain wire); vL=0v_L=0 since di/dt=0di/dt=0.
KVL equation for series RL with source V?
V=iR+L di/dtV = iR + L\,di/dt

Connections

  • RC Transient Behavior — same math, but τ=RC\tau=RC and roles of charge/current swap.
  • Time Constant — the universal exponential-settling parameter.
  • Inductor Fundamentals — where vL=L di/dtv_L=L\,di/dt and energy 12LI2\tfrac12 LI^2 come from.
  • Kirchhoff's Voltage Law — the loop equation we started from.
  • First-Order Differential Equations — the ODE technique used to derive i(t)i(t).
  • Steady-State Analysis — what the circuit looks like after 5τ5\tau.

Concept Map

voltage law

forces

creates

causes

analyzed via

separable ODE

energizing

de-energizing

governed by

governed by

after 5 tau

one tau

Inductor opposes di/dt

vL equals L di/dt

Current continuous at switch

Switch flip

RL transient

KVL loop V equals iR plus vL

Integrate and solve

i equals V/R times 1 minus exp -t/tau

i equals I0 exp -t/tau

Time constant tau equals L/R

Transient essentially over

63.2 percent of journey

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, RL circuit ka simple funda ye hai: inductor ko current ka sudden change bilkul pasand nahi. Jab tum switch on karte ho, current ekdum se V/RV/R tak jump nahi karta — inductor apni "current inertia" ki wajah se ek voltage L di/dtL\,di/dt banata hai jo change ka opposition karta hai. Isliye current dheere-dheere, smooth exponential curve me apni final value tak pahunchta hai. Poori kahani ek hi number se control hoti hai: time constant τ=L/R\tau = L/R.

τ\tau ka matlab: ek τ\tau time me current apna 63.2% safar complete kar leta hai, aur 5τ\tau ke baad practically kaam khatam (99% se zyada). Yaad rakho RR neeche hai — matlab zyada resistance = jaldi settle (kyunki energy jaldi dissipate hoti hai), aur zyada LL = slow. Rise ke liye formula i(t)=VR(1−e−t/τ)i(t)=\frac{V}{R}(1-e^{-t/\tau}), aur jab source hata do (discharge) to i(t)=I0e−t/τi(t)=I_0 e^{-t/\tau}.

Ye kyun important hai? Real hardware me — motors, relays, power supplies, switching circuits — inductor har jagah hai. Switch-off ke waqt inductor apni stored energy chhodne ki koshish me bahut bada voltage spike bana deta hai (isliye flyback diode lagate hain). Agar τ\tau samajh gaye to timing, spikes, aur settling behaviour sab predict kar sakte ho.

Ek common galti: log τ=RC\tau = RC wale RC circuit se copy karke RLRL me bhi product maan lete hain. Nahi bhai — RL me ratio hai, τ=L/R\tau=L/R. Units check karo: henry/ohm = seconds, perfect. Bas ek baar derivation khud KVL se karo, phir kabhi nahi bhoologe.

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Connections

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